This is a general question about Design of experiments. If i am not allowed
to post general questions like this here please accept my apologies and
ignore the question.
I did a DOE with six factors in eight runs. I know i cannot check for
interactions using this design, but i tried the interaction plot and it
showed me many interactions. After this I foldover the design and ran the 8
runs to learn about the interactions but i could not find any. Now my
question does the interaction plot of a resolution three design gives any
information about the possible interactions? why did the interaction plot
showed many interactions? I used Minitab for analysis, new to R so still
trying to figure out how to use R.
Thanks
Statistics novice

On 8/11/07, r-help-request@stat.math.ethz.ch <
r-help-request@stat.math.ethz.ch> wrote:
>
> Send R-help mailing list submissions to
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> than "Re: Contents of R-help digest..."
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>
> Today's Topics:
>
>    1. Re: Combining two ANOVA outputs of different lengths
>       (Peter Dalgaard)
>    2. Plot legend in margin (Daniel Brewer)
>    3. Re :  compute ROC curve? (justin bem)
>    4. Re: reading xcms files (Prof Brian Ripley)
>    5. Re: GLMM: MEEM error due to dichotomous variables
>       (lorenz.gygax@art.admin.ch)
>    6. Re: Plot legend in margin (Daniel Brewer)
>    7. Re: Plot legend in margin (Uwe Ligges)
>    8. Re: Positioning text in top left corner of plot (Daniel Brewer)
>    9. Re: Positioning text in top left corner of plot (Daniel Brewer)
>   10. Re: ordering a data.frame by average rank of multiple columns
>       (Gabor Grothendieck)
>   11. having problems with factor() (Jabez Wilson)
>   12. Re: need help with pdf-plot (Ivar Herfindal)
>   13. Re: having problems with factor() (Henrique Dallazuanna)
>   14. help with counting how many times each value occur in each
>       column (Tom Cohen)
>   15. Odp:  having problems with factor() (Petr PIKAL)
>   16. Remove redundant observations for cross-validation
>       (Eleni Rapsomaniki)
>   17. Re: Positioning text in top left corner of plot (Daniel Brewer)
>   18. Re: having problems with factor() (Rainer Hurling)
>   19. Re: [Fwd: Re: How to apply functions over rows of multiple
>       matrices] (Gabor Grothendieck)
>   20. Odp: help with counting how many times each value occur in
>       each      column (Petr PIKAL)
>   21. Re: help with counting how many times each value occur in
>       each      column (Gabor Grothendieck)
>   22. Re: [Fwd: Re: How to apply functions over rows of multiple
>       matrices] (Johannes H?sing)
>   23. Re: Odp:  having problems with factor() (Jabez Wilson)
>   24. Re: Plot legend in margin (Greg Snow)
>   25. Re: Seasonality (Roland Rau)
>   26. Re: help with counting how many times each value occur in
>       each      column (Fran?ois Pinard)
>   27. Re: help with counting how many times each value occur in
>       eachcolumn (Gasper Cankar)
>   28. Re: help with counting how many times each value occur in
>       each      column (Fran?ois Pinard)
>   29. rfImpute (Eric Turkheimer)
>   30. smoothing function for proportions (Rose Hoberman)
>   31. Re: smoothing function for proportions (Rose Hoberman)
>   32. Re: how to include bar values in a barplot? (Greg Snow)
>   33. Re: a question on lda{MASS} (Weiwei Shi)
>   34. Re: Subject: Re: how to include bar values in a barplot?
>       (Greg Snow)
>   35. Re: smoothing function for proportions (roger koenker)
>   36. GLM with tweedie: NA for AIC (laran gines)
>   37. Test (ignore) (Nordlund, Dan (DSHS/RDA))
>   38. Re: Help using gPath (Emilio Gagliardi)
>   39. write.table (Weiwei Shi)
>   40. Re: small sample techniques (Nordlund, Dan (DSHS/RDA))
>   41. Re: small sample techniques (Nordlund, Dan (DSHS/RDA))
>   42.  Cleaning up the memory (Monica Pisica)
>   43. need help to manipulate function and time interval
>       (KOITA Lassana - STAC/ACE)
>   44. Re: need help to manipulate function and time interval
>       (Henrique Dallazuanna)
>   45. Re: Cleaning up the memory (Prof Brian Ripley)
>   46. Re: write.table (Yinghai Deng)
>   47. Help wit matrices (Lanre Okusanya)
>   48. Re: Cleaning up the memory (Monica Pisica)
>   49. Re: Help wit matrices (jim holtman)
>   50. Row name of empty string issue (adiamond)
>   51. Re: Help wit matrices (Lanre Okusanya)
>   52. Re: Help wit matrices (Henrique Dallazuanna)
>   53. kde2d error message (Jennifer Dillon)
>   54. Re: Help wit matrices (Ravi Varadhan)
>   55. Re: Help wit matrices (John Kane)
>   56. half-logit and glm (again) (Richard D. Morey)
>   57. Re: write.table (Weiwei Shi)
>   58. Re: Cleaning up the memory (Prof Brian Ripley)
>   59. Re: Help wit matrices ( (Ted Harding))
>   60. Re: Help wit matrices (Roland Rau)
>   61. Re: kde2d error message (Prof Brian Ripley)
>   62. Re: need help to manipulate function and time interval
>       (Uwe Ligges)
>   63. Re: need help to manipulate function and time interval
>       (Uwe Ligges)
>   64. QUESTION ON R!!!!!!!!!!!1 (lecastil)
>   65. Re: R-excel (Peter Wickham)
>   66. Re: R-excel (Prof Brian Ripley)
>   67. Request (zahid khan)
>   68. Re: Help using gPath (Paul Murrell)
>   69. shell and shell.exec on Windows (Erich Neuwirth)
>
>
> ----------------------------------------------------------------------
>
> Message: 1
> Date: Fri, 10 Aug 2007 12:01:25 +0200
> From: Peter Dalgaard <p.dalgaard@biostat.ku.dk>
> Subject: Re: [R] Combining two ANOVA outputs of different lengths
> To: Christoph.Scherber@agr.uni-goettingen.de
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <46BC3775.5080301@biostat.ku.dk>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
> Christoph Scherber wrote:
> > Dear R users,
> >
> > I have been trying to combine two anova outputs into one single table
> > (for later publication). The outputs are of different length, and share
> > only some common explanatory variables.
> >
> > Using merge() or melt() (from the reshape package) did not work out.
> >
> > Here are the model outputs and what I would like to have:
> >
> > anova(model1)
> >              numDF denDF  F-value p-value
> > (Intercept)     1    74 0.063446  0.8018
> > days          1    74 6.613997  0.0121
> > logdiv          1    74 1.587983  0.2116
> > leg             1    74 4.425843  0.0388
> >
> > anova(model2)
> >               numDF denDF   F-value p-value
> > (Intercept)      1    73 165.94569  <.0001
> > funcgr           1    73   7.91999  0.0063
> > grass            1    73  42.16909  <.0001
> > leg              1    73   4.72108  0.0330
> > funcgr:grass     1    73   8.49068  0.0047
> >
> > #"merge(anova(model1),anova(model2),...)"
> >
> >               F-value 1       p-val1  F-value 2       p-value 2
> > (Intercept)   0.063446        0.8018  165.94569       <.0001
> > days          6.613997        0.0121  NA              NA
> > logdiv                1.587983        0.2116  NA              NA
> > leg           4.425843        0.0388  4.72108         0.033
> > funcgr                NA              NA      7.91999         0.0063
> > grass         NA              NA      42.16909        <.0001
> > funcgr:grass  NA              NA      8.49068         0.0047
> >
> >
> > I would be glad if someone would have an idea of how to do this in
> > principle.
> >
> The main problems are that the merge key is the rownames and that you
> want to keep entries that are missing in one of the analysis. There are
> ways to deal with that:
>
> > example(anova.lm)
> .....
> > merge(anova(fit2), anova(fit4), by=0, all=T)
>   Row.names Df.x  Sum Sq.x Mean Sq.x F value.x    Pr(>F).x Df.y  Sum Sq.y
> 1      ddpi   NA        NA        NA        NA          NA    1  63.05403
> 2       dpi   NA        NA        NA        NA          NA    1  12.40095
> 3     pop15    1 204.11757 204.11757 13.211166 0.000687868    1 204.11757
> 4     pop75    1  53.34271  53.34271  3.452517 0.069425385    1  53.34271
> 5 Residuals   47 726.16797  15.45038        NA          NA   45 650.71300
>   Mean Sq.y  F value.y     Pr(>F).y
> 1  63.05403  4.3604959 0.0424711387
> 2  12.40095  0.8575863 0.3593550848
> 3 204.11757 14.1157322 0.0004921955
> 4  53.34271  3.6889104 0.0611254598
> 5  14.46029         NA           NA
>
>
>
> Presumably, you can take it from here.
>
>
>
> ------------------------------
>
> Message: 2
> Date: Fri, 10 Aug 2007 11:09:32 +0100
> From: Daniel Brewer <daniel.brewer@icr.ac.uk>
> Subject: [R] Plot legend in margin
> To: r-help@stat.math.ethz.ch
> Message-ID: <46BC395C.3080109@icr.ac.uk>
> Content-Type: text/plain; charset=ISO-8859-1
>
> Hi all,
> Another plotting question I am afraid.  Is there anyway of putting a
> legend for a plot in a margin rather than within the figure.  I am
> trying to plot a 3x2 plot and I want to have:
> 1) One key along the bottom for all the plots
> 2) A label (a,b,c) for each plot (see previous emails)
>
> Is there any websites etc. that explain this sort of thing?
>
> Dan
>
> --
> **************************************************************
> Daniel Brewer, Ph.D.
>
> Institute of Cancer Research
> Email: daniel.brewer@icr.ac.uk
> **************************************************************
>
> The Institute of Cancer Research: Royal Cancer Hospital, a charitable
> Company Limited by Guarantee, Registered in England under Company No. 534147
> with its Registered Office at 123 Old Brompton Road, London SW7 3RP.
>
> This e-mail message is confidential and for use by the addre...{{dropped}}
>
>
>
> ------------------------------
>
> Message: 3
> Date: Fri, 10 Aug 2007 10:28:49 +0000 (GMT)
> From: justin bem <justin_bem@yahoo.fr>
> Subject: [R] Re :  compute ROC curve?
> To: gallon li <gallon.li@gmail.com>
> Cc: R Maillist <r-help@stat.math.ethz.ch>
> Message-ID: <156417.27358.qm@web23006.mail.ird.yahoo.com>
> Content-Type: text/plain
>
> see ROCR or accuracy package.
>
> Justin BEM
> BP 1917 Yaoundé
> Tél (237) 99597295
>     (237) 22040246
>
> ----- Message d'origine ----
> De : gallon li <gallon.li@gmail.com>
> À : r-help <r-help@stat.math.ethz.ch>
> Envoyé le : Vendredi, 10 Août 2007, 4h15mn 36s
> Objet : [R] compute ROC curve?
>
> Hello,
>
> i have continuous test results for dieased and nondiseased subjects, say X
> and Y. Both are vectors of numbers.
>
> is there any R function which can generate the step function of ROC curve
> automatically?
>
> Thanks!
>
>     [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
>
>
>
>
>
>
>       ______________________________________________________________________________
>
> ail !
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 4
> Date: Fri, 10 Aug 2007 11:47:21 +0100 (BST)
> From: Prof Brian Ripley <ripley@stats.ox.ac.uk>
> Subject: Re: [R] reading xcms files
> To: Roberto Olivares Hernandez <roh@biocentrum.dtu.dk>
> Cc: "'r-help@stat.math.ethz.ch'" <r-help@stat.math.ethz.ch>
> Message-ID: <Pine.LNX.4.64.0708101142260.20343@gannet.stats.ox.ac.uk>
> Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
>
> On Fri, 10 Aug 2007, Roberto Olivares Hernandez wrote:
>
> > Hi,
> >
> > I am using xcms library to read mass spectrum data. I generate objects
> > from CDF files using the command line
> >
> >>  SME10 <- xcmsRaw("SME_10.CDF")
> >
> > I have 50 CDF files with different name and I don't want to repeat the
> > command for each one. Is there any option to read all the files and
> > generate a corresponding object name?
>
> Something like
>
> for(f in Sys.glob("*.CDF")) assign(sub("\\.CDF$", "", f), xcmsRaw(f))
>
> (untested, of course).
>
> --
> Brian D. Ripley,                  ripley@stats.ox.ac.uk
> Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> University of Oxford,             Tel:  +44 1865 272861 (self)
> 1 South Parks Road,                     +44 1865 272866 (PA)
> Oxford OX1 3TG, UK                Fax:  +44 1865 272595
>
>
>
> ------------------------------
>
> Message: 5
> Date: Fri, 10 Aug 2007 12:55:46 +0200
> From: <lorenz.gygax@art.admin.ch>
> Subject: Re: [R] GLMM: MEEM error due to dichotomous variables
> To: <r-help@stat.math.ethz.ch>
> Cc: elva_robinson@hotmail.com
> Message-ID:
>         <
> 145C63777EF3ED41A5A99035845F7DD9D313AE@EVD-C8002.bk.evdad.admin.ch>
> Content-Type: text/plain;       charset="iso-8859-1"
>
> > I am trying to run a GLMM on some binomial data. My fixed
> > factors include 2
> > dichotomous variables, day, and distance. When I run the model:
> >
> > modelA<-glmmPQL(Leaving~Trial*Day*Dist,random=~1|Indiv,family=
> >"binomial")
> >
> > I get the error:
> >
> > iteration 1
> > Error in MEEM(object, conLin, control$niterEM) :
> >         Singularity in backsolve at level 0, block 1
> >
> > >From looking at previous help topics,(
> > http://tolstoy.newcastle.edu.au/R/help/02a/4473.html)
> > I gather this is because of the dichotomous predictor
> > variables - what approach should I take to avoid this problem?
>
> Are you sure? I have never had problems including factors in a glmmPQL so
> far. More likely, the combination of your explanatory variables leads to a
> fragmentation in your response such that each combination of your factor
> levels only contain 0s or 1s. Thus, your model is 'too good' (it has too
> many predictors given the amount of data). Try e.g. to fit a model without
> the interactions.
>
> Cheers, Lorenz
> -
> Lorenz Gygax
> Centre for proper housing of ruminants and pigs
> Agroscope Reckenholz-T?nikon Research Station ART
> T?nikon, CH-8356 Ettenhausen / Switzerland
>
>
>
> ------------------------------
>
> Message: 6
> Date: Fri, 10 Aug 2007 11:55:29 +0100
> From: Daniel Brewer <daniel.brewer@icr.ac.uk>
> Subject: Re: [R] Plot legend in margin
> To: Lauri Nikkinen <lauri.nikkinen@iki.fi>, r-help@stat.math.ethz.ch
> Message-ID: <46BC4421.2060100@icr.ac.uk>
> Content-Type: text/plain; charset=ISO-8859-1
>
> Thanks.  That got me onto the right track.  Because it is a multiplot
> and I wanted it along the bottom, I found that I had to use par(xpd=NA)
> and then position it relative to the last of the multiplots.  After a
> bit of trial and error I got there.
>
> Thanks
>
> Lauri Nikkinen wrote:
> > Very simple example:
> >
> > opar <- par(mar = c(10, 4, 4, 4))
> > plot(1:10)
> > lines(1:10)
> > par(xpd=TRUE)
> > legend(4,-1.5,lty=1, col="black", legend="straigh line")
> > par(opar)
> >
> > -Lauri
>
> --
> **************************************************************
> Daniel Brewer, Ph.D.
> Institute of Cancer Research
> Email: daniel.brewer@icr.ac.uk
> **************************************************************
>
> The Institute of Cancer Research: Royal Cancer Hospital, a charitable
> Company Limited by Guarantee, Registered in England under Company No. 534147
> with its Registered Office at 123 Old Brompton Road, London SW7 3RP.
>
> This e-mail message is confidential and for use by the addre...{{dropped}}
>
>
>
> ------------------------------
>
> Message: 7
> Date: Fri, 10 Aug 2007 13:00:23 +0200
> From: Uwe Ligges <ligges@statistik.uni-dortmund.de>
> Subject: Re: [R] Plot legend in margin
> To: Daniel Brewer <daniel.brewer@icr.ac.uk>
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <46BC4547.6020801@statistik.uni-dortmund.de>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
>
>
> Daniel Brewer wrote:
> > Hi all,
> > Another plotting question I am afraid.  Is there anyway of putting a
> > legend for a plot in a margin rather than within the figure.  I am
> > trying to plot a 3x2 plot and I want to have:
> > 1) One key along the bottom for all the plots
> > 2) A label (a,b,c) for each plot (see previous emails)
> >
> > Is there any websites etc. that explain this sort of thing?
>
> Please read the posting guide.
>
> After that, type:
>    RSiteSearch("legend margin")
>
> Currently, the fourth entry shows a solution:
> http://finzi.psych.upenn.edu/R/Rhelp02a/archive/67979.html
>
> Uwe Ligges
>
>
> > Dan
> >
>
>
>
> ------------------------------
>
> Message: 8
> Date: Fri, 10 Aug 2007 12:03:56 +0100
> From: Daniel Brewer <daniel.brewer@icr.ac.uk>
> Subject: Re: [R] Positioning text in top left corner of plot
> To: S Ellison <S.Ellison@lgc.co.uk>, r-help@stat.math.ethz.ch
> Message-ID: <46BC461C.1010308@icr.ac.uk>
> Content-Type: text/plain; charset=ISO-8859-1
>
> Thanks.  That works if it is only a single plot, but if there are
> multiple plots (e.g. par(mfrow=c(2,2))) it confusingly puts the label in
> the absolute top left always i.e. the top left of plot one.
>
> Dan
>
> S Ellison wrote:
> > Try something like
> > mtext(side=3, line=-1, text="Here again?", adj=0, outer=T)
> >
> > This puts text just inside the top left corner.
> >
> >
> >>>> Jim Lemon <jim@bitwrit.com.au> 10/08/2007 10:37:30 >>>
> > Daniel Brewer wrote:
> >> Thanks for the replies, but I still cannot get what I want.  I do not
> >> want the label inside the plot area, but in the top left of the paper,
> I
> >> suppose in the margins.  When I try to use text to do this, it does not
> >> seem to plot it outside the plot area.  I have also tried to use mtext,
> >> but that does not really cut it, as I cannot get the label in the
> >> correct position.  Ideally, it would be best if I could use legend but
> >> have it outside the plot area.
> >>
> >> Any ideas?
> >>
> > Hi Dan,
> >
> > Try this:
> >
> > plot(1:5)
> > par(xpd=TRUE)
> > text(0.5,5.5,"Outside")
> > par(xpd=FALSE)
> >
> > Jim
>
>
> The Institute of Cancer Research: Royal Cancer Hospital, a charitable
> Company Limited by Guarantee, Registered in England under Company No. 534147
> with its Registered Office at 123 Old Brompton Road, London SW7 3RP.
>
> This e-mail message is confidential and for use by the addre...{{dropped}}
>
>
>
> ------------------------------
>
> Message: 9
> Date: Fri, 10 Aug 2007 12:05:31 +0100
> From: Daniel Brewer <daniel.brewer@icr.ac.uk>
> Subject: Re: [R] Positioning text in top left corner of plot
> To: Paul Murrell <paul@stat.auckland.ac.nz>
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <46BC467B.2070309@icr.ac.uk>
> Content-Type: text/plain; charset=ISO-8859-1
>
> This works fine for one plot, but if it is a multiple plot (mfrow=c(2,2)
> say) then each individual label is placed in the same position i.e.
> absolute top left on the canvas.  I would like it top left of each
> individual plot.
>
> Thanks anyway.  Got any idea how to fix this?
>
> Dan
>
> Paul Murrell wrote:
> > Hi
> >
> >
> > Daniel Brewer wrote:
> >> Thanks for the replies, but I still cannot get what I want.  I do not
> >> want the label inside the plot area, but in the top left of the paper,
> I
> >> suppose in the margins.  When I try to use text to do this, it does not
> >> seem to plot it outside the plot area.  I have also tried to use mtext,
> >> but that does not really cut it, as I cannot get the label in the
> >> correct position.  Ideally, it would be best if I could use legend but
> >> have it outside the plot area.
> >>
> >> Any ideas?
> >
> >
> > plot(1:10)
> > library(grid)
> > grid.text("What do we want?  Text in the corner!\nWhere do we want it?
> > Here!",
> >           x=unit(2, "mm"), y=unit(1, "npc") - unit(2, "mm"),
> >           just=c("left", "top"))
> >
> > Paul
> >
> >
> >> Thanks
> >>
> >> Benilton Carvalho wrote:
> >>> maybe this is what you want?
> >>>
> >>> plot(rnorm(10))
> >>> legend("topleft", "A)", bty="n")
> >>>
> >>> ?
> >>>
> >>> b
> >>>
> >>> On Aug 7, 2007, at 11:08 AM, Daniel Brewer wrote:
> >>>
> >>>> Simple question how can you position text in the top left hand
> >>>> corner of
> >>>> a plot?  I am plotting multiple plots using par(mfrow=c(2,3)) and all
> I
> >>>> want to do is label these plots a), b), c) etc.  I have been fiddling
> >>>> around with both text and mtext but without much luck.  text is fine
> >>>> but
> >>>>  each plot has a different scale on the axis and so this makes it
> >>>> problematic.  What is the best way to do this?
> >>>>
> >>>> Many thanks
> >>>>
> >>>> Dan
> >
> >
>
>
> The Institute of Cancer Research: Royal Cancer Hospital, a charitable
> Company Limited by Guarantee, Registered in England under Company No. 534147
> with its Registered Office at 123 Old Brompton Road, London SW7 3RP.
>
> This e-mail message is confidential and for use by the addre...{{dropped}}
>
>
>
> ------------------------------
>
> Message: 10
> Date: Fri, 10 Aug 2007 07:36:11 -0400
> From: "Gabor Grothendieck" <ggrothendieck@gmail.com>
> Subject: Re: [R] ordering a data.frame by average rank of multiple
>         columns
> To: "Tom.O" <tom.olsson@dnbnor.com>
> Cc: r-help@stat.math.ethz.ch
> Message-ID:
>         <971536df0708100436p3efc4d19me89f4a83ffd546d6@mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> Try this:
>
> positions <- order(ranks)
>
> On 8/10/07, Tom.O <tom.olsson@dnbnor.com> wrote:
> >
> > Hi
> >
> > I have run into a problem and i wonder if anyone has a smart way of
> doing
> > this.
> >
> > For example i have this data frame for 5 different test groups:
> >
> > Res1 <- c(1,5,4,-0.5,3)
> > Res2 <- c(-1,8,2,0,3)
> > Mean <- c(0.5,1,1.5,-.5,2)
> > MyFrame <- data.frame(Res1,Res2,Mean,row.names=c
> ("G1","G2","G3","G4","G5"))
> >
> > where the first two columns are the results of two different tests, the
> > third column is the mean of the group.
> >
> > I want to order this data.frame by the combined rank of Res1 & Res2, but
> > where weigths are assigned to the importeance av each column. Lets
> assume
> > that Res1 is twice as important and lower values rank better.
> >
> > MyRanks<-data.frame
> (Rank1=rank(MyFrame[,"Res1"]),Rank2=rank(MyFrame[,"Res2"]),CombR=2*rank(MyFrame[,"Res1"])+rank(MyFrame[,"Res2"]),
> row.names=c("G1","G2","G3","G4","G5"))
> >
> >    Rank1 Rank2 CombR
> > G1     2     1     5
> > G2     5     5    15
> > G3     4     3    11
> > G4     1     2     4
> > G5     3     4    10
> >
> >
> > and the rank of the combined is 2,5,4,1,3 , but to be able to sort
> MyFrame
> > in that order I need to enter this vector of positions c(4,1,5,3,2) but
> do
> > anyone have a smart way of converting ranks to positions?
> >
> > Tom
> >
> >
> > --
> > View this message in context:
> http://www.nabble.com/ordering-a-data.frame-by-average-rank-of-multiple-columns-tf4247393.html#a12087498
> > Sent from the R help mailing list archive at Nabble.com.
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>
>
> ------------------------------
>
> Message: 11
> Date: Fri, 10 Aug 2007 11:41:53 +0000 (GMT)
> From: Jabez Wilson <jabezwuk@yahoo.co.uk>
> Subject: [R] having problems with factor()
> To: R-Help <r-help@stat.math.ethz.ch>
> Message-ID: <266689.93457.qm@web27410.mail.ukl.yahoo.com>
> Content-Type: text/plain
>
> Dear R Help,
> I have a set of data of heights of trees described by area that they are
> in. The areas are numerical (0 to 7).
>
>     ht    area
> 1   320   3
> 2   410   4
> 3   230   2
> 4   360   3
> 5   126   1
> 6   280   2
> 7   260   2
> 8   280   2
> 9   280   2
> 10  260   2
> .......
> 180 450   4
> 181  90   1
> 182 120   1
> 183 440   4
> 184 210   2
> 185 330   3
> 186 210   2
> 187 100   1
> 188   0   0
>
> I want to convert the area column values to factors, to do an anova.
> However, if I use:
>
> df$areaf <- factor(df$area,
> labels=c("0","I","II","III","IV","V","VI","VII"))
>
> it gives the following message:
>
> Error in factor(df$area, labels = c("0", "I", "II", "III", "IV", "V",
> "VI",  :
>         invalid labels; length 8 should be 1 or 7
>
> Can anyone help?
>
> Jabez
>
>
>       ___________________________________________________________
>
> now.
>
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 12
> Date: Fri, 10 Aug 2007 13:41:38 +0200
> From: Ivar Herfindal <ivar.herfindal@bio.ntnu.no>
> Subject: Re: [R] need help with pdf-plot
> To: Antje <antje.niederlein@yahoo.de>
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <46BC4EF2.7060504@bio.ntnu.no>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
> Dear Antje
>
> I cannot see that you have got any replies yet, so I will make and
> attempt. However, I am sure other have more formally correct solutions.
>
> When you call the pdf(), you can set paper="a4" (or "a4r" for
> landscape). However, the width and the height of your plot should then
> not exceed the size of the paper (which is approximately 8.27*11.69
> inches for "a4"). Try (I have only tested on windows XP, R 2.5.0):
>
> pdf("test1.pdf", width=10, heigh=5, paper="a4r")
> par(mfrow=c(1,3), pty="s") #pty="s" gives square plotting regions
> plot(rnorm(100))
> plot(rnorm(100))
> plot(rnorm(100))
> dev.off()
>
> Hope this helps
>
> Ivar
>
>
> Antje skrev:
> > I still have this problem. Does anybody know any solution?
> >
> > Antje
> >
> > Antje schrieb:
> >
> >> Hello,
> >>
> >> I'm trying to plot a set of barplots like a matrix (2 rows, 10 columns
> >> from"reduced_mat") to a pdf. It works with the following parameters:
> >>
> >> pdf("test.pdf",width=ncol(reduced_mat)*2, height=nrow(reduced_mat)*2,
> pointsize
> >> = 12)
> >>
> >> par(mfcol = c(nrow(reduced_mat),ncol(reduced_mat)), oma = c(0,0,0,0),
> >> lwd=48/96, cex.axis = 0.5, las = 2, cex.main = 1.0)
> >>
> >> The I get a long narrow page format with the quadratic barplots.
> >>
> >> But I would like to have a A4 format in the end and the plots not
> filling the
> >> whole page (they should stay somehow quadratic and not be
> stretched...).
> >>
> >> What shall I look for to achieve this?
> >>
> >> Antje
> >>
> >> ______________________________________________
> >> R-help@stat.math.ethz.ch mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> >>
> >>
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>
>
> ------------------------------
>
> Message: 13
> Date: Fri, 10 Aug 2007 08:50:44 -0300
> From: "Henrique Dallazuanna" <wwwhsd@gmail.com>
> Subject: Re: [R] having problems with factor()
> To: "Jabez Wilson" <jabezwuk@yahoo.co.uk>
> Cc: R-Help <r-help@stat.math.ethz.ch>
> Message-ID:
>         <da79af330708100450h44bc9541mf584d4e0c56a9e26@mail.gmail.com>
> Content-Type: text/plain
>
> Hi,
>
> df
>     ht area
> 1  320    3
> 2  410    4
> 3  230    2
> 4  360    3
> 5  126    1
> 6  280    2
> 7  260    2
> 8  280    2
> 9  280    2
> 10 260    2
>
> df$area <- as.factor(df$area)
> levels(df$area) <- c("I", "II", "III", "IV")
>
>
>
> --
> Henrique Dallazuanna
> Curitiba-Paraná-Brasil
> 25° 25' 40" S 49° 16' 22" O
>
> On 10/08/07, Jabez Wilson <jabezwuk@yahoo.co.uk> wrote:
> >
> > Dear R Help,
> > I have a set of data of heights of trees described by area that they are
> > in. The areas are numerical (0 to 7).
> >
> >     ht    area
> > 1   320   3
> > 2   410   4
> > 3   230   2
> > 4   360   3
> > 5   126   1
> > 6   280   2
> > 7   260   2
> > 8   280   2
> > 9   280   2
> > 10  260   2
> > .......
> > 180 450   4
> > 181  90   1
> > 182 120   1
> > 183 440   4
> > 184 210   2
> > 185 330   3
> > 186 210   2
> > 187 100   1
> > 188   0   0
> >
> > I want to convert the area column values to factors, to do an anova.
> > However, if I use:
> >
> > df$areaf <- factor(df$area,
> > labels=c("0","I","II","III","IV","V","VI","VII"))
> >
> > it gives the following message:
> >
> > Error in factor(df$area, labels = c("0", "I", "II", "III", "IV", "V",
> > "VI",  :
> >         invalid labels; length 8 should be 1 or 7
> >
> > Can anyone help?
> >
> > Jabez
> >
> >
> >       ___________________________________________________________
> >
> > now.
> >
> >         [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 14
> Date: Fri, 10 Aug 2007 14:01:44 +0200 (CEST)
> From: Tom Cohen <tom.cohen78@yahoo.se>
> Subject: [R] help with counting how many times each value occur in
>         each    column
> To: r-help@stat.math.ethz.ch
> Message-ID: <570265.73780.qm@web23006.mail.ird.yahoo.com>
> Content-Type: text/plain
>
> Dear list,
>   I have the following dataset and want to know how many times each value
> occur in each column.
>    >data
>         [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> [1,] -100 -100 -100    0    0    0    0    0    0  -100
> [2,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [3,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [4,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [5,] -100 -100 -100 -100 -100 -100 -100 -100 -100   -50
> [6,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [7,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [8,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [9,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [10,] -100 -100 -100  -50 -100 -100 -100 -100 -100  -100
> [11,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [12,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [13,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [14,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [15,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [16,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [17,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [18,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [19,] -100 -100 -100    0    0    0    0    0    0  -100
> [20,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
>   The result matrix should look like
>        -100 0 -50
> [1]   20
> [2]   20
> [3]   20
> [4]   17
> [5]   18
> [6]   18
> [7]   18  and so on
> [8]
> [9]
> [10]
>
> How can I do this in R ?
>   Thanks alot for your help,
> Tom
>
>
> ---------------------------------
>
> Jämför pris på flygbiljetter och hotellrum:
> http://shopping.yahoo.se/c-169901-resor-biljetter.html
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 15
> Date: Fri, 10 Aug 2007 14:02:21 +0200
> From: Petr PIKAL <petr.pikal@precheza.cz>
> Subject: [R] Odp:  having problems with factor()
> To: Jabez Wilson <jabezwuk@yahoo.co.uk>
> Cc: R-Help <r-help@stat.math.ethz.ch>
> Message-ID:
>         <
> OF53CE58BC.6CC27AD5-ONC1257333.0041ED7A-C1257333.00421CE9@precheza.cz>
>
> Content-Type: text/plain; charset="US-ASCII"
>
> Hi
> r-help-bounces@stat.math.ethz.ch napsal dne 10.08.2007 13:41:53:
>
> > Dear R Help,
> > I have a set of data of heights of trees described by area that they are
> in.
> > The areas are numerical (0 to 7).
> >
> >     ht    area
> > 1   320   3
> > 2   410   4
> > 3   230   2
> > 4   360   3
> > 5   126   1
> > 6   280   2
> > 7   260   2
> > 8   280   2
> > 9   280   2
> > 10  260   2
> > .......
> > 180 450   4
> > 181  90   1
> > 182 120   1
> > 183 440   4
> > 184 210   2
> > 185 330   3
> > 186 210   2
> > 187 100   1
> > 188   0   0
> >
> > I want to convert the area column values to factors, to do an anova.
> However, if I use:
> >
> > df$areaf <- factor(df$area,
> labels=c("0","I","II","III","IV","V","VI","VII"))
> >
> > it gives the following message:
> >
>
> Hm, maybe some of the values are missing
>
> > num<-sample(1:3, 10, replace=T)
> > num
> [1] 1 3 1 2 3 3 1 3 3 3
> > factor(num, labels=c("O", "I", "II"))
> [1] O  II O  I  II II O  II II II
> Levels: O I II
>
> > factor(num, labels=c("O", "I", "II", "III"))
> Error in factor(num, labels = c("O", "I", "II", "III")) :
>         invalid labels; length 4 should be 1 or 3
> >
>
> try
>
> table(df$area)
> to see what level you really have
>
> Regards
> Petr
>
>
> > Error in factor(df$area, labels = c("0", "I", "II", "III", "IV", "V",
> "VI",  :
> >         invalid labels; length 8 should be 1 or 7
> >
> > Can anyone help?
> >
> > Jabez
> >
> >
> >       ___________________________________________________________
> >
> > now.
> >
> >    [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
>
> ------------------------------
>
> Message: 16
> Date: Fri, 10 Aug 2007 13:03:49 +0100
> From: Eleni Rapsomaniki <e.rapsomaniki@mail.cryst.bbk.ac.uk>
> Subject: [R] Remove redundant observations for cross-validation
> To: r-help@stat.math.ethz.ch
> Message-ID: <1186747429.46bc54250e417@webmail.cryst.bbk.ac.uk>
> Content-Type: text/plain; charset=ISO-8859-1
>
>
>
> Hi,
>
> This is a general statistics question that I believe occurs often so may
> have
> some R functions/packages dedicated to it.
> Suppose you want to check the accuracy of a classifier using a large
> training
> data-set where each row represents an observation. Is there a simple
> approach
> for removing redundant rows (rows with very similar values for all
> columns)
> from the training data so as to obtain a realistic classification
> performance
> upon x-validation? The only one I can think of is clustering the data into
> an
> arbitary number of clusters and selecting one observation from each
> cluster.
>
> e.g
> library(cluster)
> x <- rbind(cbind(rnorm(10,0,0.5), rnorm(10,0,0.5)),
>            cbind(rnorm(10,5,2.5), rnorm(15,5,2.5)),
>            cbind(rnorm(10,15,0.5), rnorm(15,15,0.5)),
>            cbind(rnorm(5,5,0.1), rnorm(5,5,0.1)))
>
> pamx <- pam(x, 15)
>
> y=array(NA, dim=c(15,ncol(x)))
> for(i in 1:15){
>         y[i,]=x[sample(which(pamx$clustering==i), 1),]
> }
>
> This seems a bit subjective though... Any better ideas?
>
> Eleni Rapsomaniki
>
>
>
> ------------------------------
>
> Message: 17
> Date: Fri, 10 Aug 2007 13:06:11 +0100
> From: Daniel Brewer <daniel.brewer@icr.ac.uk>
> Subject: Re: [R] Positioning text in top left corner of plot
> To: Jim Lemon <jim@bitwrit.com.au>
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <46BC54B3.3090506@icr.ac.uk>
> Content-Type: text/plain; charset=ISO-8859-1
>
> Jim Lemon wrote:
> > Daniel Brewer wrote:
> >> Thanks for the replies, but I still cannot get what I want.  I do not
> >> want the label inside the plot area, but in the top left of the paper,
> I
> >> suppose in the margins.  When I try to use text to do this, it does not
> >> seem to plot it outside the plot area.  I have also tried to use mtext,
> >> but that does not really cut it, as I cannot get the label in the
> >> correct position.  Ideally, it would be best if I could use legend but
> >> have it outside the plot area.
> >>
> >> Any ideas?
> >>
> > Hi Dan,
> >
> > Try this:
> >
> > plot(1:5)
> > par(xpd=TRUE)
> > text(0.5,5.5,"Outside")
> > par(xpd=FALSE)
> >
> > Jim
>
> Here is what I used in the end:
> par(xpd=T)
> text(-0.15*
> (par("usr")[2]-par("usr")[1]),par("usr")[4]+0.14*(par("usr")[4]-par("usr")[3]),labels[i],cex=
> 1.5)
> par(xpd=F)
>
> Ans that worked a treat.
> Thanks
>
> Dan
>
> --
> **************************************************************
> Daniel Brewer, Ph.D.
> Institute of Cancer Research
> Email: daniel.brewer@icr.ac.uk
> **************************************************************
>
> The Institute of Cancer Research: Royal Cancer Hospital, a charitable
> Company Limited by Guarantee, Registered in England under Company No. 534147
> with its Registered Office at 123 Old Brompton Road, London SW7 3RP.
>
> This e-mail message is confidential and for use by the addre...{{dropped}}
>
>
>
> ------------------------------
>
> Message: 18
> Date: Fri, 10 Aug 2007 14:11:37 +0200
> From: Rainer Hurling <rhurlin@gwdg.de>
> Subject: Re: [R] having problems with factor()
> To: Jabez Wilson <jabezwuk@yahoo.co.uk>,        Henrique Dallazuanna
>         <wwwhsd@gmail.com>
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <46BC55F9.1020601@gwdg.de>
> Content-Type: text/plain; charset=ISO-8859-15; format=flowed
>
> I am afraid the above example will not work. In original dataset of
> Jabez Wilson numerical range is from 0..7.
>
> So try this one:
>
> df<-as.factor(c("0","I","II","III","IV","V","VI","VII")[df$area+1])
>
> Hope this is what you want,
> Rainer
>
>
> Henrique Dallazuanna schrieb:
> > Hi,
> >
> > df
> >     ht area
> > 1  320    3
> > 2  410    4
> > 3  230    2
> > 4  360    3
> > 5  126    1
> > 6  280    2
> > 7  260    2
> > 8  280    2
> > 9  280    2
> > 10 260    2
> >
> > df$area <- as.factor(df$area)
> > levels(df$area) <- c("I", "II", "III", "IV")
> >
> >
> > On 10/08/07, Jabez Wilson <jabezwuk@yahoo.co.uk> wrote:
> >> Dear R Help,
> >> I have a set of data of heights of trees described by area that they
> >> are in. The areas are numerical (0 to 7).
> >>
> >>     ht    area
> >> 1   320   3
> >> 2   410   4
> >> 3   230   2
> >> 4   360   3
> >> 5   126   1
> >> 6   280   2
> >> 7   260   2
> >> 8   280   2
> >> 9   280   2
> >> 10  260   2
> >> .......
> >> 180 450   4
> >> 181  90   1
> >> 182 120   1
> >> 183 440   4
> >> 184 210   2
> >> 185 330   3
> >> 186 210   2
> >> 187 100   1
> >> 188   0   0
> >>
> >> I want to convert the area column values to factors, to do an anova.
> >> However, if I use:
> >>
> >> df$areaf <- factor(df$area,
> >> labels=c("0","I","II","III","IV","V","VI","VII"))
> >>
> >> it gives the following message:
> >>
> >> Error in factor(df$area, labels = c("0", "I", "II", "III", "IV", "V",
> >> "VI",  :
> >>         invalid labels; length 8 should be 1 or 7
> >>
> >> Can anyone help?
> >>
> >> Jabez
>
>
>
> ------------------------------
>
> Message: 19
> Date: Fri, 10 Aug 2007 08:16:05 -0400
> From: "Gabor Grothendieck" <ggrothendieck@gmail.com>
> Subject: Re: [R] [Fwd: Re: How to apply functions over rows of
>         multiple        matrices]
> To: johannes@huesing.name
> Cc: r-help@stat.math.ethz.ch
> Message-ID:
>         <971536df0708100516l2b6638b4rb9eb4d2682157074@mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> 1. matrices are stored columnwise so R is better at column-wise operations
> than row-wise.
>
> 2. Here is one way to do it (although I am not sure its better than the
> index approach):
>
>    row.apply <- function(f, a, b)
>       t(mapply(f, as.data.frame(t(a)), as.data.frame(t(b))))
>
> 3. The code for the example in this post could be simplified to:
>
>     first.1 <- apply(cbind(goldstandard, 1), 1, which.max)
>     ifelse(col(newtest) > first.1, NA, newtest)
>
> 4. given that both examples did not inherently need row by row operations
>    I wonder if that is the wrong generalization in the first place?
>
>
> On 8/10/07, Johannes H?sing <johannes@huesing.name> wrote:
> > [Apologies to Gabor, who I sent a personal copy of the reply
> > erroneously instead of posting to List directly]
> >
> > [...]
> > >  Perhaps what you really intend is to
> > > take the average over those elements in each row of the first matrix
> > which correspond to 1's in the second in the corresponding
> > > row of the second.  In that case its just:
> > >
> > > rowSums(newtest * goldstandard) / rowSums(goldstandard)
> > >
> >
> > Thank you for clearing my thoughts about the particular example.
> > My question was a bit more general though, as I have different
> > functions which are applied row-wise to multiple matrices. An
> > example that sets all values of a row of matrix A to NA after the
> > first occurrence of TRUE in matrix B.
> >
> > fillfrom <- function(applvec, testvec=NULL) {
> >  if (is.null(testvec)) testvec <- applvec
> >  if (length(testvec) != length(applvec)) {
> >    stop("applvec and testvec have to be of same length!")
> >  } else if(any(testvec, na.rm=TRUE)) {
> >    applvec[min(which(testvec)) : length(applvec)] <- NA
> >  }
> >  applvec
> > }
> >
> > fillafter <- function(applvec, testvec=NULL) {
> >  if (is.null(testvec)) testvec <- applvec
> >  fillfrom(applvec, c(FALSE, testvec[-length(testvec)]))
> > }
> >
> > numtest <- 6
> > numsubj <- 20
> >
> > newtest <- array(rbinom(numtest*numsubj, 1, .5),
> >    dim=c(numsubj, numtest))
> > goldstandard <- array(rbinom(numtest*numsubj, 1, .5),
> >    dim=c(numsubj, numtest))
> >
> > newtest.NA <- t(sapply(1:nrow(newtest), function(i) {
> >  fillafter(newtest[i,], goldstandard[i,]==1)}))
> >
> > My general question is if R provides some syntactic sugar
> > for the awkward sapply(1:nrow(A)) expression. Maybe in this
> > case there is also a way to bypass the apply mechanism and
> > my way of thinking about the problem has to be adapted. But
> > as the *apply calls are galore in R, I feel this is a standard
> > way of dealing with vectors and matrices.
> >
> >
> >
> >
> >
> > --
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>
>
> ------------------------------
>
> Message: 20
> Date: Fri, 10 Aug 2007 14:42:16 +0200
> From: Petr PIKAL <petr.pikal@precheza.cz>
> Subject: [R] Odp: help with counting how many times each value occur
>         in      each    column
> To: Tom Cohen <tom.cohen78@yahoo.se>
> Cc: r-help@stat.math.ethz.ch
> Message-ID:
>         <
> OFB69CE2C1.78D8A4B4-ONC1257333.00459B57-C1257333.0045C459@precheza.cz>
>
> Content-Type: text/plain; charset="ISO-8859-2"
>
> Hi
>
> > mat<-sample(c(-50,0,-100), 100,replace=T)
> > dim(mat)<-c(10,10)
> > mat
>       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> [1,]    0    0    0    0  -50    0    0    0    0     0
> [2,] -100 -100  -50  -50    0    0 -100  -50 -100   -50
> [3,]    0  -50 -100 -100    0  -50 -100    0    0  -100
> [4,]    0 -100    0  -50 -100 -100  -50  -50    0  -100
> [5,]  -50  -50    0    0    0 -100 -100 -100    0  -100
> [6,]    0    0  -50  -50    0    0 -100 -100  -50  -100
> [7,] -100 -100 -100  -50 -100    0 -100 -100    0  -100
> [8,] -100    0    0    0    0 -100    0 -100    0  -100
> [9,] -100    0  -50 -100  -50    0    0  -50    0  -100
> [10,]  -50 -100    0    0  -50  -50  -50  -50 -100  -100
>
> > apply(mat, 2, table)
>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> -100    4    4    2    2    2    3    5    4    2     8
> -50     2    2    3    4    3    2    2    4    1     1
> 0       4    4    5    4    5    5    3    2    7     1
>
> Transposing and ordering columns is up to you.
>
> Regards
> Petr
>
> r-help-bounces@stat.math.ethz.ch napsal dne 10.08.2007 14:01:44:
>
> > Dear list,
> >   I have the following dataset and want to know how many times each
> value
> > occur in each column.
> >    >data
> >         [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> >  [1,] -100 -100 -100    0    0    0    0    0    0  -100
> >  [2,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [3,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [4,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [5,] -100 -100 -100 -100 -100 -100 -100 -100 -100   -50
> >  [6,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [7,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [8,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [9,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [10,] -100 -100 -100  -50 -100 -100 -100 -100 -100  -100
> > [11,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [12,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [13,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [14,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [15,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [16,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [17,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [18,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [19,] -100 -100 -100    0    0    0    0    0    0  -100
> > [20,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >   The result matrix should look like
> >        -100 0 -50
> > [1]   20
> > [2]   20
> > [3]   20
> > [4]   17
> > [5]   18
> > [6]   18
> > [7]   18  and so on
> > [8]
> > [9]
> > [10]
> >
> > How can I do this in R ?
> >   Thanks alot for your help,
> > Tom
> >
> >
> > ---------------------------------
> >
> > J?mf?r pris p? flygbiljetter och hotellrum:
> http://shopping.yahoo.se/c-169901-
> > resor-biljetter.html
> >    [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
>
> ------------------------------
>
> Message: 21
> Date: Fri, 10 Aug 2007 09:23:38 -0400
> From: "Gabor Grothendieck" <ggrothendieck@gmail.com>
> Subject: Re: [R] help with counting how many times each value occur in
>         each    column
> To: "Tom Cohen" <tom.cohen78@yahoo.se>
> Cc: r-help@stat.math.ethz.ch
> Message-ID:
>         <971536df0708100623o2ebd4845w7ae398fdca97c98e@mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> Try this where we have constructed the example to illustrate that
> it does handle the case where not all values are in each column:
>
>    mat <- matrix(rep(1:6, each = 4), 6)
>
>    table(col(mat), mat)
>
> On 8/10/07, Tom Cohen <tom.cohen78@yahoo.se> wrote:
> > Dear list,
> >  I have the following dataset and want to know how many times each value
> occur in each column.
> >   >data
> >        [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> >  [1,] -100 -100 -100    0    0    0    0    0    0  -100
> >  [2,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [3,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [4,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [5,] -100 -100 -100 -100 -100 -100 -100 -100 -100   -50
> >  [6,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [7,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [8,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  [9,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [10,] -100 -100 -100  -50 -100 -100 -100 -100 -100  -100
> > [11,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [12,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [13,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [14,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [15,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [16,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [17,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [18,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [19,] -100 -100 -100    0    0    0    0    0    0  -100
> > [20,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  The result matrix should look like
> >       -100 0 -50
> > [1]   20
> > [2]   20
> > [3]   20
> > [4]   17
> > [5]   18
> > [6]   18
> > [7]   18  and so on
> > [8]
> > [9]
> > [10]
> >
> > How can I do this in R ?
> >  Thanks alot for your help,
> > Tom
> >
> >
> > ---------------------------------
> >
> > J?mf?r pris p? flygbiljetter och hotellrum:
> http://shopping.yahoo.se/c-169901-resor-biljetter.html
> >        [[alternative HTML version deleted]]
> >
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
> >
>
>
>
> ------------------------------
>
> Message: 22
> Date: Fri, 10 Aug 2007 15:24:34 +0200 (CEST)
> From: Johannes H?sing <johannes@huesing.name>
> Subject: Re: [R] [Fwd: Re: How to apply functions over rows of
>         multiple matrices]
> To: r-help@stat.math.ethz.ch
> Message-ID: <20473.129.206.90.2.1186752274.squirrel@mail.panix.com>
> Content-Type: text/plain;charset=iso-8859-1
>
> > 1. matrices are stored columnwise so R is better at column-wise
> operations
> > than row-wise.
>
> I am seeing this by my code which contains more t() than
> what seems healthy. However, the summaries are patient-wise
> over repeated measurements. Out of convention, I am storing
> patients in rows and measurements in columns.
>
> >
> > 2. Here is one way to do it (although I am not sure its better than the
> > index approach):
> >
> >    row.apply <- function(f, a, b)
> >       t(mapply(f, as.data.frame(t(a)), as.data.frame(t(b))))
> >
>
> Ah, thank you so much. I'll take the generalization to N arguments
> ? la mapply() as an exercise for the reader.
>
> > 3. The code for the example in this post could be simplified to:
> >
> >     first.1 <- apply(cbind(goldstandard, 1), 1, which.max)
> >     ifelse(col(newtest) > first.1, NA, newtest)
> >
>
> Ouch! Consider this scholar slapped.
>
> > 4. given that both examples did not inherently need row by row
> operations
> >    I wonder if that is the wrong generalization in the first place?
> >
>
> Given that you managed to squeeze my 20 lines of code into 2 lines
> AND that row.apply() does not exist in base without many people
> missing it, I'll have to concede this point and eliminate the
> craving for row.apply() in favour of the whole-object approach.
>
>
>
> ------------------------------
>
> Message: 23
> Date: Fri, 10 Aug 2007 13:52:18 +0000 (GMT)
> From: Jabez Wilson <jabezwuk@yahoo.co.uk>
> Subject: Re: [R] Odp:  having problems with factor()
> To: Petr PIKAL <petr.pikal@precheza.cz>
> Cc: R-Help <r-help@stat.math.ethz.ch>
> Message-ID: <396309.57812.qm@web27408.mail.ukl.yahoo.com>
> Content-Type: text/plain
>
> You've spotted it!
>
> table(df$area)
> 0  1  2  3  4  5  7
> 21 27 71 46 19  3  1
>
> There are no values in area 6.
>
> Thank you very much.
>
> Jabez
>
>
> ----- Original Message ----
> From: Petr PIKAL <petr.pikal@precheza.cz>
> To: Jabez Wilson <jabezwuk@yahoo.co.uk>
> Cc: R-Help <r-help@stat.math.ethz.ch>
> Sent: Friday, 10 August, 2007 1:02:21 PM
> Subject: Odp: [R] having problems with factor()
>
>
> Hi
> r-help-bounces@stat.math.ethz.ch napsal dne 10.08.2007 13:41:53:
>
> > Dear R Help,
> > I have a set of data of heights of trees described by area that they are
> in.
> > The areas are numerical (0 to 7).
> >
> >     ht    area
> > 1   320   3
> > 2   410   4
> > 3   230   2
> > 4   360   3
> > 5   126   1
> > 6   280   2
> > 7   260   2
> > 8   280   2
> > 9   280   2
> > 10  260   2
> > .......
> > 180 450   4
> > 181  90   1
> > 182 120   1
> > 183 440   4
> > 184 210   2
> > 185 330   3
> > 186 210   2
> > 187 100   1
> > 188   0   0
> >
> > I want to convert the area column values to factors, to do an anova.
> However, if I use:
> >
> > df$areaf <- factor(df$area,
> labels=c("0","I","II","III","IV","V","VI","VII"))
> >
> > it gives the following message:
> >
>
> Hm, maybe some of the values are missing
>
> > num<-sample(1:3, 10, replace=T)
> > num
> [1] 1 3 1 2 3 3 1 3 3 3
> > factor(num, labels=c("O", "I", "II"))
> [1] O  II O  I  II II O  II II II
> Levels: O I II
>
> > factor(num, labels=c("O", "I", "II", "III"))
> Error in factor(num, labels = c("O", "I", "II", "III")) :
>         invalid labels; length 4 should be 1 or 3
> >
>
> try
>
> table(df$area)
> to see what level you really have
>
> Regards
> Petr
>
>
> > Error in factor(df$area, labels = c("0", "I", "II", "III", "IV", "V",
> "VI",  :
> >         invalid labels; length 8 should be 1 or 7
> >
> > Can anyone help?
> >
> > Jabez
> >
> >
> >       ___________________________________________________________
> >
> > now.
> >
> >    [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
>       ___________________________________________________________
>
> now.
>
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 24
> Date: Fri, 10 Aug 2007 08:28:07 -0600
> From: "Greg Snow" <Greg.Snow@intermountainmail.org>
> Subject: Re: [R] Plot legend in margin
> To: "Daniel Brewer" <daniel.brewer@icr.ac.uk>,  "Lauri Nikkinen"
>         <lauri.nikkinen@iki.fi>, r-help@stat.math.ethz.ch
> Message-ID:
>         <07E228A5BE53C24CAD490193A7381BBBB2F677@LP-EXCHVS07.CO.IHC.COM>
> Content-Type: text/plain; charset=us-ascii
>
> Another couple of things to think about:
>
> You could use the layout function to set up your multiple plots and
> include an extra plotting area at the bottom to place the legend in.
>
> If you stick with the solution below then the cnvrt.coords function from
> the TeachingDemos package may be useful (will help you find the
> coordinates relative to the last plot).
>
> Hope this helps,
>
> --
> Gregory (Greg) L. Snow Ph.D.
> Statistical Data Center
> Intermountain Healthcare
> greg.snow@intermountainmail.org
> (801) 408-8111
>
>
>
> > -----Original Message-----
> > From: r-help-bounces@stat.math.ethz.ch
> > [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Daniel Brewer
> > Sent: Friday, August 10, 2007 4:55 AM
> > To: Lauri Nikkinen; r-help@stat.math.ethz.ch
> > Subject: Re: [R] Plot legend in margin
> >
> > Thanks.  That got me onto the right track.  Because it is a
> > multiplot and I wanted it along the bottom, I found that I
> > had to use par(xpd=NA) and then position it relative to the
> > last of the multiplots.  After a bit of trial and error I got there.
> >
> > Thanks
> >
> > Lauri Nikkinen wrote:
> > > Very simple example:
> > >
> > > opar <- par(mar = c(10, 4, 4, 4))
> > > plot(1:10)
> > > lines(1:10)
> > > par(xpd=TRUE)
> > > legend(4,-1.5,lty=1, col="black", legend="straigh line")
> > > par(opar)
> > >
> > > -Lauri
> >
> > --
> > **************************************************************
> > Daniel Brewer, Ph.D.
> > Institute of Cancer Research
> > Email: daniel.brewer@icr.ac.uk
> > **************************************************************
> >
> > The Institute of Cancer Research: Royal Cancer Hospital, a
> > charitable Company Limited by Guarantee, Registered in
> > England under Company No. 534147 with its Registered Office
> > at 123 Old Brompton Road, London SW7 3RP.
> >
> > This e-mail message is confidential and for use by the\ > ...{{dropped}}
>
>
>
> ------------------------------
>
> Message: 25
> Date: Fri, 10 Aug 2007 10:28:25 -0400
> From: Roland Rau <roland.rproject@gmail.com>
> Subject: Re: [R] Seasonality
> To: Alberto Monteiro <albmont@centroin.com.br>
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <46BC7609.4020906@gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
> Alberto Monteiro wrote:
> > I have a time series x = f(t), where t is taken for each
> > month. What is the best function to detect if _x_ has a seasonal
> > variation? If there is such seasonal effect, what is the
> > best function to estimate it?
> >
> From my own experience, I had the impression that there is nothing like
> a best approach to estimate the seasonal component of time series data.
>
> Maybe it is possible for you to simulate the assumed nature of your data
> (variable trend? variable seasonal pattern? count data with
> overdispersion? maybe a bimodal pattern every year?) and then try
> various of these methods and check if they can extract your input
> approximately correctly?
>
> Best,
> Roland
>
>
>
> ------------------------------
>
> Message: 26
> Date: Fri, 10 Aug 2007 10:47:49 -0400
> From: Fran?ois Pinard <pinard@iro.umontreal.ca>
> Subject: Re: [R] help with counting how many times each value occur in
>         each    column
> To: Tom Cohen <tom.cohen78@yahoo.se>
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <20070810144749.GA5766@phenix.progiciels-bpi.ca>
> Content-Type: text/plain; charset=iso-8859-1; format=flowed
>
> [Tom Cohen]
>
> >  I have the following dataset and want to know how many times each value
> occur in each column.
>
> >   >data
> >        [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> > [1,] -100 -100 -100    0    0    0    0    0    0  -100
> > [2,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [3,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [4,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [5,] -100 -100 -100 -100 -100 -100 -100 -100 -100   -50
> > [6,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [7,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [8,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> > [9,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >[10,] -100 -100 -100  -50 -100 -100 -100 -100 -100  -100
> >[11,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >[12,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >[13,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >[14,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >[15,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >[16,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >[17,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >[18,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >[19,] -100 -100 -100    0    0    0    0    0    0  -100
> >[20,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> >  The result matrix should look like
> >       -100 0 -50
> >[1]   20
> >[2]   20
> >[3]   20
> >[4]   17
> >[5]   18
> >[6]   18
> >[7]   18  and so on
> >[8]
> >[9]
> >[10]
>
> Presuming that "data" is a matrix, one could try a sequence like this:
>
> dataf <- factor(data)
> dim(dataf) <- dim(data)
> result <- t(apply(dataf, 2, tabulate, nlevels(dataf)))
> colnames(result) <- levels(dataf)
> result
>
> If you want the columns sorted, you might decide the order of the levels
> on the "factor()" call, or explicitly reorder columns afterwards.
>
> --
> Fran?ois Pinard   http://pinard.progiciels-bpi.ca
>
>
>
> ------------------------------
>
> Message: 27
> Date: Fri, 10 Aug 2007 14:47:45 +0200
> From: "Gasper Cankar" <gcanka@guest.arnes.si>
> Subject: Re: [R] help with counting how many times each value occur in
>         eachcolumn
> To: <r-help@stat.math.ethz.ch>
> Message-ID: <001301c7db4c$a699c040$5200a8c0@ric.si>
> Content-Type: text/plain;       charset="ISO-8859-1"
>
>
> Tom,
>
> If all values (-100,0,-50) would be in every column then simple
>
> apply(data,2,table)
>
> would work. Even if there aren0t all values in every column you could
> correct that and insert additional lines with all values for all columns
> like
>
> data <- cbind(data,matrix(ncol=10,nrow=3,rep(c(-100,0,-50),10)))
>
> and then do
>
> apply(data,2,table)-1
>
> to get correct results. But someone on a list can probably make much more
> elegant solution.
>
> Bye,
>
> Gasper Cankar, PhD
> Researcher
> National Examinations Centre
> Slovenia
>
> -----Original Message-----
> From: Tom Cohen [mailto:tom.cohen78@yahoo.se]
> Sent: Friday, August 10, 2007 2:02 PM
> To: r-help@stat.math.ethz.ch
> Subject: [R] help with counting how many times each value occur in
> eachcolumn
>
> Dear list,
>   I have the following dataset and want to know how many times each value
> occur in each column.
>    >data
>         [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> [1,] -100 -100 -100    0    0    0    0    0    0  -100
> [2,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [3,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [4,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [5,] -100 -100 -100 -100 -100 -100 -100 -100 -100   -50
> [6,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [7,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [8,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [9,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [10,] -100 -100 -100  -50 -100 -100 -100 -100 -100  -100
> [11,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [12,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [13,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [14,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [15,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [16,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [17,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [18,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
> [19,] -100 -100 -100    0    0    0    0    0    0  -100
> [20,] -100 -100 -100 -100 -100 -100 -100 -100 -100  -100
>   The result matrix should look like
>        -100 0 -50
> [1]   20
> [2]   20
> [3]   20
> [4]   17
> [5]   18
> [6]   18
> [7]   18  and so on
> [8]
> [9]
> [10]
>
> How can I do this in R ?
>   Thanks alot for your help,
> Tom
>
>
> ---------------------------------
>
> J?mf?r pris p? flygbiljetter och hotellrum:
> http://shopping.yahoo.se/c-169901-resor-biljetter.html
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 28
> Date: Fri, 10 Aug 2007 10:57:50 -0400
> From: Fran?ois Pinard <pinard@iro.umontreal.ca>
> Subject: Re: [R] help with counting how many times each value occur in
>         each    column
> To: Gabor Grothendieck <ggrothendieck@gmail.com>
> Cc: r-help@stat.math.ethz.ch, Tom Cohen <tom.cohen78@yahoo.se>
> Message-ID: <20070810145750.GB5766@phenix.progiciels-bpi.ca>
> Content-Type: text/plain; charset=iso-8859-1; format=flowed
>
> [Gabor Grothendieck]
>
> >   table(col(mat), mat)
>
> Clever, simple, and elegant! :-)
>
> --
> Fran?ois Pinard   http://pinard.progiciels-bpi.ca
>
>
>
> ------------------------------
>
> Message: 29
> Date: Fri, 10 Aug 2007 11:17:57 -0400
> From: "Eric Turkheimer" <ent3c@virginia.edu>
> Subject: [R] rfImpute
> To: r-help@stat.math.ethz.ch
> Message-ID:
>         <a42e7d400708100817v4eef5495o8b9e1c925d4720d7@mail.gmail.com>
> Content-Type: text/plain
>
> I am having trouble with the rfImpute function in the randomForest
> package.
> Here is a sample...
>
> clunk.roughfix<-na.roughfix(clunk)
> >
> > clunk.impute<-rfImpute(CONVERT~.,data=clunk)
> ntree      OOB      1      2
>   300:  26.80%  3.83% 85.37%
> ntree      OOB      1      2
>   300:  18.56%  5.74% 51.22%
> Error in randomForest.default(xf, y, ntree = ntree, ..., do.trace = ntree,
> :
>         NA not permitted in predictors
>
> So roughFix works, but rfImpute doesn't....
>
> Thanks,
> Eric
> ent3c *at* virginia.edu
>
> --
> Eric Turkheimer, PhD
> Department of Psychology
> University of Virginia
> PO Box 400400
> Charlottesville, VA  22904-4400
>
> 434-982-4732
> 434-982-4766 (FAX)
>
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 30
> Date: Fri, 10 Aug 2007 11:18:38 -0400
> From: "Rose Hoberman" <roseh@cs.cmu.edu>
> Subject: [R] smoothing function for proportions
> To: r-help@stat.math.ethz.ch
> Message-ID:
>         <64c985560708100818w1d78e72ftfc11a3e8916f0b@mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> I am looking for a function that can fit a smooth function to a vector
> of estimated proportions, such that the smoothed value is within
> specified confidence bounds of each proportion.  In other words, given
> a small number of trials and large confidence intervals, I would
> prefer the function to vary smoothly, but given a large number of
> trials and small confidence intervals, I would prefer the function to
> lie within the confidence intervals, even if it is not smooth.
>
> I have attached a postscript file illustrating a data set I would like
> to smooth.  As the figure shows, for large values of x, I have few
> data points, and so the ML estimate of the proportion varies widely,
> and the confidence intervals are very large.  When I use the
> smooth.spline function with a large value of spar (the red line), the
> function is not as smooth as desired for large values of x.  When I
> use a smaller value of spar (the green line), the function fails to
> stay within the confidence bounds of the proportions.   Is there a
> smoothing function for which I can specify upper and lower limits for
> the y value for specific values of x?
>
> Thanks for any suggestions,
>
> Rose
>
>
>
> ------------------------------
>
> Message: 31
> Date: Fri, 10 Aug 2007 11:23:09 -0400
> From: "Rose Hoberman" <roseh@cs.cmu.edu>
> Subject: Re: [R] smoothing function for proportions
> To: r-help@stat.math.ethz.ch
> Message-ID:
>         <64c985560708100823h35e041a4i6301b8ee8f2917dd@mail.gmail.com>
> Content-Type: text/plain; charset="iso-8859-1"
>
> Sorry, forgot to attach the graph.
>
> On 8/10/07, Rose Hoberman <roseh@cs.cmu.edu> wrote:
> > I am looking for a function that can fit a smooth function to a vector
> > of estimated proportions, such that the smoothed value is within
> > specified confidence bounds of each proportion.  In other words, given
> > a small number of trials and large confidence intervals, I would
> > prefer the function to vary smoothly, but given a large number of
> > trials and small confidence intervals, I would prefer the function to
> > lie within the confidence intervals, even if it is not smooth.
> >
> > I have attached a postscript file illustrating a data set I would like
> > to smooth.  As the figure shows, for large values of x, I have few
> > data points, and so the ML estimate of the proportion varies widely,
> > and the confidence intervals are very large.  When I use the
> > smooth.spline function with a large value of spar (the red line), the
> > function is not as smooth as desired for large values of x.  When I
> > use a smaller value of spar (the green line), the function fails to
> > stay within the confidence bounds of the proportions.   Is there a
> > smoothing function for which I can specify upper and lower limits for
> > the y value for specific values of x?
> >
> > Thanks for any suggestions,
> >
> > Rose
> >
> -------------- next part --------------
> A non-text attachment was scrubbed...
> Name: smoothProportions.ps
> Type: application/postscript
> Size: 11419 bytes
> Desc: not available
> Url :
> https://stat.ethz.ch/pipermail/r-help/attachments/20070810/09031ae4/attachment-0001.ps
>
> ------------------------------
>
> Message: 32
> Date: Fri, 10 Aug 2007 09:30:42 -0600
> From: "Greg Snow" <Greg.Snow@intermountainmail.org>
> Subject: Re: [R] how to include bar values in a barplot?
> To: "Donatas G." <dgvirtual@akl.lt>, r-help@stat.math.ethz.ch
> Message-ID:
>         <07E228A5BE53C24CAD490193A7381BBBB2F698@LP-EXCHVS07.CO.IHC.COM>
> Content-Type: text/plain; charset=us-ascii
>
> Welcome to the world of R.
>
> I'm glad that you found the discussion enlightening, now that you have
> thought about things a bit, here is some code to try out that shows some
> of the alternatives to the original plots you provided (which is best
> depends on your audience and what your main question of interest is
> (which comparisons are most important):
>
> tmp <- c(34,22,77)
>
> tmp2 <- barplot(tmp, names=LETTERS[1:3])
>
> # put numbers at bottom of bars
> axis(1, at=tmp2, labels=as.character(tmp), tick=FALSE, line = -1)
>
> # put numbers at top of plot
> axis(3, at=tmp2, labels=as.character(tmp), tick=FALSE)
>
>
> # horizontal boxplot
>
> op <- par(mar=c(5,6,4,6)+0.1)
> tmp2 <- barplot(tmp, names=LETTERS[1:3], horiz=TRUE)
>
> # put numbers on the right
> axis(4, at=tmp2, labels=as.character(tmp), tick=FALSE, las=1)
>
> par(op)
>
> # the dotplot
> library(Hmisc)
> dotchart2(tmp, labels=LETTERS[1:3], auxdata=tmp, xlim=range(0,tmp))
>
>
> # alternatives to stacked bars
>
> tmp1 <- c(8, 22, 60, 10,  10, 21, 59, 10)
> tmp2 <- factor(rep(c('A','B'), each=4))
> tmp3 <- factor(rep(1:4, 2))
>
> dotchart2(tmp1, groups=tmp2, labels=tmp3, xlim=range(0,tmp1))
> dotchart2(tmp1, groups=tmp3, labels=tmp2, xlim=range(0,tmp1))
>
> library(lattice)
> tmp <- data.frame( tmp1=tmp1, tmp2=tmp2, tmp3=tmp3 )
>
> dotplot( tmp2~tmp1, data=tmp, groups=tmp3, pch=levels(tmp3),
>         scales=list(x=list(limits=range(0,tmp1))) )
> dotplot( tmp3~tmp1, data=tmp, groups=tmp2, pch=levels(tmp2),
>         xlim=range(0,tmp1) )
>
>
>
>
>
> --
> Gregory (Greg) L. Snow Ph.D.
> Statistical Data Center
> Intermountain Healthcare
> greg.snow@intermountainmail.org
> (801) 408-8111
>
>
>
> > -----Original Message-----
> > From: r-help-bounces@stat.math.ethz.ch
> > [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Donatas G.
> > Sent: Friday, August 10, 2007 2:15 AM
> > To: r-help@stat.math.ethz.ch
> > Subject: Re: [R] how to include bar values in a barplot?
> >
> > Quoting Greg Snow <Greg.Snow@intermountainmail.org>:
> >
> > > My original intent was to get the original posters out of
> > the mode of
> > > thinking they want to match what the spreadsheet does and into
> > > thinking about what message they are trying to get across.  To get
> > > them (and possibly others) thinking I made the statements a
> > bit more
> > > bold than my actual position (I did include a couple of qualifiers).
> >
> > As an original poster (and a brand new user of R), I would
> > like to comment on the educational experience I have just received. ;)
> >
> > The discussion was interesting and enlightening, and gives
> > some good ideas about the ways (tables, graphs, graphs with
> > numbers etc.) to get the data accross to the ones one is
> > presenting to. I see some of you guys do feel quite strongly
> > about it, which is fine for me. I do not. I usually care for
> > barplot aesthetics and informativeness more than for visual
> > simplicity. That may change in time :)
> >
> > I see R graphical capabilities are huge but hard to access at
> > times - that is when spreadsheet seems preferrable. For
> > example, as a user of Linux I still cannot figure out why the
> > fonts (and graphics in general) look much more ugly on R in
> > Linux than they do in R on Windows - no smoothing, sub-pixell
> > hinting, anything like that. That is what my next free time
> > homework on R will be about
> > :)
> >
> > Sincerely
> >
> > Donatas Glodenis
> > PhD candidate
> > Department of Sociology of the Faculty of Philosophy Vilnius
> > University Lithuania
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>
>
> ------------------------------
>
> Message: 33
> Date: Fri, 10 Aug 2007 11:37:30 -0400
> From: "Weiwei Shi" <helprhelp@gmail.com>
> Subject: Re: [R] a question on lda{MASS}
> To: "r-help@stat.math.ethz.ch" <R-help@stat.math.ethz.ch>
> Message-ID:
>         <cdf817830708100837s444bd626g422844e24fd29460@mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> hi,
>
> maybe I should re-phrase my question a bit:
>
> is there a way to get explicit formulae like Y ~ sum of CiXi from the
> model build by lda{MASS} to calculate $x (value) ?
>
> I assume scaling is the coeff and Xi is from test data and Y is $x
> called LD1. But I want to confirm this.
>
> Thanks.
>
> Weiwei
>
> On 8/9/07, Weiwei Shi <helprhelp@gmail.com> wrote:
> > hi,
> >
> > assume
> > val is the test data while m is lda model value by using CV=F
> >
> > x = predict(m, val)
> >
> > val2 = val[, 1:(ncol(val)-1)] # the last column is class label
> >
> > # col is sample, row is variable
> >
> > then I am wondering if
> >
> > x$x == (apply(val2*m$scaling), 2, sum)
> >
> > i.e., the scaling (is it coeff vector?) times val data and sum is the
> > discrimant result $x?
> >
> > Thanks.
> >
> > --
> > Weiwei Shi, Ph.D
> > Research Scientist
> > GeneGO, Inc.
> >
> > "Did you always know?"
> > "No, I did not. But I believed..."
> > ---Matrix III
> >
>
>
> --
> Weiwei Shi, Ph.D
> Research Scientist
> GeneGO, Inc.
>
> "Did you always know?"
> "No, I did not. But I believed..."
> ---Matrix III
>
>
>
> ------------------------------
>
> Message: 34
> Date: Fri, 10 Aug 2007 09:46:52 -0600
> From: "Greg Snow" <Greg.Snow@intermountainmail.org>
> Subject: Re: [R] Subject: Re: how to include bar values in a barplot?
> To: "Jim Lemon" <jim@bitwrit.com.au>,   "Gabor Grothendieck"
>         <ggrothendieck@gmail.com>
> Cc: r-help@stat.math.ethz.ch
> Message-ID:
>         <07E228A5BE53C24CAD490193A7381BBBB2F6A2@LP-EXCHVS07.CO.IHC.COM>
> Content-Type: text/plain; charset=us-ascii
>
>
>
> Jim Lemon Wrote:
>
> > I also greatly enjoyed Ted's rebuttal of the "Bar charts are
> > evil and must be banned" argument. If bar charts are
> > appropriate for the audience, give 'em bar charts. One great
> > way to turn off your customers is to tell them what they can
> > and can't do with your product.
>
> I don't remember anyone saying that barcharts are evil or that they
> should be banned (3-D bar charts and pie charts on the other hand ...).
>
> I think that a variation on fortune(108) applies here.  While barcharts
> may be appropriate for some audiences, it is also appropriate to educate
> our audiences to better alternatives.
>
>
> --
> Gregory (Greg) L. Snow Ph.D.
> Statistical Data Center
> Intermountain Healthcare
> greg.snow@intermountainmail.org
> (801) 408-8111
>
>
>
> ------------------------------
>
> Message: 35
> Date: Fri, 10 Aug 2007 10:51:42 -0500
> From: roger koenker <rkoenker@uiuc.edu>
> Subject: Re: [R] smoothing function for proportions
> To: roseh@cs.cmu.edu
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <C151C902-F6E5-45C4-A2F7-85D07856393F@uiuc.edu>
> Content-Type: text/plain; charset=US-ASCII; delsp=yes; format=flowed
>
> It is not entirely clear what you are using for y values in
> smooth.spline,
> but it would appear that it is just the point estimates.  I would
> suggest
> using instead -- at each x value -- a few equally spaced quantiles of
> the estimated proportions.  Implicitly, smooth.spline expects to be
> fitting
> a mean curve to data that has constant variance, so you might also
> consider reweighting to approximate this, as well.
>
>
> url:    www.econ.uiuc.edu/~roger            Roger Koenker
> email    rkoenker@uiuc.edu            Department of Economics
> vox:     217-333-4558                University of Illinois
> fax:       217-244-6678                Champaign, IL 61820
>
>
> On Aug 10, 2007, at 10:23 AM, Rose Hoberman wrote:
>
> > Sorry, forgot to attach the graph.
> >
> > On 8/10/07, Rose Hoberman <roseh@cs.cmu.edu> wrote:
> >> I am looking for a function that can fit a smooth function to a
> >> vector
> >> of estimated proportions, such that the smoothed value is within
> >> specified confidence bounds of each proportion.  In other words,
> >> given
> >> a small number of trials and large confidence intervals, I would
> >> prefer the function to vary smoothly, but given a large number of
> >> trials and small confidence intervals, I would prefer the function to
> >> lie within the confidence intervals, even if it is not smooth.
> >>
> >> I have attached a postscript file illustrating a data set I would
> >> like
> >> to smooth.  As the figure shows, for large values of x, I have few
> >> data points, and so the ML estimate of the proportion varies widely,
> >> and the confidence intervals are very large.  When I use the
> >> smooth.spline function with a large value of spar (the red line), the
> >> function is not as smooth as desired for large values of x.  When I
> >> use a smaller value of spar (the green line), the function fails to
> >> stay within the confidence bounds of the proportions.   Is there a
> >> smoothing function for which I can specify upper and lower limits for
> >> the y value for specific values of x?
> >>
> >> Thanks for any suggestions,
> >>
> >> Rose
> >>
> >> <smoothProportions.ps>
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-
> > guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
>
> ------------------------------
>
> Message: 36
> Date: Fri, 10 Aug 2007 16:26:29 +0000
> From: laran gines <larangines@hotmail.com>
> Subject: [R] GLM with tweedie: NA for AIC
> To: <r-help@stat.math.ethz.ch>
> Message-ID: <BAY110-W13908E5B3D08A6B034E36C9E10@phx.gbl>
> Content-Type: text/plain
>
> Dear R users;
>
> I am modelling densities of some species of birds, so I have a problem
> with a great ammount of zeros.
> I have decided to try GLMs with the tweedie family, but in all the models
> I have tried  I got an NA for the AIC value.
> Just  to check the problem I've compared the a glm using the Gaussian
> family with the identity link and a glm using the tweedie family with
> var.power=0 and link.power=1. These are equal, as expected, except the
> fact that the tweedie output gives me an NA for the AIC.
> Can anyone help me with this problem?
> Below you can find the two outputs I refer.
>
> Best Wishes;
>
> Catarina
>
> > summary(glm(formula=ACIN~DIST_REF+DIST_H2O+DIST_OST+
> COTA+H2O_SUP+vasa,family=gaussian(link="identity")))
> Call:glm(formula = ACIN ~ DIST_REF + DIST_H2O + DIST_OST + COTA +
> H2O_SUP + vasa, family = gaussian(link = "identity"))
> Deviance Residuals:       Min         1Q     Median
> 3Q        Max  -0.112792  -0.042860  -0.021113  -0.006311   1.551824
> Coefficients:              Estimate Std. Error t value
> Pr(>|t|)  (Intercept) -6.625e-02  5.454e-02  -1.215   0.2256  DIST_REF
> 3.581e-06  1.336e-05   0.268   0.7889  DIST_H2O    -3.168e-05  1.527e-05
>   -2.074   0.0391 *DIST_OST    -1.799e-05  1.953e-05  -0.921   0.3579  COTA
> 5.648e-04  2.470e-04   2.287   0.0230 *H2O_SUP     -2.172e-04  3.994e-04
>   -0.544   0.5870  vasa         3.695e-02  4.573e-02   0.808   0.4199  ---Signif.
> codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
> (Dispersion parameter for gaussian family taken to be 0.02151985)
>     Null deviance: 5.6028  on 257  degrees of freedomResidual deviance:
> 5.4015  on 251  degrees of freedomAIC: -249.33
> Number of Fisher Scoring iterations: 2
>
>
> > summary(glm(formula=ACIN~DIST_REF+DIST_H2O+DIST_OST+
> COTA+H2O_SUP+vasa,control=glm.control(maxit=750),family=tweedie(
> var.power=0, link.power=1)))
> Call:glm(formula = ACIN ~ DIST_REF + DIST_H2O + DIST_OST + COTA +
> H2O_SUP + vasa, family = tweedie(var.power = 0, link.power = 1),
> control = glm.control(maxit = 750))
> Deviance Residuals:       Min         1Q     Median
> 3Q        Max  -0.112792  -0.042860  -0.021113  -0.006311   1.551824
> Coefficients:              Estimate Std. Error t value
> Pr(>|t|)  (Intercept) -6.625e-02  5.454e-02  -1.215   0.2256  DIST_REF
> 3.581e-06  1.336e-05   0.268   0.7889  DIST_H2O    -3.168e-05  1.527e-05
>   -2.074   0.0391 *DIST_OST    -1.799e-05  1.953e-05  -0.921   0.3579  COTA
> 5.648e-04  2.470e-04   2.287   0.0230 *H2O_SUP     -2.172e-04  3.994e-04
>   -0.544   0.5870  vasa         3.695e-02  4.573e-02   0.808   0.4199  ---Signif.
> codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
> (Dispersion parameter for Tweedie family taken to be 0.02151985)
>     Null deviance: 5.6028  on 257  degrees of freedomResidual deviance:
> 5.4015  on 251  degrees of freedomAIC: NA
> Number of Fisher Scoring iterations: 2
>
> _________________________________________________________________
> Conheça o Windows Live Spaces, a rede de relacionamentos conectada ao
> Messenger!
>
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 37
> Date: Thu, 9 Aug 2007 21:56:24 -0700
> From: "Nordlund, Dan (DSHS/RDA)" <NordlDJ@dshs.wa.gov>
> Subject: [R] Test (ignore)
> To: r-help@stat.math.ethz.ch
> Message-ID:
>         <941871A13165C2418EC144ACB212BDB04E1354@dshsmxoly1504g.dshs.wa.lcl
> >
> Content-Type: text/plain; charset=utf-8
>
>
>
> Daniel J. Nordlund
> Research and Data Analysis
> Washington State Department of Social and Health Services
> Olympia, WA  98504-5204
>
>
>
> ------------------------------
>
> Message: 38
> Date: Fri, 10 Aug 2007 10:35:23 -0600
> From: "Emilio Gagliardi" <emilio@ualberta.ca>
> Subject: Re: [R] Help using gPath
> To: "Paul Murrell" <paul@stat.auckland.ac.nz>
> Cc: General R-Help <r-help@stat.math.ethz.ch>
> Message-ID:
>         <3a9ebf340708100935m43ff3eabp942c49a444387169@mail.gmail.com>
> Content-Type: text/plain
>
> haha Paul,
>
>
> It's important not only to post code, but also to make sure that other
> > people can run it (i.e., include real data or have the code generate
> > data or use one of R's predefined data sets).
>
>
> Oh, I hadn't thought of using the predefined datasets, thats a good idea!
>
> Also, isn't this "next time" ? :)
>
>
> By next time I meant, when I ask a question in the future, I didn't think
> you'd respond!
>
> So here is some code!
>
> library(reshape)
> library(ggplot2)
>
> theme_t <- list(grid.fill="white",grid.colour="lightgrey",
> background.colour=
> "black",axis.colour="dimgrey")
> ggtheme(theme_t)
>
> grp <-
>
> c(2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3)
> time <-
>
> c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2)
> cc <- c(0.7271,0.7563,0.6979,0.8208,0.7521,0.7875,0.7563,0.7771,0.8208,
> 0.7938,0.8083,0.7188,0.7521,0.7854,0.7979,0.7583,0.7646,0.6938,0.6813,
> 0.7708
> ,0.7375,0.8104,0.8104,0.7792,0.7833,0.8083,0.8021,0.7313,0.7958,0.7021,
> 0.8167,0.8167,0.7583,0.7167,0.6563,0.6896,0.7333,0.8208,0.7396,0.8063,
> 0.7083
> ,0.6708,0.7292,0.7646,0.7667,0.775,0.8021,0.8125,0.7646,0.6917,0.7458,
> 0.7833
> ,0.7396,0.7229,0.7708,0.7729,0.8083,0.7771,0.6854,0.8417,0.7667,0.7063,
> 0.75,
> 0.7813,0.8271,0.7896,0.7979,0.625,0.7938,0.7583,0.7396,0.7583,0.7938,
> 0.7333,
> 0.7875,0.8146)
>
> data <- as.data.frame(cbind(time,grp,cc))
> data$grp <- factor(data$grp,labels=c("Group A","Group B"))
> data$time <- factor(data$time,labels=c("Pre-test","Post-test"))
> boxplot <- qplot(grp, cc, data=data, geom="boxplot",
> orientation="horizontal", ylim=c(0.5,1), main="Hello World!", xlab="Label
> X", ylab="Label Y", facets=.~time, colour="red", size=2)
> boxplot + geom_jitter(aes(colour="steelblue")) + scale_colour_identity() +
> scale_size_identity()
> grid.gedit("ylabel", gp=gpar(fontsize=16))
>
>
> > There's a book that provides a full explanation and the (basic) grid
> > chapter is online (see
> > http://www.stat.auckland.ac.nz/~paul/RGraphics/rgraphics.html)
>
>
> Awesome, I'll check that out.
>
> Yep, the facilities for investigating the viewport and grob tree are
> > basically inadequate.  Based on some work Hadley did for ggplot, the
> > development version of R has a slightly better tool called grid.ls()
> > that can show how the grob tree and the viewport tree intertwine.  That
> > would allow you to see which viewport each grob was drawn in, which
> > would help you, for example, to know which viewport you had to go to to
> > replace a rectangle you want to remove.
>
>
> okie dokie, I'm ready to be amazed! hehe.
> emilio
>
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 39
> Date: Fri, 10 Aug 2007 12:41:24 -0400
> From: "Weiwei Shi" <helprhelp@gmail.com>
> Subject: [R] write.table
> To: "r-help@stat.math.ethz.ch" <R-help@stat.math.ethz.ch>
> Message-ID:
>         <cdf817830708100941n188c386as3c65fe3e6b7ca948@mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> Hi,
>
> I am always with this qustion when I tried to write a data.frame with
> row.names and col.names. I have to re-make the data frame to let its
> first column be the rownames and let row.names=F so that I can align
> the colnames correctly.
>
> Is there a way or option in write.table to automatically do that?
>
> thanks.
>
> --
> Weiwei Shi, Ph.D
> Research Scientist
> GeneGO, Inc.
>
> "Did you always know?"
> "No, I did not. But I believed..."
> ---Matrix III
>
>
>
> ------------------------------
>
> Message: 40
> Date: Thu, 9 Aug 2007 12:59:16 -0700
> From: "Nordlund, Dan (DSHS/RDA)" <NordlDJ@dshs.wa.gov>
> Subject: Re: [R] small sample techniques
> To: r-help@stat.math.ethz.ch
> Message-ID:
>         <941871A13165C2418EC144ACB212BDB04E1350@dshsmxoly1504g.dshs.wa.lcl
> >
> Content-Type: text/plain; charset=iso-8859-1
>
> > -----Original Message-----
> > From: r-help-bounces@stat.math.ethz.ch
> > [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Nair,
> > Murlidharan T
> > Sent: Thursday, August 09, 2007 12:02 PM
> > To: Nordlund, Dan (DSHS/RDA); r-help@stat.math.ethz.ch
> > Subject: Re: [R] small sample techniques
> >
> > n=300
> > 30% taking A relief from pain
> > 23% taking B relief from pain
> > Question; If there is no difference are we likely to get a 7%
> > difference?
> >
> > Hypothesis
> > H0: p1-p2=0
> > H1: p1-p2!=0 (not equal to)
> >
> > 1>Weighed average of two sample proportion
> >     300(0.30)+300(0.23)
> >     ------------------- = 0.265
> >       300+300
> > 2>Std Error estimate of the difference between two
> > independent proportions
> >       sqrt((0.265 *0.735)*((1/300)+(1/300))) = 0.03603
> >
> > 3>Evaluation of the difference between sample proportion as a
> > deviation from the hypothesized difference of zero
> >          ((0.30-0.23)-(0))/0.03603 = 1.94
> >
> >
> > z did not approach 1.96 hence H0 is not rejected.
> >
> > This is what I was trying to do using prop.test.
> >
> > prop.test(c(30,23),c(300,300))
> >
> > What function should I use?
> >
> >
>
> The proportion test above indicates that p1=0.1 and p2=0.07666667.  But in
> your t-test you specify p1=0.3 and p2=0.23.  Which is correct?  If p1=0.3and p2=
> 0.23, then use
>
> prop.test(c(.30*300,.23*300),c(300,300))
>
> Hope this is helpful,
>
> Dan
>
> Daniel J. Nordlund
> Research and Data Analysis
> Washington State Department of Social and Health Services
> Olympia, WA  98504-5204
>
>
>
> ------------------------------
>
> Message: 41
> Date: Thu, 9 Aug 2007 12:59:16 -0700
> From: "Nordlund, Dan (DSHS/RDA)" <NordlDJ@dshs.wa.gov>
> Subject: Re: [R] small sample techniques
> To: r-help@stat.math.ethz.ch
> Message-ID:
>         <941871A13165C2418EC144ACB212BDB04E1350@dshsmxoly1504g.dshs.wa.lcl
> >
> Content-Type: text/plain; charset=iso-8859-1
>
> > -----Original Message-----
> > From: r-help-bounces@stat.math.ethz.ch
> > [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Nair,
> > Murlidharan T
> > Sent: Thursday, August 09, 2007 12:02 PM
> > To: Nordlund, Dan (DSHS/RDA); r-help@stat.math.ethz.ch
> > Subject: Re: [R] small sample techniques
> >
> > n=300
> > 30% taking A relief from pain
> > 23% taking B relief from pain
> > Question; If there is no difference are we likely to get a 7%
> > difference?
> >
> > Hypothesis
> > H0: p1-p2=0
> > H1: p1-p2!=0 (not equal to)
> >
> > 1>Weighed average of two sample proportion
> >     300(0.30)+300(0.23)
> >     ------------------- = 0.265
> >       300+300
> > 2>Std Error estimate of the difference between two
> > independent proportions
> >       sqrt((0.265 *0.735)*((1/300)+(1/300))) = 0.03603
> >
> > 3>Evaluation of the difference between sample proportion as a
> > deviation from the hypothesized difference of zero
> >          ((0.30-0.23)-(0))/0.03603 = 1.94
> >
> >
> > z did not approach 1.96 hence H0 is not rejected.
> >
> > This is what I was trying to do using prop.test.
> >
> > prop.test(c(30,23),c(300,300))
> >
> > What function should I use?
> >
> >
>
> The proportion test above indicates that p1=0.1 and p2=0.07666667.  But in
> your t-test you specify p1=0.3 and p2=0.23.  Which is correct?  If p1=0.3and p2=
> 0.23, then use
>
> prop.test(c(.30*300,.23*300),c(300,300))
>
> Hope this is helpful,
>
> Dan
>
> Daniel J. Nordlund
> Research and Data Analysis
> Washington State Department of Social and Health Services
> Olympia, WA  98504-5204
>
>
>
> ------------------------------
>
> Message: 42
> Date: Fri, 10 Aug 2007 17:00:34 +0000
> From: Monica Pisica <pisicandru@hotmail.com>
> Subject: [R]  Cleaning up the memory
> To: <r-help@stat.math.ethz.ch>
> Message-ID: <BAY104-W14C5EC6532FFCAD5372821C3E10@phx.gbl>
> Content-Type: text/plain
>
>
> Hi,
>
> I have 4 huge tables on which i want to do a PCA analysis and a kmean
> clustering. If i run each table individually i have no problems, but if i
> want to run it in a for loop i exceed the memory alocation after the second
> table, even if i save the results as a csv table and i clean up all the big
> objects with rm command. To me it seems that even if i don't have the
> objects anymore, the memory these objects used to occupy is not cleared. Is
> there any way to clear up the memory as well? I don't want to close R and
> start it up again. Also i am running R under Windows.
>
> thanks,
>
> Monica
> _________________________________________________________________
> [[trailing spam removed]]
>
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 43
> Date: Fri, 10 Aug 2007 19:05:26 +0200
> From: "KOITA Lassana - STAC/ACE"
>         <lassana.koita@aviation-civile.gouv.fr>
> Subject: [R] need help to manipulate function and time interval
> To: r-help@stat.math.ethz.ch
> Message-ID:
>         <
> OF757EAC49.8C2AD1F7-ONC1257333.005876DC-C1257333.005DC714@aviation-civile.gouv.fr
> >
>
> Content-Type: text/plain
>
> Hi R-users,
>
> I have to define a noise level function L and its energy in the various
> moment of the day by:
>
> if time is between 18:00:00 and 23:59:59 then  L[j] <- L[j]+5 and W <-
> 10^((L+5)/10)
>
> if time is between 22:00:00 and 05:59:59 ==> L <- L+10  and W <-
> 10^((L+10)/10)
> else
> L=L and W = W
>
> Could someone help me to realize this function please? You will find my
> following proposal  code, but my main problem is to handle the time
> interval.
>
> Best regard
>
> ###########################
> myfunc <- function(mytab, Time, Level)
> {
> vect <- rep(0, length(mytab))
> for(i in 1:length(vect))
>    {
>       for(j in 1:length(Time))
>
>             if(time[j] is between 18:00:00 and 23:59:59)
>
>             L[i] <- L[j]+5
>
>            vect[i] <- 10^((L[i])/10
>
>             if (time[j] is between 22:00:00 and 05:59:59)
>
>             L[i] <- L[j]+10
>
>              vect[i] <- 10^((L[i])/10
>
>             else
>
>            L[i] = L[j]
>
>           vect[i] <- 10^((L[i])/10
>      }
> }
>
> #######################
>
> Lassana KOITA
> Chargé d'Etudes de Sécurité Aéroportuaire et d'Analyse Statistique  /
> Project Engineer Airport Safety Studies & Statistical analysis
> Service Technique de l'Aviation Civile (STAC) / Civil Aviation Technical
> Department
> Direction Générale de l'Aviation Civile (DGAC) / French Civil Aviation
> Headquarters
> Tel: 01 49 56 80 60
> Fax: 01 49 56 82 14
> E-mail: Lassana.Koita@aviation-civile.gouv.fr
> http://www.stac.aviation-civile.gouv.fr/
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 44
> Date: Fri, 10 Aug 2007 14:22:00 -0300
> From: "Henrique Dallazuanna" <wwwhsd@gmail.com>
> Subject: Re: [R] need help to manipulate function and time interval
> To: "KOITA Lassana - STAC/ACE" <lassana.koita@aviation-civile.gouv.fr>
> Cc: r-help@stat.math.ethz.ch
> Message-ID:
>         <da79af330708101022l6c2ef266sfd2913cfd55742de@mail.gmail.com>
> Content-Type: text/plain
>
> Hi,
>
> Try whit:
>
> if(time[j] >= "18:00:00" & < "23:59:59")
> ...
> ...
>
> --
> Henrique Dallazuanna
> Curitiba-Paraná-Brasil
> 25° 25' 40" S 49° 16' 22" O
>
> On 10/08/07, KOITA Lassana - STAC/ACE <
> lassana.koita@aviation-civile.gouv.fr>
> wrote:
> >
> > Hi R-users,
> >
> > I have to define a noise level function L and its energy in the various
> > moment of the day by:
> >
> > if time is between 18:00:00 and 23:59:59 then  L[j] <- L[j]+5 and W <-
> > 10^((L+5)/10)
> >
> > if time is between 22:00:00 and 05:59:59 ==> L <- L+10  and W <-
> > 10^((L+10)/10)
> > else
> > L=L and W = W
> >
> > Could someone help me to realize this function please? You will find my
> > following proposal  code, but my main problem is to handle the time
> > interval.
> >
> > Best regard
> >
> > ###########################
> > myfunc <- function(mytab, Time, Level)
> > {
> > vect <- rep(0, length(mytab))
> > for(i in 1:length(vect))
> >    {
> >       for(j in 1:length(Time))
> >
> >             if(time[j] is between 18:00:00 and 23:59:59)
> >
> >             L[i] <- L[j]+5
> >
> >            vect[i] <- 10^((L[i])/10
> >
> >             if (time[j] is between 22:00:00 and 05:59:59)
> >
> >             L[i] <- L[j]+10
> >
> >              vect[i] <- 10^((L[i])/10
> >
> >             else
> >
> >            L[i] = L[j]
> >
> >           vect[i] <- 10^((L[i])/10
> >      }
> > }
> >
> > #######################
> >
> > Lassana KOITA
> > Chargé d'Etudes de Sécurité Aéroportuaire et d'Analyse Statistique  /
> > Project Engineer Airport Safety Studies & Statistical analysis
> > Service Technique de l'Aviation Civile (STAC) / Civil Aviation Technical
> > Department
> > Direction Générale de l'Aviation Civile (DGAC) / French Civil Aviation
> > Headquarters
> > Tel: 01 49 56 80 60
> > Fax: 01 49 56 82 14
> > E-mail: Lassana.Koita@aviation-civile.gouv.fr
> > http://www.stac.aviation-civile.gouv.fr/
> >         [[alternative HTML version deleted]]
> >
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
> >
>
>         [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 45
> Date: Fri, 10 Aug 2007 18:28:07 +0100 (BST)
> From: Prof Brian Ripley <ripley@stats.ox.ac.uk>
> Subject: Re: [R] Cleaning up the memory
> To: Monica Pisica <pisicandru@hotmail.com>
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <Pine.LNX.4.64.0708101822140.6256@gannet.stats.ox.ac.uk>
> Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
>
> On Fri, 10 Aug 2007, Monica Pisica wrote:
>
> >
> > Hi,
> >
> > I have 4 huge tables on which i want to do a PCA analysis and a kmean
> > clustering. If i run each table individually i have no problems, but if
> > i want to run it in a for loop i exceed the memory alocation after the
> > second table, even if i save the results as a csv table and i clean up
> > all the big objects with rm command. To me it seems that even if i don't
> > have the objects anymore, the memory these objects used to occupy is not
> > cleared. Is there any way to clear up the memory as well? I don't want
> > to close R and start it up again. Also i am running R under Windows.
>
> See ?gc, which does the clearing.
>
> However, unless you study the memory allocation in detail (which you
> cannot do from R code), you don't actually know that this is the problem.
> More likely is that you have fragmentation of your 32-bit address space:
> see ?"Memory-limits".
>
> Without any idea what memory you have and what 'huge' means, we can only
> make wild guesses.  It might be worth raising the memory limit (the
> --max-mem-size flag).
>
> >
> > thanks,
> >
> > Monica
> > _________________________________________________________________
> > [[trailing spam removed]]
> >
> >       [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
> --
> Brian D. Ripley,                  ripley@stats.ox.ac.uk
> Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> University of Oxford,             Tel:  +44 1865 272861 (self)
> 1 South Parks Road,                     +44 1865 272866 (PA)
> Oxford OX1 3TG, UK                Fax:  +44 1865 272595
>
>
>
> ------------------------------
>
> Message: 46
> Date: Fri, 10 Aug 2007 13:58:27 -0400
> From: "Yinghai Deng" <Yinghai.Deng@bri.nrc.ca>
> Subject: Re: [R] write.table
> To: "Weiwei Shi" <helprhelp@gmail.com>, "r-help@stat.math.ethz.ch"
>         <R-help@stat.math.ethz.ch>
> Message-ID: <KNEDKNFOAOJGKIMAMOMDKEKKCBAA.Yinghai.Deng@bri.nrc.ca>
> Content-Type: text/plain;       charset="us-ascii"
>
> write.table(mydata.frame, "mydata", col.names=NA, quote=F, sep="\t") will
> solve the problem.
> Deng
> -----Original Message-----
> From: r-help-bounces@stat.math.ethz.ch
> [mailto:r-help-bounces@stat.math.ethz.ch]On Behalf Of Weiwei Shi
> Sent: August 10, 2007 12:41 PM
> To: r-help@stat.math.ethz.ch
> Subject: [R] write.table
>
>
> Hi,
>
> I am always with this qustion when I tried to write a data.frame with
> row.names and col.names. I have to re-make the data frame to let its
> first column be the rownames and let row.names=F so that I can align
> the colnames correctly.
>
> Is there a way or option in write.table to automatically do that?
>
> thanks.
>
> --
> Weiwei Shi, Ph.D
> Research Scientist
> GeneGO, Inc.
>
> "Did you always know?"
> "No, I did not. But I believed..."
> ---Matrix III
>
> ______________________________________________
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
>
> ------------------------------
>
> Message: 47
> Date: Fri, 10 Aug 2007 14:05:50 -0400
> From: "Lanre Okusanya" <lanre.okusanya@gmail.com>
> Subject: [R] Help wit matrices
> To: "'r-help@stat.math.ethz.ch'" <r-help@stat.math.ethz.ch>
> Message-ID:
>         <6e25bb420708101105o17e80f36re816c031777e85c7@mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> Hello all,
>
> I am working with a 1000x1000 matrix, and I would like to return a
> 1000x1000 matrix that tells me which value in the matrix is greater
> than a theshold value (1 or 0 indicator).
> i have tried
>   mat2<-as.matrix(as.numeric(mat1>0.25))
> but that returns a 1:100000 matrix.
> I have also tried for loops, but they are grossly inefficient.
>
> THanks for all your help in advance.
>
> Lanre
>
>
>
> ------------------------------
>
> Message: 48
> Date: Fri, 10 Aug 2007 18:09:58 +0000
> From: Monica Pisica <pisicandru@hotmail.com>
> Subject: Re: [R] Cleaning up the memory
> To: Prof Brian Ripley <ripley@stats.ox.ac.uk>
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <BAY104-W21EDAF4F804B3715CCF814C3E10@phx.gbl>
> Content-Type: text/plain
>
>
> Thanks! I will look into ...
>
> I have 4 GB RAM, and i was monitoring the memory with Windows task manager
> so i was looking how R "gets" more and more memory allocation from less than
> 100Mb to .... 1500Mb .....
>
> My initial tables are between 30 to 80 Mb and the resulting tables that
> incorporate the initial tables plus PCA and kmeans results are inbetween 50
> to 200MB or thereabouts!
>
> And yes, i don't really care about memory allocation in detail - what i
> want is to free that memory after every cycle ;-)
>
> Although, after i didn't do anything in R and it was idle for more than 30
> min. the memory allocation according to Task manager dropped to 15 Mb .....
> which is good - but i cannot wait inbetween cycles half an hour though .....
>
> Again thanks,
>
> Monica> Date: Fri, 10 Aug 2007 18:28:07 +0100> From: ripley@stats.ox.ac.uk>
> To: pisicandru@hotmail.com> CC: r-help@stat.math.ethz.ch> Subject: Re: [R]
> Cleaning up the memory> > On Fri, 10 Aug 2007, Monica Pisica wrote:> > >> >
> Hi,> >> > I have 4 huge tables on which i want to do a PCA analysis and a
> kmean > > clustering. If i run each table individually i have no problems,
> but if > > i want to run it in a for loop i exceed the memory alocation
> after the > > second table, even if i save the results as a csv table and i
> clean up > > all the big objects with rm command. To me it seems that even
> if i don't > > have the objects anymore, the memory these objects used to
> occupy is not > > cleared. Is there any way to clear up the memory as well?
> I don't want > > to close R and start it up again. Also i am running R under
> Windows.> > See ?gc, which does the clearing.> > However, unless you study
> the memory allocation in detail (which you > cannot do from R code), you
> don't actually !
> know that this is the problem. > More likely is that you have
> fragmentation of your 32-bit address space: > see ?"Memory-limits".> >
> Without any idea what memory you have and what 'huge' means, we can only >
> make wild guesses. It might be worth raising the memory limit (the >
> --max-mem-size flag).> > >> > thanks,> >> > Monica> >
> _________________________________________________________________> >
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> ______________________________________________> > R-help@stat.math.ethz.chmailing list> >
> https://stat.ethz.ch/mailman/listinfo/r-help> > PLEASE do read the posting
> guide http://www.R-project.org/posting-guide.html> > and provide
> commented, minimal, self-contained, reproducible code.> >> > -- > Brian D.
> Ripley, ripley@stats.ox.ac.uk> Professor of Applied Statistics,
> http://www.stats.ox.ac.uk/~ripley/> University of Oxford, Tel: +44 1865
> 272861 (self)> 1 South Parks Road, +44 1865 272866 (PA)> Oxford OX1 3TG!
> , UK Fax: +44 1865 272595
> _________________________________________________________________
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>
> ------------------------------
>
> Message: 49
> Date: Fri, 10 Aug 2007 14:12:43 -0400
> From: "jim holtman" <jholtman@gmail.com>
> Subject: Re: [R] Help wit matrices
> To: "Lanre Okusanya" <lanre.okusanya@gmail.com>
> Cc: "r-help@stat.math.ethz.ch" <r-help@stat.math.ethz.ch>
> Message-ID:
>         <644e1f320708101112g270aeeafub6211ce98cee644d@mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> Is this what you want:
>
> > x <- matrix(runif(100), 10)
> > round(x, 3)
>        [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8]  [,9] [,10]
> [1,] 0.268 0.961 0.262 0.347 0.306 0.762 0.524 0.062 0.028 0.226
> [2,] 0.219 0.100 0.165 0.131 0.578 0.933 0.317 0.109 0.527 0.131
> [3,] 0.517 0.763 0.322 0.374 0.910 0.471 0.278 0.382 0.880 0.982
> [4,] 0.269 0.948 0.510 0.631 0.143 0.604 0.788 0.169 0.373 0.327
> [5,] 0.181 0.819 0.924 0.390 0.415 0.485 0.702 0.299 0.048 0.507
> [6,] 0.519 0.308 0.511 0.690 0.211 0.109 0.165 0.192 0.139 0.681
> [7,] 0.563 0.650 0.258 0.689 0.429 0.248 0.064 0.257 0.321 0.099
> [8,] 0.129 0.953 0.046 0.555 0.133 0.499 0.755 0.181 0.155 0.119
> [9,] 0.256 0.954 0.418 0.430 0.460 0.373 0.620 0.477 0.132 0.050
> [10,] 0.718 0.340 0.854 0.453 0.943 0.935 0.170 0.771 0.221 0.929
> > ifelse(x > .5, 1, 0)
>       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> [1,]    0    1    0    0    0    1    1    0    0     0
> [2,]    0    0    0    0    1    1    0    0    1     0
> [3,]    1    1    0    0    1    0    0    0    1     1
> [4,]    0    1    1    1    0    1    1    0    0     0
> [5,]    0    1    1    0    0    0    1    0    0     1
> [6,]    1    0    1    1    0    0    0    0    0     1
> [7,]    1    1    0    1    0    0    0    0    0     0
> [8,]    0    1    0    1    0    0    1    0    0     0
> [9,]    0    1    0    0    0    0    1    0    0     0
> [10,]    1    0    1    0    1    1    0    1    0     1
>
>
> On 8/10/07, Lanre Okusanya <lanre.okusanya@gmail.com> wrote:
> > Hello all,
> >
> > I am working with a 1000x1000 matrix, and I would like to return a
> > 1000x1000 matrix that tells me which value in the matrix is greater
> > than a theshold value (1 or 0 indicator).
> > i have tried
> >  mat2<-as.matrix(as.numeric(mat1>0.25))
> > but that returns a 1:100000 matrix.
> > I have also tried for loops, but they are grossly inefficient.
> >
> > THanks for all your help in advance.
> >
> > Lanre
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem you are trying to solve?
>
>
>
> ------------------------------
>
> Message: 50
> Date: Fri, 10 Aug 2007 11:12:47 -0700 (PDT)
> From: adiamond <adiamond@csidentity.com>
> Subject: [R] Row name of empty string issue
> To: r-help@stat.math.ethz.ch
> Message-ID: <12096455.post@talk.nabble.com>
> Content-Type: text/plain; charset=us-ascii
>
>
> I have a data.frame with rownames taken from a database.  Unfortunately,
> one
> of the rownames (automatically obtained from the DB) is an empty
> string.  I
> often do computations on the DB s.t. the answers (rows) are indexed with
> respect to the rownames so a computation on a DB record might necessitate
> the indexing into the data.frame by the emtpy row name.  Unfortunately,
> that
> doesn't seem to work.  Explicitly:
> I have this statement (in a loop going through sourceNames which are
> rownames of the data.frames CurrentRecordBlankFieldsCountSums and
> BlankFieldsCount):
>
> BlankFieldsCount[sourceNamei,]= BlankFieldsCount[sourceNamei,] +
> CurrentRecordBlankFieldsCountSums[sourceNamei ,];
>
> if sourceNamei is any name other than "" it works fine but otherwise
> CurrentRecordBlankFieldsCountSums[sourceNamei ,] returns a bunch of NAs
> because apparently it didn't fine a row named "".  IMHO, if R lets you
> name
> a row "", then it should let you index it with the name "".
>
> Anyway, as further proof of the setup:
> > rownames(CurrentRecordBlankFieldsCountSums)[1]
> [1] ""
> # So the first rowname of CurrentRecordBlankFieldsCountSums is an empty
> string ""
>
> > CurrentRecordBlankFieldsCountSums[1 ,]
> IDCaseNumber Category SSN LastName FirstName
>             0        0   1        0         0
> # So, the first row has some data (not just NAs as would be returned if
> that
> row didn't exist)
>
> But ff I index that same row using the rowname it doesn't find the row:
> > CurrentRecordBlankFieldsCountSums[rownames(
> > CurrentRecordBlankFieldsCountSums)[1] ,]
>    IDCaseNumber Category SSN LastName
> NA           NA       NA  NA
> # I get the same result if I do this:CurrentRecordBlankFieldsCountSums[""
> ,]
>
> As a sanity check:
> > "" == rownames(CurrentRecordBlankFieldsCountSums)[1]
> [1] TRUE
>
> For other rows, (rownames that aren't "", there's no problem):
> > rownames(CurrentRecordBlankFieldsCountSums)[2]
> [1] "FRED"
> > CurrentRecordBlankFieldsCountSums[rownames(
> > CurrentRecordBlankFieldsCountSums)[2] ,]
>                       IDCaseNumber Category SSN LastName FirstName
> FRED            0        0   0        0
>
> --
> View this message in context:
> http://www.nabble.com/Row-name-of-empty-string-issue-tf4250291.html#a12096455
> Sent from the R help mailing list archive at Nabble.com.
>
>
>
> ------------------------------
>
> Message: 51
> Date: Fri, 10 Aug 2007 14:19:57 -0400
> From: "Lanre Okusanya" <lanre.okusanya@gmail.com>
> Subject: Re: [R] Help wit matrices
> To: "jim holtman" <jholtman@gmail.com>
> Cc: "r-help@stat.math.ethz.ch" <r-help@stat.math.ethz.ch>
> Message-ID:
>         <6e25bb420708101119u55f04e5aq977a718c873e1557@mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1
>
> that was ridiculously simple. duh.
>
> THanks
>
> Lanre
>
> On 8/10/07, jim holtman <jholtman@gmail.com> wrote:
> > Is this what you want:
> >
> > > x <- matrix(runif(100), 10)
> > > round(x, 3)
> >        [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8]  [,9] [,10]
> >  [1,] 0.268 0.961 0.262 0.347 0.306 0.762 0.524 0.062 0.028 0.226
> >  [2,] 0.219 0.100 0.165 0.131 0.578 0.933 0.317 0.109 0.527 0.131
> >  [3,] 0.517 0.763 0.322 0.374 0.910 0.471 0.278 0.382 0.880 0.982
> >  [4,] 0.269 0.948 0.510 0.631 0.143 0.604 0.788 0.169 0.373 0.327
> >  [5,] 0.181 0.819 0.924 0.390 0.415 0.485 0.702 0.299 0.048 ...
>
> [Message clipped]

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