[BioC] asking for some code

[zhihua li]

Ya, it works!  maybe I made some mistakes writing the if...else clause 
yesterday. Thanks!



>From: "STKH (Steen Krogsgaard)" <StKH at novozymes.com>
>To: "zhihua li" <lzhtom at hotmail.com>
>Subject: RE: [BioC] asking for some code
>Date: Thu, 10 Mar 2005 14:20:28 +0100
>
>Is there anything wrong with
>
>if (clause1) {
>...
>} else if (clause2) {
>...
>} else if (clause3) {
>...
>} else {
>...
>}
>
>That's what you do in Pascal!
>
>/Steen
>
>-----Original Message-----
>From: bioconductor-bounces at stat.math.ethz.ch
>[mailto:bioconductor-bounces at stat.math.ethz.ch] On Behalf Of zhihua li
>Sent: 10. marts 2005 14:12
>To: ramasamy at cancer.org.uk
>Cc: bioconductor at stat.math.ethz.ch
>Subject: Re: [BioC] asking for some code
>
>
>Thanks for all the help! I'm on the border of resolving the problem
>but.........
>
>for the if-else clause, I alwayse got error message when using if..else
>if....else.....statement. I looked it up in the R mannual and found that
>R only have a "if....else..." structure, but doesn't have something like
>
>"if....elseif....elseif......else....."? Does it mean that only two
>conditions can be selected in the conditional execution grammer of R?
>
>for the cut() and switch() clause, I found it's very efficient in
>application to a vector.
>but when I wanna use it in a matrix row by row, it's less convenient (or
>I
>didn't write the
>smart code).
>
>
>
> >From: Adaikalavan Ramasamy <ramasamy at cancer.org.uk>
> >Reply-To: ramasamy at cancer.org.uk
> >To: Sean Davis <sdavis2 at mail.nih.gov>
> >CC: zhihua li <lzhtom at hotmail.com>, BioConductor mailing list
><bioconductor at stat.math.ethz.ch>
> >Subject: Re: [BioC] asking for some code
> >Date: Wed, 09 Mar 2005 14:41:36 +0000
> >
> >A slightly more intuitive way is to cut() the variable returning
> >labels, which you can then feed into switch().
> >
> ># simulate data
> >  set.seed(1)
> >  x <- runif(5)
> >  x
> >[1] 0.2655087 0.3721239 0.5728534 0.9082078 0.2016819
> >
> >
> >Let us say that n=0.25 and m=0.6.
> >
> >my.switch.fn <- function(x ){
> >
> >    y <- cut( x, breaks=c(0, 0.25, 0.6, 1), labels=FALSE )
> >
> >    z <- switch(y,
> >                "1" = x,
> >                "2" = x + 1,
> >                "3" = x + 2
> >                )
> >    return(z)
> >}
> >
> >sapply( x, my.switch.fn )
> >[1] 1.2655087 1.3721239 1.5728534 2.9082078 0.2016819
> >
> >
> >In the case of 2 breakpoints, my example does not look much shorter
> >than the if-else clause below. The drawback of my suggestion is that
> >you need to know the min and max occuring values values (in this case 0
>
> >and 1) and switch does not have a default value.
> >
> >Regards, Adai
> >
> >
> >On Wed, 2005-03-09 at 05:56 -0500, Sean Davis wrote:
> > > Taking this apart:
> > >
> > > to find the Q5(quantile):
> > >    See ?quantile
> > >
> > > for each row:
> > >    See ?apply
> > >
> > > if x<n {
> > >    do stuff
> > > } else if ((x>n) & (x<m)) {
> > >    do other stuff
> > > } else {
> > >    do other other stuff
> > > }
> > >
> > > Sean
> > >
> > > On Mar 9, 2005, at 4:15 AM, zhihua li wrote:
> > >
> > > > Dear netters,
> > > >
> > > > Now I have a matrix at hand. For each row, I want to find the
> > > > Q5(the 5th quantile) and the Q95, and put them in a function
> > > > fyn(), which will return two numbers n and m(n<m). Then I'll take
> > > > each element
>(say
> > > > x) in the row one by one, and perform a judgment: if x<n, do
> > > > sub1(x); if n<=x<=m, do sub2(x); if x>m, do sub3(x). The process
> > > > goes on through each row in the matrix. This program isnot very
> > > > complicated, but hard enough for me...........So could anybody
> > > > give me a prototype of the code? Thanks a lot!
> > > >
> > > > _______________________________________________
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> > >
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> > >
> >
>
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