[R] a vectorized solution to some simple dataframe math?

[Petr PIKAL]

Hi


r-help-bounces at r-project.org napsal dne 27.03.2010 11:53:30:

> Hi:
> 
> Does this do what you want?
> 
> # Create some fake data...
> 
> df <- data.frame(id = factor(rep(c('cell1', 'cell2'), each = 10)),
>                   cond = factor(rep(rep(c('A', 'B'), each = 5), 2)),
>                   time = round(rnorm(20, 350, 10), 2))
> 
> # Create a function to subtract each element of a vector from its mean
>  f <- function(x) x - mean(x)
> # Load the plyr package, which contains the function ddply():
> library(plyr)
> df2 <- ddply(df, .(id, cond), transform, dev = f(time))

another option is ave

df2 <- data.frame(df, dev=ave(df$time, df$id, df$cond, FUN=f))

Regards
Petr

> 
> # output
> > df2
>       id cond   time     dev
> 1  cell1    A 353.01   7.226
> 2  cell1    A 351.06   5.276
> 3  cell1    A 343.59  -2.194
> 4  cell1    A 341.50  -4.284
> 5  cell1    A 339.76  -6.024
> 6  cell1    B 351.18   0.644
> 7  cell1    B 340.53 -10.006
> 8  cell1    B 345.09  -5.446
> 9  cell1    B 347.44  -3.096
> 10 cell1    B 368.44  17.904
> 11 cell2    A 343.48  -3.776
> 12 cell2    A 352.35   5.094
> 13 cell2    A 350.78   3.524
> 14 cell2    A 340.38  -6.876
> 15 cell2    A 349.29   2.034
> 16 cell2    B 364.45  15.524
> 17 cell2    B 354.52   5.594
> 18 cell2    B 350.41   1.484
> 19 cell2    B 345.78  -3.146
> 20 cell2    B 329.47 -19.456
> 
> # cell means
> > with(df, aggregate(time, list(id = id, cond = cond), mean))
>      id cond       x
> 1 cell1    A 345.784
> 2 cell2    A 347.256
> 3 cell1    B 350.536
> 4 cell2    B 348.926
> 
> HTH,
> Dennis
> 
> On Fri, Mar 26, 2010 at 1:31 PM, Dgnn <sharkbrainpdx at gmail.com> wrote:
> 
> >
> > I have a data frame containing the results of time measurements taken 
from
> > several cells. Each cell was measured in conditions A and B, and there 
are
> > an arbitrary number of measurements in each condition. I am trying to
> > calculate the difference of each measurement from the mean of a given 
cell
> > in a given condition without relying on loops.
> >
> > >my.df
> >           id       cond    time
> > 1         cell1     A       343.5
> > 2         cell1     A       355.2
> > ...
> > 768      cell1     B       454.0
> > ...
> > 2106    cell2     A       433.9
> > ...
> >
> > as a first approach I tried:
> >
> > > mews<-aggregate(my.df$time, list(cond=data$id, id=data$cond), mean)
> > id      cond      time
> > cell1    A         352
> > cell1    B         446
> > cell2    A         244
> > cell2    B         ...
> >
> > I then tried to use %in% to match id and cond of mews with my.df, but 
I
> > haven't been able to get it to work.
> > Am I on the right track? What are some other solutions?
> >
> > Thanks for any help.
> >
> > jason
> >
> >
> > --
> > View this message in context:
> > 
http://n4.nabble.com/a-vectorized-solution-to-some-simple-dataframe-math-
> tp1692810p1692810.html
> > Sent from the R help mailing list archive at Nabble.com.
> >
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