I would recommend that you add a column to the dataset which indicates
what function needs to be evaluated in the apply loop.
innerFun <- function(row) {
if (row[1] == 1){
result <- mean(row[-1])
} else {
result <- median(row[-1])
}
return(result)
}
myDf <- cbind(action = rep(c(0, 1), dim(myDf)[2]/2), myDf)
apply(myDf, 1, innerFun)
Cheers,
Thierry
--------------------------------------------------------
ir. Thierry Onkelinx
Research Institute for Nature and Forest
Team Biometrics, methodology and quality assurance
Gaverstraat 4
B-9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
Thierry.Onkelinx@inbo.be
-----Original Message-----
From: s-news-owner@lists.biostat.wustl.edu
[mailto:s-news-owner@lists.biostat.wustl.edu] On Behalf Of
Rich@Mango-Solutions.com
Sent: Saturday, February 11, 2006 1:00 PM
To: 'Schwarz,Paul'; s-news@lists.biostat.wustl.edu
Subject: Re: [S] identifying which column/row is passed in calls to
apply()
You could use sapply to pass in a vector of numbers (for column numbers)
or
characters (for column names) and index the columns of the input data.
Something like this:-
> innerFun <- function(i, df) {
if (i == 1) mean(df[[i]])
else median(df[[i]])
}
> sapply(1:length(myDf), innerFun, df=myDf)
Not as efficient as apply for larger data structures I'd imagine, since
you're passing the entire dataset in each time ...
Rich.
-----Original Message-----
From: s-news-owner@lists.biostat.wustl.edu
[mailto:s-news-owner@lists.biostat.wustl.edu] On Behalf Of Schwarz,Paul
Sent: 11 February 2006 02:19
To: s-news@lists.biostat.wustl.edu
Subject: [S] identifying which column/row is passed in calls to apply()
S-News readers,
How would I identify which column (or row) is being operated on in a
call to the function specified in apply()? That is, if I wanted to
perform some column-specific (or row-specific) operation, how do I
identify which column/row is being passed to the function that's
specified in the call to apply(), where the number of columns ranges
from 1 to ncol(mymat)?
apply( mymat, 2, function(x){ ? } )
I suspect that this is easy, but I'm drawing a blank on how to do this.
Thanks a lot everyone.
-Paul
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