I am making some changes to the permax library (so that it will accept
NA's). This function performs a permutation analysis to identify
discriminating attributes distinguishing two groups of observations. It takes
the form (at its most simplistic):
permax(data, ig1)
where ig1 is one group of interest. The other group (if not specified) is
assumed to be the remaining observations, namely, -ig1.
The problem is that I would like to be able to specify ig1 not as a vector of
numeric indicies, but as a vector of column labels (as specified by data <-
as.matrix(read.table("file", header=T....)). But if I do this:
ig1 <- c('patient1', 'patient2')
then, while I can do this:
data[,ig1]
I can no longer do this:
data[,-ig1]
because the unary operator does not like working with column labels. Is there a
way to go backwards and determine the column indices from the column labels, or
perhaps and even more simple way to obtain a vector of complementary column
labels given a subset of labels?
Thanks,
Eric
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Kort, Eric <Eric.Kort at vai.org> writes:> ig1 <- c('patient1', 'patient2') > then, while I can do this: > data[,ig1] > I can no longer do this: > data[,-ig1] > because the unary operator does not like working with column labels. Is > there a way to go backwards and determine the column indices from the column > labels...?If you know "ig1" is indeed a column of the matrix "data", this should work: R> data[ , -match(ig1, colnames(data))] If you don't know that for sure, and/or you want to preserve the matrix structure of the result (drop=F), then here's a fancier version: R> data[ , setdiff(colnames(data), ig1), drop=F] -- -- David Brahm (brahm at alum.mit.edu) -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
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