I’ve often selected columns or rows of a data frame using grep or which, based on some property. That is inherently sound, but the trouble comes when you wish to remove rows or columns based on that grep or which call, e.g.,
dat <- dat[,-grep('\\.1', names(dat))]
which would remove columns with a .1 in the name. This is fine the first time around, but if you forget and re-run the code, grep('\\.1',names(dat)) gives a vector of length 0, and hence dat becomes a data.frame with 0 columns. The function which also has similar pitfalls, as demonstrated in a recent R-help posting by David Winsemius. I find a more reliable method is to do
dat <- dat[,setdiff(1:ncol(dat),grep('\\.1',names(dat)))]
which will always give the right number of columns. Other suggestions for getting around this issue are welcomed in the comments.
use grepl instead.
e.g.
colSlice <- function(dat,pattern,…){
dat[,grepl(pattern,names(dat),...)]
}
In general (beyond “grep” and “which”) you can avoid this sort of problem by creating new temporary variables rather than overwriting the existing variable. This is a minor pain because you end up with variables dat1, dat2, dat3, … representing the steps of the analysis, but may be worth it in the long run.
I always just use grepl like Dhruv recommends.