I am going through one of my .R files and by cleaning it up a little bit I am trying to get more familiar with writing the code the r-ight way. As a beginner, one of my favorite starting points is to get rid of the for() loops and try to transform the expression into a functional programming form.
So here is the scenario:
I am assembling a bunch of data.frames into a list for later usage.
dataList <- list (dataA,
dataB,
dataC,
dataD,
dataE
)
Now I like to take a look at each data.frame's column names and substitute certain character strings. Eg I like to substitute each "foo" and "bar" with "baz". At the moment I am getting the job done with a for() loop which looks a bit awkward.
colnames(dataList[[1]])
[1] "foo" "code" "lp15" "bar" "lh15"
colnames(dataList[[2]])
[1] "a" "code" "lp50" "ls50" "foo"
matchVec <- c("foo", "bar")
for (i in seq(dataList)) {
for (j in seq(matchVec)) {
colnames (dataList[[i]])[grep(pattern=matchVec[j], x=colnames (dataList[[i]]))] <- c("baz")
}
}
Since I am working here with a list I thought about the lapply function. My attempts handling the job with the lapply function all seem to look alright but only at first sight. If I write
f <- function(i, xList) {
gsub(pattern=c("foo"), replacement=c("baz"), x=colnames(xList[[i]]))
}
lapply(seq(dataList), f, xList=dataList)
the last line prints out almost what I am looking for. However, if i take another look at the actual names of the data.frames in dataList:
lapply (dataList, colnames)
I see that no changes have been made to the initial character strings.
So how can I rewrite the for() loop and transform it into a functional programming form?
And how do I substitute both strings, "foo" and "bar", in an efficient way? Since the gsub() function takes as its pattern argument only a character vector of length one.
