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.