How to use lapply to change the colnames like this in R?

Question

?Following is the R's interactive envirment's output.

I try to change the colnames of P300.H1,P300.GM and Suz12.H1, but didn't work.

> lapply(list(P300.H1,P300.GM,Suz12.H1), function(x) {colnames(x) <- c("chrom","start", "end", "name", "score")})
[[1]]
[1] "chrom" "start" "end"   "name"  "score"

[[2]]
[1] "chrom" "start" "end"   "name"  "score"

[[3]]
[1] "chrom" "start" "end"   "name"  "score"

> colnames(P300.H1)
[1] "V1" "V2" "V3" "V4" "V5"

I think the problem may be about the assignment, but I'm still confused about that.

Can anybody explain to me the reason about that?

check this related, simple answer using setNames http://stackoverflow.com/questions/19942762/assign-column-names-to-list-of-dataframes — ano, Oct 15 '15 at 18:34

score 2 · Accepted Answer · answered Jun 29 '12 at 20:48

There may be a better way, and using assign like this can get you into trouble:

d1 <- d2 <- data.frame(x=1, y=2)

myfun <- function(x) {
  x.df <- get(x)
  colnames(x.df) <- c('n1', 'n2')
  assign(x, x.df, env=.GlobalEnv)
}

lapply(c('d1', 'd2'), myfun)

score 1 · Answer 2 · answered Jun 29 '12 at 20:51

If you are really changing the column names of each to the identical set of names, here's a sort of trivial way (that's only appropriate for a small number of data frames):

colnames(P300.H1) <- colnames(P300.GM) <- colnames(Suz12.H1) <- c("chrom","start", "end", "name", "score")

score 0 · Answer 3 · answered Jun 29 '12 at 20:49

Using a for loop here may prove easier and/or safer than using lapply (or atleast that's what came to mind at 4PM on a Friday):

DF <- data.frame(x = c(1, 2, 3,NA), y = c(1,0, 10, NA), z=c(43,NA, 33, NA))
#make your list
yourList <- list(DF,DF)

for (i in seq(yourList)){
  colnames(yourList[[i]]) <- c("foo", "bar", "baz")
}
#Confirm output is right
> lapply(yourList, colnames)
[[1]]
[1] "foo" "bar" "baz"

[[2]]
[1] "foo" "bar" "baz"

David Robinson · Answer 4 · 2012-06-29T20:48:32.077

R passes by value, not by reference, so changing the column names of a data frame within a function won't change it outside the function.

As a simple, reproducible example:

m <- matrix(rnorm(12), ncol=3)
colnames(m) <- c("A", "B", "C")

change.names = function(x) {
    colnames(x) <- c("X", "Y", "Z")
}

change.names(m)
print(colnames(m))

The column names will still be A, B, C. The x within that function is not the matrix m: it is just a copy of it, and any changes you make to it won't apply to the original.

How to use lapply to change the colnames like this in R?

4 Answers4