Rename one named column in R - r

Question

I want to update one column of a dataframe, referencing it using its original name, is this possible? For example say I had the table 'data'
a b c
1 2 2
3 2 3
4 1 2
and I wanted to update the name of column b to 'd'. I know I could use
colnames(data)[2] <- 'd'
but can I make the change by specifically referencing b, i.e. something like
colnames(data)['b'] <- 'd'
so that if the column ordering of the dataframe changes the correct column name will still be updated.
Thanks in advance

Answer 1

There is a function setnames built into package data.table for exactly that.
setnames(DT, "b", "d")
It changes the names by reference with no copy at all. Any other method using names(data)<- or names(data)[i]<- or similar will copy the entire object, usually several times. Even though all you're doing is changing a column name.
DT must be type data.table for setnames to work, though. So you'd need to switch to data.table or convert using as.data.table, to use it.
Here is the extract from ?setnames. The intention is that you run example(setnames) at the prompt and then the comments relate to the copies you see being reported by tracemem.
DF = data.frame(a=1:2,b=3:4) # base data.frame to demo copies
tracemem(DF)
colnames(DF)[1] <- "A" # 4 copies of entire object
names(DF)[1] <- "A" # 3 copies of entire object
names(DF) <- c("A", "b") # 2 copies of entire object
`names<-`(DF,c("A","b")) # 1 copy of entire object
x=`names<-`(DF,c("A","b")) # still 1 copy (so not print method)
# What if DF is large, say 10GB in RAM. Copy 10GB just to change a column name?
DT = data.table(a=1:2,b=3:4,c=5:6)
tracemem(DT)
setnames(DT,"b","B") # by name; no match() needed. No copy.
setnames(DT,3,"C") # by position. No copy.
setnames(DT,2:3,c("D","E")) # multiple. No copy.
setnames(DT,c("a","E"),c("A","F")) # multiple by name. No copy.
setnames(DT,c("X","Y","Z")) # replace all. No copy.

Answer 2

As of October 2014 this can now be done easily in the dplyr package:
rename(data, d = b)

Answer 3

This seems like a hack, but the first thing that came to mind was to use grepl() with a sufficiently detailed enough search string to only get the column you want. I'm sure there are better options:
dat <- data.frame(a = 1:3, b = 1:3, c = 1:3)
colnames(dat)[grepl("b", colnames(dat))] <- "foo"
dat
#------
a foo c
1 1 1 1
2 2 2 2
3 3 3 3
As Joran points out below, I overcomplicated things...no need for a regex at all. This saves a few characters on the typing too.
colnames(dat)[colnames(dat) == "foo"] <- "bar"
#------
a bar c
1 1 1 1
2 2 2 2
3 3 3 3

Answer 4

Yes but it's more difficult (as far as I know) than numeric indexing. I'm going to provide a dirty function that will do this and if you want to see how to do it just tear the function apart line by line:
rename <- function(df, column, new){
x <- names(df) #Did this to avoid typing twice
if (is.numeric(column)) column <- x[column] #Take numeric input by indexing
names(df)[x %in% column] <- new #What you're interested in
return(df)
}
#try it out
rename(mtcars, 'mpg', 'NEW')
rename(mtcars, 1, 'NEW')

Answer 5

I disagree with #Chase - the grepl solution ain't the luckiest one. I'd say: go with simple ==. Here's why:
d <- data.frame(matrix(rnorm(100), 10))
colnames(d) <- replicate(10, paste(sample(letters[1:5], size = 5, replace=TRUE, prob=c(.1, .6, .1, .1, .1)), collapse = ""))
Now try doing grepl("b", colnames(d)). Either pass fixed = TRUE, or even better do simple colnames(d) == "b" like #joran suggested. Regex matching will always be slower than ==, so for simple tasks like this you may want to use simple ==.