Chop off the first letter of every variable name

Question

I have some data that looks like this:

    country agdp apop
1        US  100  100
2 Australia   50   50

The variable names are agdp and apop, but I would like them to be gdp and pop. My real data has many, many variables that all need that transformation.

And this is what my desired outcome:

 country gdp pop
1        US  100  100
2 Australia   50   50

Reproducible code below:

df <- data.frame(stringsAsFactors=FALSE,
     country = c("US", "Australia"),
        agdp = c(100, 50),
        apop = c(100, 50)

desired_df <- data.frame(stringsAsFactors=FALSE,
     country = c("US", "Australia"),
        gdp = c(100, 50),
        pop = c(100, 50)

Possible duplicate https://stackoverflow.com/q/35113553/680068 and https://stackoverflow.com/questions/46954165/rename-a-set-of-columns-in-r-by-position — zx8754, Jun 11 '19 at 07:10

score 4 · Answer 1 · answered Jun 11 '19 at 06:28

4

Using regex we can extract everything other than first character and assign the names.

names(df)[-1] <- sub("^.(.*)$", "\\1", names(df)[-1])

df
#    country gdp pop
#1        US 100 100
#2 Australia  50  50

answered Jun 11 '19 at 06:28

Ronak Shah

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tmfmnk · Answer 2 · 2019-06-11T07:54:02.880

3

One dplyr possibility could be:

df %>%
 rename_at(2:length(.), list(~ substr(., 2, nchar(.))))

    country gdp pop
1        US 100 100
2 Australia  50  50

The same with base R:

names(df)[-1] <- substr(names(df)[-1], 2, nchar(names(df)[-1]))

edited Jun 11 '19 at 07:54

answered Jun 11 '19 at 06:26

tmfmnk

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4
47
67

score 3 · Answer 3 · edited Jun 11 '19 at 06:41

3

Here is one approach

library(stringr)

names(df)[-1] = str_sub(names(df)[-1], 2)

edited Jun 11 '19 at 06:41

Sotos

51,121
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answered Jun 11 '19 at 06:32

Theo

575
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Adding to this, you could swap out `str_sub` with `substring` if you wanted to use a base function. – Andrew Jun 11 '19 at 12:13

score 1 · Answer 4 · answered Jun 11 '19 at 06:54

One could also do:

Purely base(can use setdiff or %in% to "automate" selection.):

sapply(names(df), function(x) ifelse(x=="country",x,substring(x,2,nchar(x))))

Less elegant with tidyverse since rename_at has been shown:

names(df)<-unlist(names(df) %>% 
  map(.,function(x) ifelse(x=="country",x,substring(x,2,nchar(x)))))

Chop off the first letter of every variable name

4 Answers4

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