R: Set column name from dataframe name in list of dataframes

Question

I have a large list of dataframes like the following:

> head(lst)
$Set1
  ID Value
1  A     1
2  B     1
3  C     1

$Set2
  ID Value
1  A     1
2  D     1
3  E     1

$Set3
  ID Value
1  B     1
2  C     1

I would like to change the name of the column "Value" in each dataframe to be similar to the name of the dataframe, so that the list of dataframes looks like this:

> head(lst)
$Set1
  ID Set1
1  A     1
2  B     1
3  C     1

$Set2
  ID Set2
1  A     1
2  D     1
3  E     1

$Set3
  ID Set3
1  B     1
2  C     1

Can anyone think of a function that takes the name of each dataframe in the list and names the column accordingly? My original list has >400 dataframes, so I was hoping to automate this somehow. Sorry if this is a naive question, but I'm somehow stuck...

Thanks so much!

Here is an example of a list of dfs:

lst <- list(
  data.frame(ID = c("A", "B", "C"), Value = c(1, 1, 1)),
  data.frame(ID = c("A", "D", "E"), Value = c(1, 1, 1)),
  data.frame(ID = c("B", "C"), Value = c(1, 1)),
  data.frame(ID = c("B", "C"), Value = c(1, 1)),
  data.frame(ID = c("B", "C"), Value = c(1, 1)),
  data.frame(ID = c("B", "C"), Value = c(1, 1)))
lst_names <- c("Set1", "Set2", "Set3", "Set4", "Set5","Set6")
names(lst) <- lst_names

An almost identical question was asked here eariler today - see stackoverflow.com/questions/71645690/… - does that answer your question? — Allan Cameron
– Allan Cameron, Commented Mar 28, 2022 at 12:48
Thanks, I think the issue here is slightly simpler and the answer below by TimTeaFan is more efficient than the ones in the other thread. — yakumo
– yakumo, Commented Mar 28, 2022 at 13:11

TimTeaFan · Accepted Answer · 2022-03-28 12:59:13Z

5

In the tidyverse we can use purrr::imap and dplyr::rename:

library(purrr)
library(dplyr)

lst %>% 
  imap(~ rename(.x, "{.y}" := Value))
#> $Set1
#>   ID Set1
#> 1  A    1
#> 2  B    1
#> 3  C    1
#> 
#> $Set2
#>   ID Set2
#> 1  A    1
#> 2  D    1
#> 3  E    1
#> 
#> $Set3
#>   ID Set3
#> 1  B    1
#> 2  C    1
#> 
#> $Set4
#>   ID Set4
#> 1  B    1
#> 2  C    1
#> 
#> $Set5
#>   ID Set5
#> 1  B    1
#> 2  C    1
#> 
#> $Set6
#>   ID Set6
#> 1  B    1
#> 2  C    1

^{Created on 2022-03-28 by the reprex package (v2.0.1)}

answered Mar 28, 2022 at 12:59

TimTeaFan

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2 Comments

r_user Over a year ago

I'm getting a could not find function ":=" error when I use this.

TimTeaFan Over a year ago

@r_user: In this case there seems to be a problem with your installed packages. {dplyr} depends on {rlang} from which the walrus operator := comes from. Try reinstalling rlang.

Nad Pat · Accepted Answer · 2022-03-28 12:49:59Z

2

We can do,

lapply(
  names(lst), 
  function(x) setNames(lst[[x]], c(names(lst[[x]])[2], x))
)

[[1]]
  Value Set1
1     A    1
2     B    1
3     C    1

[[2]]
  Value Set2
1     A    1
2     D    1
3     E    1

answered Mar 28, 2022 at 12:49

Nad Pat

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Comments

juarpasi · Accepted Answer · 2023-05-24 00:01:15Z

0

You can also make a function to change one column from a datafram

changeOneColumName <- function(df, oldName, newName){
  names(df)[names(df) == oldName] <- newName
  return(df)
}

And then, apply that function with the Map

Map(changeOneColumName, lst, oldName = 'Value', newName = names(lst))

answered May 24, 2023 at 0:01

juarpasi

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