'How to take a substring of dataframes' names inside a list to replace all values in a column with this string?

I have a long list of data frames. I want to get the last two strings of each data frame's name and use it to replace all values in certain column of each data frame within the list.

Toy example:

df_stringPA <- data.frame(name="MU",
                  col1= c(1,2,3),
                  col2= c(1,2,3))

df_stringCA <- data.frame(name="MU",
                  col1= c(1,2,3),
                  col2= c(1,2,3))

test_list <- list(PA=df_PA, CA=df_CA)

### desired output:

output_df_PA <- data.frame(name="PA",
                    col1= c(1,2,3),
                    col2= c(1,2,3))


output_df_CA <- data.frame(name="CA",
                    col1= c(1,2,3),
                    col2= c(1,2,3))

If I do it for an individual dataframe within the list, it works:

### doing it separately for each df within the list:
test_list$PA$name <- str_sub(deparse(substitute(test_list$PA)), start= -2)

But it is a long list, so doing it manually is not the best option. I have tested:

test_list <- lapply(test_list, function(x) x$name <- str_sub(deparse(substitute(x)), start= -2))

But I get the data frames within the list replaced by ']]'

rstringlistdataframe


Solution 1:[1]

You can use Map:

 Map(function(a, b) {a$name <- b; a}, a = test_list, b = names(test_list))
$PA
  name col1 col2
1   PA    1    1
2   PA    2    2
3   PA    3    3

$CA
  name col1 col2
1   CA    1    1
2   CA    2    2
3   CA    3    3

Solution 2:[2]

We could use imap

library(purrr)
library(dplyr)
imap(test_list, ~ .x %>%
     mutate(name = .y))
$PA
  name col1 col2
1   PA    1    1
2   PA    2    2
3   PA    3    3

$CA
  name col1 col2
1   CA    1    1
2   CA    2    2
3   CA    3    3

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Allan Cameron
Solution 2 akrun

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