通过提取数据框列表的括号中的内容来重命名多个列

Question

Given data and code as follows, I'm able to generate a list of dataframes dfs :

library(data.table)
library(purrr)
library(glue)
library(gt)
library(tidyverse)
library(purrr)

df <- structure(list(id = c("M0000607", "M0000609", "M0000612"), `2021-08(actual)` = c(12.6, 
                                                                                       19.2, 8.3), `2021-09(actual)` = c(10.3, 17.3, 6.4), `2021-10(actual)` = c(8.9, 
                                                                                                                                                                 15.7, 5.3), `2021-11(actual)` = c(7.3, 14.8, 3.1), `2021-12(actual)` = c(6.1, 
                                                                                                                                                                                                                                          14.2, 3.5), `2021-08(pred)` = c(11.65443222, 14.31674997, 7.084180415
                                                                                                                                                                                                                                          ), `2021-09(pred)` = c(12.29810914, 17.7143733, 6.057927385), 
                     `2021-10(pred)` = c(9.619846116, 15.54553601, 6.525992602
                     ), `2021-11(pred)` = c(8.352097939, 13.97318204, 3.164682627
                     ), `2021-12(pred)` = c(6.113631596, 14.16243166, 3.288372517
                     ), `2021-08(error)` = c(2.082307066, 1.146759554, 0.687406723
                     ), `2021-09(error)` = c(1.631350383, 2.753457736, 2.952737781
                     ), `2021-10(error)` = c(0.945567783, 4.883250027, 1.215819585
                     ), `2021-11(error)` = c(1.998109138, 0.414373304, 0.342072615
                     ), `2021-12(error)` = c(0.719846116, 0.154463985, 1.225992602
                     )), class = "data.frame", row.names = c(NA, -3L))

year_months <- c('2021-12', '2021-11', '2021-10')  
curr <- lubridate::ym(year_months)
prev <- curr - months(2L)
dfs <- mapply(function(x, y) {
  df[c(
    "id", 
    format(seq.Date(y, x, by = "month"), "%Y-%m(actual)"), 
    format(x, "%Y-%m(pred)"), 
    format(x, "%Y-%m(error)")
  )]
}, curr, prev, SIMPLIFY = FALSE)

I want to rename last two column by extracting content in parentheses as new column names:

[[1]]
        id 2021-10(actual) 2021-11(actual) 2021-12(actual)          pred          error
1 M0000607             8.9             7.3             6.1      6.113632      0.7198461
2 M0000609            15.7            14.8            14.2     14.162432      0.1544640
3 M0000612             5.3             3.1             3.5      3.288373      1.2259926

[[2]]
        id 2021-09(actual) 2021-10(actual) 2021-11(actual)          pred          error
1 M0000607            10.3             8.9             7.3      8.352098      1.9981091
2 M0000609            17.3            15.7            14.8     13.973182      0.4143733
3 M0000612             6.4             5.3             3.1      3.164683      0.3420726

[[3]]
        id 2021-08(actual) 2021-09(actual) 2021-10(actual)          pred          error
1 M0000607            12.6            10.3             8.9      9.619846      0.9455678
2 M0000609            19.2            17.3            15.7     15.545536      4.8832500
3 M0000612             8.3             6.4             5.3      6.525993      1.2158196

How could I do that using R? Thanks.

Reference link:

Extract info inside all parenthesis in R

Answer 1

Here's a tidyverse alternative -

library(dplyr)
library(purrr)

map(dfs, ~.x %>%rename_with(~sub('.*\\((.*)\\)$', '\\1', .x), last_col(c(0, 1))))

#[[1]]
#        id 2021-10(actual) 2021-11(actual) 2021-12(actual)      pred     error
#1 M0000607             8.9             7.3             6.1  6.113632 0.7198461
#2 M0000609            15.7            14.8            14.2 14.162432 0.1544640
#3 M0000612             5.3             3.1             3.5  3.288373 1.2259926

#[[2]]
#        id 2021-09(actual) 2021-10(actual) 2021-11(actual)      pred     error
#1 M0000607            10.3             8.9             7.3  8.352098 1.9981091
#2 M0000609            17.3            15.7            14.8 13.973182 0.4143733
#3 M0000612             6.4             5.3             3.1  3.164683 0.3420726

#[[3]]
#        id 2021-08(actual) 2021-09(actual) 2021-10(actual)      pred     error
#1 M0000607            12.6            10.3             8.9  9.619846 0.9455678
#2 M0000609            19.2            17.3            15.7 15.545536 4.8832500
#3 M0000612             8.3             6.4             5.3  6.525993 1.2158196

Answer 2

Using lapply to manipulate the elements of the list dfs , we can use some regex to do the job.

dfs <- lapply(dfs, function(x) {
  col_num <- grep('pred|error', colnames(x))
  colnames(x)[col_num] <- gsub('.*\\(|\\)', '', colnames(x)[col_num]); x
})

Output

> dfs
[[1]]
        id 2021-10(actual) 2021-11(actual) 2021-12(actual)      pred     error
1 M0000607             8.9             7.3             6.1  6.113632 0.7198461
2 M0000609            15.7            14.8            14.2 14.162432 0.1544640
3 M0000612             5.3             3.1             3.5  3.288373 1.2259926

[[2]]
        id 2021-09(actual) 2021-10(actual) 2021-11(actual)      pred     error
1 M0000607            10.3             8.9             7.3  8.352098 1.9981091
2 M0000609            17.3            15.7            14.8 13.973182 0.4143733
3 M0000612             6.4             5.3             3.1  3.164683 0.3420726

[[3]]
        id 2021-08(actual) 2021-09(actual) 2021-10(actual)      pred     error
1 M0000607            12.6            10.3             8.9  9.619846 0.9455678
2 M0000609            19.2            17.3            15.7 15.545536 4.8832500
3 M0000612             8.3             6.4             5.3  6.525993 1.2158196