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
CodePudding user response：
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
CodePudding user response：
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