R : convert year in column name into new rows-CodePudding

How can I R to identify in columns other than an ID column, the year specified at the end of each column (which follows a variety of characters), then place corresponding rows in a new data frame with that year expressed as a new column, and finally have a new column that simply removes the year from the original column names and eliminates any underscore that now appears at the end of the column name?

For example, I want to convert a data frame of 2 records with 5 columns named col1, col2_1980, col2_1981, col3_1980, and col3_1981 in which col1 is a character value (either "a", or "b") into a data frame with 4 records that has col1 = "a" for 2 records, and col1 = "b" for 2 records, and then col2 = "1980" for 1 record per col1 value and col2 = "1981" for 1 record per col1 value.

Uses of

CodePudding user response：

pivot_longer() can handle patterns / separators in names.
With updated dataset:

library(dplyr)
library(tidyr)

set.seed(3)
df1 <-
  rbind(
    data.frame(
      col1 = "a", 
      col2_1_1980 = runif(1), 
      col2_1_1981 = runif(1), 
      col3_1_1980 = runif(1), 
      col3_1_1981 = runif(1)),
    data.frame(  
      col1 = "b", 
      col2_1_1980 = runif(1), 
      col2_1_1981 = runif(1), 
      col3_1_1980 = runif(1), 
      col3_1_1981 = runif(1)))

df1 %>% pivot_longer(
  cols = contains("_"), 
  names_pattern = "(.*)_(\\d )$", 
  names_to = c(".value", "year"))
#> # A tibble: 4 × 4
#>   col1  year  col2_1 col3_1
#>   <chr> <chr>  <dbl>  <dbl>
#> 1 a     1980   0.168  0.385
#> 2 a     1981   0.808  0.328
#> 3 b     1980   0.602  0.125
#> 4 b     1981   0.604  0.295

Original sample data in question included column names like col2_1980 & col2_1981 , for those names_sep param serves well:


df1 %>% pivot_longer(
  cols = contains("_"), 
  names_sep = "_", 
  names_to = c(".value", "year"))
#> # A tibble: 4 × 4
#>   col1  year   col2  col3
#>   <chr> <chr> <dbl> <dbl>
#> 1 a     1980  0.168 0.385
#> 2 a     1981  0.808 0.328
#> 3 b     1980  0.602 0.125
#> 4 b     1981  0.604 0.295

^{Created on 2023-01-18 with reprex v2.0.2}

CodePudding user response：

Please try the below code , accomplished the expected result using pivot_longer, pivot_wider

code

library(dplyr)

df2 <- df1 %>% pivot_longer(c(contains('_'))) %>% 
mutate(year=str_extract(name,'(?<=\\_)\\d.*'), name=str_extract(name,'^.*(?=\\_)')) %>% 
pivot_wider(c(col1,year), names_from = 'name', values_from = 'value')