*** Edited I am using R tidyverse and trying to parse specific column names as follows: The column names in question contain characters and a number (representing a year), like "var 1975", "var 1976", etc. I would like to parse these names as to become: "var_75", "var_76", etc. In other words, I want to parse the number representing the year and remove the first two digits, then add the year as two digits to the characters, separated by an underscore.
I am trying this:
library(tidyverse)
df <- tibble("var 1975" = c(1:5),
"var 1976" = c(3,2,1,1,1),
"age" = c(25,41,39,60,36) ,
"satisfaction" = c(5,3,2,5,4)
)
# Output
# var 1975 var 1976 age satisfaction
# 1 1 25 5
# 2 2 41 3
# 3 3 39 2
# 4 4 60 5
# 5 5 36 4
df <- df %>%
rename_with( .fn = function(.x){paste0(.x, "_",
parse_number(cols) -1900)},
.cols=(contains("var") )) %>% #add year as suffix
rename_with(.fn = ~gsub("([^\\d]+)\\d+_(\\d+)", "\\1_\\2", .x),
.cols=contains("var"))
rename_with(df, ~ str_replace(., " \\d{2}", "_"), contains("var"))
Explanation: " \\d{2}" is a regex string which looks for a space, followed by two digits. It's replaced by an underscore.
Output:
# A tibble: 5 × 4
var_75 var_76 age satisfaction
<int> <dbl> <dbl> <dbl>
1 1 3 25 5
2 2 2 41 3
3 3 1 39 2
4 4 1 60 5
5 5 1 36 4