inspired by hadley's nifty gather approach in this answer I tried to use tidyr's gather() and spread() in combination with a regular expression, regex, but I seem to get it wrong on the regex.
I did study several regex questions; this one, this one, and also at regex101.com. I tried to circumvent the regex by using starts_with(), ends_with() and matches() inspired by this question, but with no luck.
I am asking here in the hope that someone can show where I get it wrong and I can solve it, preferably using, the select helpers from tidyselect.
I need to select 2 regex-groups one up to the last . and one consisting of what comes after the last ., I made this two example below, one where my code s working and one where I am stuck.
First the example that is working,
# install.packages(c("tidyverse"), dependencies = TRUE)
require(tidyverse)
The first data set, that work, looks like this,
myData1 <- tibble(
  id = 1:10,
  Wage.1997.1 = c(NA, 32:38, NA, NA),
  Wage.1997.2 = c(NA, 12:18, NA, NA),
  Wage.1998.1 = c(NA, 42:48, NA, NA),
  Wage.1998.2 = c(NA, 2:8, NA, NA),  
  Wage.1998.3 =  c(NA, 42:48, NA, NA),    
  Job.Type.1997.1 = NA,
  Job.Type.1997.2 = c(NA, rep(c('A', 'B'), 4), NA),
  Job.Type.1998.1 = c(NA, rep(c('A', 'B'), 4), NA),
  Job.Type.1998.2 = c(NA, rep(c('A', 'B'), 4), NA)  
)
and this is how I gather() it,
myData1 %>% gather(key, value, -id) %>%  
   extract(col = key, into = c("variable", "id.job"), regex = "(.*?\\..*?)\\.(.)$") %>% 
   spread(variable, value)
#> # A tibble: 30 x 6
#>       id id.job Job.Type.1997 Job.Type.1998 Wage.1997 Wage.1998
#>    <int> <chr>  <chr>         <chr>         <chr>     <chr>    
#>  1     1 1      <NA>          <NA>          <NA>      <NA>     
#>  2     1 2      <NA>          <NA>          <NA>      <NA>     
#>  3     1 3      <NA>          <NA>          <NA>      <NA>     
#>  4     2 1      <NA>          A             32        42       
#>  5     2 2      A             A             12        2        
#>  6     2 3      <NA>          <NA>          <NA>      42       
#>  7     3 1      <NA>          B             33        43       
#>  8     3 2      B             B             13        3        
#>  9     3 3      <NA>          <NA>          <NA>      43       
#> 10     4 1      <NA>          A             34        44       
#> # ... with 20 more rows
It works, I suspect I overdoing it with the regex, but it works. However, my real data can have either one or two digest at the end, i.e.
The second data, where I get stuck,
myData2 <- tibble(
  id = 1:10,
  Wage.1997.1 = c(NA, 32:38, NA, NA),
  Wage.1997.12 = c(NA, 12:18, NA, NA),
  Wage.1998.1 = c(NA, 42:48, NA, NA),
  Wage.1998.12 = c(NA, 2:8, NA, NA),  
  Wage.1998.13 =  c(NA, 42:48, NA, NA),    
  Job.Type.1997.1 = NA,
  Job.Type.1997.12 = c(NA, rep(c('A', 'B'), 4), NA),
  Job.Type.1998.1 = c(NA, rep(c('A', 'B'), 4), NA),
  Job.Type.1998.12 = c(NA, rep(c('A', 'B'), 4), NA)  
)
Now, this is where I use (0[0-1]|1[0-9])$ for the second group, I also tried thing like \d{1}|\d{2}, but did that not work either.
myData2 %>% gather(key, value, -id) %>% 
     extract(col = key, into = c("variable", "id.job"), 
             regex = "(.*?\\..*?)\\.(0[0-1]|1[0-9])$") %>%  
     spread(variable, value)
The expected output would be something like this,
#> # A tibble: 30 x 6
#>       id id.job Job.Type.1997 Job.Type.1998 Wage.1997 Wage.1998
#>    <int> <chr>  <chr>         <chr>         <chr>     <chr>    
#>  1     1 1      <NA>          <NA>          <NA>      <NA>     
#>  2     1 12     <NA>          <NA>          <NA>      <NA>     
#>  3     1 13     <NA>          <NA>          <NA>      <NA>     
#>  4     2 1      <NA>          A             32        42       
#>  5     2 12     A             A             12        2        
#>  6     2 13     <NA>          <NA>          <NA>      42       
#>  7     3 1      <NA>          B             33        43       
#>  8     3 12     B             B             13        3        
#>  9     3 13     <NA>          <NA>          <NA>      43       
#> 10     4 1      <NA>          A             34        44       
#> # ... with 20 more rows
A simply solution à la t this question using select helpers like starts_with(), ends_with(), matches(), etc. would be appreciated.
 
     
    