Associate one column to another to create a third one in R-CodePudding

1  1          2         3      740      780      780      780
2  1          4         0      890      890      890      890
3  2          3         5      550      550      550      550
4  2          5        10      890      250      250      400
  wage_may wage_jun wage_jul wage_aug wage_sep wage_oct wage_nov
1      780      780      780      780      780      780      780
2      890      890      890      890      890      790      250
3      550      550      550      550      550      550      550
4      500      890      600      750      890      300      300
  wage_dec
1      780
2      300
3      550
4      300

structure(list(id = c(1L, 1L, 2L, 2L), hire_month = c(2L, 4L, 
3L, 5L), sep_month = c(3L, 0L, 5L, 10L), wage_jan = c(740L, 890L, 
550L, 890L), wage_feb = c(780L, 890L, 550L, 250L), wage_mar = c(780L, 
890L, 550L, 250L), wage_apr = c(780L, 890L, 550L, 400L), wage_may = c(780L, 
890L, 550L, 500L), wage_jun = c(780L, 890L, 550L, 890L), wage_jul = c(780L, 
890L, 550L, 600L), wage_aug = c(780L, 890L, 550L, 750L), wage_sep = c(780L, 
890L, 550L, 890L), wage_oct = c(780L, 790L, 550L, 300L), wage_nov = c(780L, 
250L, 550L, 300L), wage_dec = c(780L, 300L, 550L, 300L)), class = "data.frame", row.names = c(NA, 
-4L))

I would like to create a column for last wage in the following way: If sep_month is 3, I would like this new column to give last_wage as the one in wage_mar; if sep_month is 4, I would like this column to give last_wage as the one in wage_apr and so forth.

I basically want to associate the sep_month to its respective wage depending on the month.

CodePudding user response：

You could do this with mutate() and case_when() from the dplyr library:

df %>% 
  mutate(last_wage = 
    case_when(
      sep_month == 0 ~ wage_dec,
      
      sep_month == 1 ~ wage_jan, 
      sep_month == 2 ~ wage_feb, 
      sep_month == 3 ~ wage_mar, 
      sep_month == 4 ~ wage_apr,
      sep_month == 5 ~ wage_may,
      sep_month == 6 ~ wage_jun,
      sep_month == 7 ~ wage_jul,
      sep_month == 8 ~ wage_aug,
      sep_month == 9 ~ wage_sep,
      sep_month == 10 ~ wage_oct,
      sep_month == 11 ~ wage_nov,
      sep_month == 12 ~ wage_dec,
      
      TRUE ~ NA_integer_
    )
  )

CodePudding user response：

Another option would be to pivot to long form and make the calculation then pivot back to wide form with tidyverse. I also added in converting 0 to 12, but then converted it back at the end.

library(tidyverse)

df %>%
  mutate(rowid = row_number()) %>%
  pivot_longer(-c(id, hire_month, sep_month, rowid), names_to = c(".value", "month"), names_sep = "_") %>%
  mutate(sep_month = ifelse(sep_month == 0, 12, sep_month),
         month_num = match(str_to_title(month),month.abb),
         last_wage = ifelse(sep_month == month_num, wage, NA) ) %>%
  select(-month_num) %>%
  pivot_wider(id_cols = c(id, hire_month, sep_month, last_wage), names_from = "month", names_glue = "{.value}_{month}", values_from = wage) %>%
  group_by(id, hire_month, sep_month) %>%
  fill(everything(), .direction = "downup") %>%
  slice(1) %>% 
  mutate(sep_month = ifelse(sep_month == 12, 0, sep_month))

Output

     id hire_month sep_month last_wage wage_jan wage_feb wage_mar wage_apr wage_may wage_jun wage_jul wage_aug wage_sep wage_oct wage_nov wage_dec
  <int>      <int>     <dbl>     <int>    <int>    <int>    <int>    <int>    <int>    <int>    <int>    <int>    <int>    <int>    <int>    <int>
1     1          2         3       780      740      780      780      780      780      780      780      780      780      780      780      780
2     1          4        12       300      890      890      890      890      890      890      890      890      890      790      250      300
3     2          3         5       550      550      550      550      550      550      550      550      550      550      550      550      550
4     2          5        10       300      890      250      250      400      500      890      600      750      890      300      300      300