具有不同行（相同标识符）的日期之间的月份差异（在 R 中）

Question

 cpf sep_month hire_month hire_year sep_day  hire_date   sep_date
4 123         4          2      2012       1 2012-02-01 2013-04-01
5 123         0          4      2013       1 2013-04-01       <NA>
6 122        10          9      2012       1 2012-09-01 2013-10-01
7 122         0         12      2013       1 2013-12-01       <NA>

structure(list(cpf = c(123L, 123L, 122L, 122L), sep_month = c(4L, 
0L, 10L, 0L), hire_month = c(2L, 4L, 9L, 12L), hire_year = c(2012L, 
2013L, 2012L, 2013L), sep_day = c(1L, 1L, 1L, 1L), hire_date = structure(c(15371, 
15796, 15584, 16040), class = "Date"), sep_date = structure(c(15796, 
NA, 15979, NA), class = "Date")), row.names = 4:7, class = "data.frame")

In my dataset, each row is a job contract. I want to see the difference in months between sep_date and hire_date across different rows for the same CPF (identifier). For example, individual 123 separated from its job on 2013-04. On the following row, he was hired (under a different contract/job) on 2013-04. My goal is to create a dummy equal to 1 for individuals who separated from a contract and found a job in the same or following month. That would be the case for individual 123, but not for individual 122.

I appreciate any help.

Answer 1

dplyr

library(dplyr)
dat %>%
  group_by(cpf) %>%
  mutate(dummy = sapply(sep_date, function(z) any(!is.na(z) & z <= hire_date))) %>%
  ungroup()
# # A tibble: 4 x 8
#     cpf sep_month hire_month hire_year sep_day hire_date  sep_date   dummy
#   <int>     <int>      <int>     <int>   <int> <date>     <date>     <lgl>
# 1   123         4          2      2012       1 2012-02-01 2013-04-01 TRUE 
# 2   123         0          4      2013       1 2013-04-01 NA         FALSE
# 3   122        10          9      2012       1 2012-09-01 2013-10-01 TRUE 
# 4   122         0         12      2013       1 2013-12-01 NA         FALSE

If you want dummy to be truly just 1 s and 0 s, then add a + before it, as in dummy = +sapply(...) ( + is a shortcut to convert logical to integer ).

base R

dat$dummy <- ave(
  seq_along(dat$cpf), dat$cpf,
  FUN = function(i) sapply(dat$sep_date[i], function(z) any(!is.na(z) & z <= dat$hire_date[i])))
dat
#   cpf sep_month hire_month hire_year sep_day  hire_date   sep_date dummy
# 4 123         4          2      2012       1 2012-02-01 2013-04-01     1
# 5 123         0          4      2013       1 2013-04-01       <NA>     0
# 6 122        10          9      2012       1 2012-09-01 2013-10-01     1
# 7 122         0         12      2013       1 2013-12-01       <NA>     0

This is a little uglier because ave doesn't like doing multi-column (multi-vector) stuff, but it gets the same results.