计算 R 中不同操作的日期之间的差异

Question

I have a dataframe like this:

d=data.frame('Id'=c(101,101,101,101,103,103,103),
             'Action'=c('hire','terminate','promoted','promoted','hire','promoted','terminate'),
             'date'=c('02/22/2017','12/11/2020','11/11/2018','03/22/2019','02/23/2016','01/12/2018','03/21/2019'))

I want to create a new column time_spent which calculates the day between hire date and terminate date, and the hire date and promoted date.

df$date=mdy(df$date)

df %>%
    mutate(date = as.Date(EFFDT)) %>%
    group_by(EMPLID) %>%
    summarise(time_spent = as.numeric(difftime(EFFDT[ACTION == 'TER'], EFFDT[ACTION == 'HIR'], units = 'days')))%>%
    inner_join(df,by='EMPLID')

Above code calculates the time but it's between hire and terminate. How can we do for hire and promoted?

Also if we see that for Id 101 promotion happens twice we have two different dates. So if we apply the above code for hire and promoted it only calculates the days between hire and the first occurrence of promoted Action for Id 101, and not all promoted Action which happened for Id 101.

Answer 1

Maybe this might be helpful. Instead of summarise and inner_join you can use mutate and have the new column time_spent be the time difference between that row's date and date when the person was hired.

library(tidyverse)

d %>%
  mutate(date = as.Date(date, format = "%m/%d/%Y")) %>%
  arrange(Id, date) %>%
  group_by(Id) %>%
  mutate(time_spent = difftime(date, date[Action == "hire"], units = "days"))

Output

     Id Action    date       time_spent
  <dbl> <chr>     <date>     <drtn>    
1   101 hire      2017-02-22    0 days 
2   101 promoted  2018-11-11  627 days 
3   101 promoted  2019-03-22  758 days 
4   101 terminate 2020-12-11 1388 days 
5   103 hire      2016-02-23    0 days 
6   103 promoted  2018-01-12  689 days 
7   103 terminate 2019-03-21 1122 days

Edit : If you want to include NA when no "hire" date is available, you can filter to include Id that has any "hire" and then rejoin data again. Just make sure data frame has date in correct format first.

d$date <- as.Date(d$date, format = "%m/%d/%Y")

d %>%
  arrange(Id, date) %>%
  group_by(Id) %>%
  filter(any(Action == "hire")) %>%
  mutate(time_spent = difftime(date, date[Action == "hire"], units = "days")) %>%
  right_join(d)

Data

d <- structure(list(Id = c(101, 101, 101, 101, 103, 103, 103), Action = c("hire", 
"terminate", "promoted", "promoted", "hire", "promoted", "terminate"
), date = c("02/22/2017", "12/11/2020", "11/11/2018", "03/22/2019", 
"02/23/2016", "01/12/2018", "03/21/2019")), class = "data.frame", row.names = c(NA, 
-7L))