填充 R 中缺失的日期

Question

I would like some help regarding a data frame transformation required for an analysis. My data consists of a large amount of individuals with all their historic employment. "EX" is a code representing the reason for ending employment. Something like this:

id  Date_start    Date_end       EX
13  "2001-02-01"  "2001-05-30"   A
13  "2002-03-01"  "2010-06-02"   B
14  ...           ...
...

So what I would like to do is to "fill in the gaps". This may not be easy but its even more difficult because I want it aggregated by id and each new row should have the EX value of the row before, like this:

id  Date_start    Date_end       EX
13  "2001-02-01"  "2001-05-30"   A
13  "2001-05-31"  "2002-02-28"   A
13  "2002-03-01"  "2010-06-02"   B
14  ...           ...
...

I believe the trick would be some kind of lag and aggregate but I'm totally lost.

Answer 1

This is a little bit tricky, and you can mainly utilize the dplyr package to do the manipulation and lubridate packages to convert the date format(you can use as.Date() for sure, but lubridate makes it easier).

library(dplyr)
library(lubridate)

1.Creating the sample data you provided.

names <- c("id", "Date_start",    "Date_end",       "EX")
row1 <- c(13 , "2001-02-01" , "2001-05-30" ,  "A")
row2 <- c(13 , "2002-03-01" , "2010-06-02" ,  "B")


testdata <- rbind(row1,row2) %>% data.frame(stringsAsFactors = F)
row.names(testdata) <- NULL

names(testdata) <- names

testdata$Date_start <- testdata$Date_start %>% as_date()
testdata$Date_end <- testdata$Date_end %>% as_date()
testdata

2.Creating a new data set that has the data you want to add.

id : we are using the same id value since it is grouping by id.
Date_start : we are creating the Date_start with a value if there is gap, otherwise "" (empty column, and we are filtering them out).
Date_end : Same logic for Date_end.
EX : we are using the second last EX value as you stated.

  new_data <- test_data %>% 
  group_by(id) %>% 
  mutate(Date_start1 = ifelse(Date_start-lag(Date_end) == 1,0,lag(Date_end)+1),
         Date_end1 = ifelse(Date_start-lag(Date_end) == 1,0,Date_start-1),
         EX=first(EX)) %>% 
  filter(!Date_start1 ==0) %>% 
  select(id, Date_start=Date_start1,Date_end=Date_end1,EX) %>% 
  distinct() %>% 
  ungroup()

3.Since we want to fill the gap days, mutate made it into numeric value, and we are using as_date() from lubriate to convert it into date format.

new_data$Date_start <- as_date(new_data$Date_start)
new_data$Date_end <- as_date(new_data$Date_end)

4.Combine it with your sample data and arrange it by Date_state.

final <- rbind(testdata,new_data) %>% data.frame() %>% arrange(Date_start)
final

Your final result is as below.