The sample dataset is given as below:

v = data.frame(group = c(1,1,2,3,3),date = as.Date(c('01-01-2000','01-01-2001','01-05-2000','02-07-2000','01-01-2008'), "%d-%m-%Y"))
  
v%>% group_by(group ) %>% mutate(difference_day = ifelse(n() == 2, 
                                                         c(0,diff(date )),
                                                         difftime(date ,as.Date('31-12-2021', "%d-%m-%Y"),units='days')))

My desirable result is :

In the above code, if the length of groups is equal to one, then the days_difference will be
difftime(date ,as.Date('31-12-2021', "%d-%m-%Y"),units='days')).

However,
the output of the code was:

# A tibble: 5 × 3
# Groups:   group [3]
  group date       difference_day
  <dbl> <date>              <dbl>
1     1 2000-01-01              0
2     1 2001-01-01              0
3     2 2000-05-01          -7914
4     3 2000-07-02              0
5     3 2008-01-01              0

which was very strange.
Please give me some suggestions, thank you!

>Solution :

Since you want to replace either the first vector or the second vector, use if instead of if_else. (That is, your conditional is external to the vectors, not an element-by-element conditional, where if_else would be more appropriate.)

v %>%
  group_by(group) %>%
  mutate(d = if (n() == 2L) diff(c(date[1], date)) else difftime(as.Date("2021-12-31"), date, units = "days")) %>%
  ungroup()
# # A tibble: 5 × 3
#   group date       d        
#   <dbl> <date>     <drtn>   
# 1     1 2000-01-01    0 days
# 2     1 2001-01-01  366 days
# 3     2 2000-05-01 7914 days
# 4     3 2000-07-02    0 days
# 5     3 2008-01-01 2739 days

There are some differences of +/- 1 from your expected output, not sure if that was a typo or some other intent outside of a traditional diff.

The return from both diff and difftime here are class "difftime", which prints naturally with ". days"; they are still number-enough that math or such still works on them. If you prefer not, just wrap with as.integer(.) or as.numeric(.).