将指示符== 1的全天更改为指示符== 1的第一天

df%>%
  group_by(transaction)%>%
  mutate(day=case_when(indicator==0~day,
                       T~head(day[indicator==1],1)))%>%
  group_by(transaction,day)%>%
  summarise(indicator=sum(indicator))%>%
  ungroup

  transaction   day indicator
        <int> <int>     <int>
1           1     1         0
2           1     2         0
3           1     3         0
4           1     4         3
5           2     1         0
6           2     2         0
7           2     3         0
8           2     4         0
9           2     5         2

请尝试以下代码
编号

df <- bind_rows(df1, df2) %>% group_by(transaction) %>% 
mutate(cumsum=cumsum(indicator), cumsum2=ifelse(cumsum==1, day, NA)) %>% 
fill(cumsum2) %>% 
mutate(day=ifelse(!is.na(cumsum2), cumsum2, day)) %>% 
group_by(transaction, day) %>% slice_tail(n=1) %>% select(-cumsum2)

创建于2023年1月19日，使用reprex v2.0.2

输出

# A tibble: 8 × 4
# Groups:   transaction, day [8]
  transaction   day indicator cumsum
        <dbl> <int>     <dbl>  <dbl>
1           1     1         0      0
2           1     2         0      0
3           1     3         0      0
4           1     4         1      3
5           2     1         0      0
6           2     2         0      0
7           2     3         0      0
8           2     4         1      2

另一种尝试方法。按transaction分组后，将indicator更改为0（相同）或indicator的sum。最后，保留或filter以前的行，其中indicator的cumall（所有累积）值为0。使用lag将提供包含总和的最后一行。

library(tidyverse)

df %>%
  group_by(transaction) %>%
  mutate(indicator = ifelse(indicator == 0, 0, sum(indicator))) %>%
  filter(cumall(lag(indicator, default = 0) == 0))

• • 产出**

transaction   day indicator
        <int> <int>     <dbl>
1           1     1         0
2           1     2         0
3           1     3         0
4           1     4         3
5           2     1         0
6           2     2         0
7           2     3         0
8           2     4         0
9           2     5         2