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How to use dplyr's coalesce function with group_by() to create one row per person with all values filled in?（2个答案）
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我正在使用r，我遇到的问题是，我希望每个id和date都有唯一的time 1和time 2值，但对于某些行，它们是分开的。我想通过组合列的值来合并组合某些列中具有相同值的行。
具体来说，如果“id”和“date”匹配，我想将两行减少为一行。

id <- c('A', 'B', 'C', 'B', 'C'); date <- c('2023-10-09', "2023-10-10", "2023-10-10", "2023-10-10", "2023-10-10")
time1 <- c('2023-10-09 05:55:55', '2023-10-10 05:55:55', NA, NA, '2023-10-10 10:55:55')
time2 <- c('2023-10-09 06:10:55', NA, '2023-10-10 20:55:55', '2023-10-10 20:59:55', NA)
df <- data.frame(id, date, time1, time2)
> df
  id       date               time1               time2
1  A 2023-10-09 2023-10-09 05:55:55 2023-10-09 06:10:55
2  B 2023-10-10 2023-10-10 05:55:55                <NA>
3  C 2023-10-10                <NA> 2023-10-10 20:55:55
4  B 2023-10-10                <NA> 2023-10-10 20:59:55

我想要的是

id <- c('A', 'B', 'C'); date <- c('2023-10-09', "2023-10-10", "2023-10-10")
time1 <- c('2023-10-09 05:55:55', '2023-10-10 05:55:55','2023-10-10 10:55:55')
time2 <- c('2023-10-09 06:10:55', '2023-10-10 20:55:55', '2023-10-10 20:59:55')
df <- data.frame(id, date, time1, time2)
> df
  id       date               time1               time2
1  A 2023-10-09 2023-10-09 05:55:55 2023-10-09 06:10:55
2  B 2023-10-10 2023-10-10 05:55:55 2023-10-10 20:55:55
3  C 2023-10-10 2023-10-10 10:55:55 2023-10-10 20:59:55

任何输入都将有帮助！