R 将两个变量与传统行分组

我有一个看起来与此类似的数据帧：

         date      uid duration
1  29.03.2020 0zOs6ZS9        1
2  29.03.2020 0zOs6ZS9        5
3  29.03.2020 0zOs6ZS9        2
4  31.03.2020 0zOs6ZS9        6
5  01.04.2020 0zOs6ZS9        7
6  01.04.2020 0zOs6ZS9        4
7  29.03.2020 0zOs6ZS9        3
8  29.03.2020   3jtMiD        2
9  30.03.2020   3jtMiD        7
10 30.03.2020   3jtMiD        5
11 31.03.2020   3jtMiD        1
12 02.04.2020   3jtMiD        2

我的目标是对日期和uid的持续时间求和，但如果有一个用户的度量值，而不是另一个用户的度量值，我实际上希望此行的持续时间为0

因此，与我通过

df2 <- df1 %>%
  group_by (date,uid) %>%
  summarise(duration =sum(duration, na.rm = TRUE))

  date       uid      duration
  <date>     <chr>       <dbl>
1 2020-03-29 0zOs6ZS9       11
2 2020-03-29 3jtMiD          2
3 2020-03-30 3jtMiD         12
4 2020-03-31 0zOs6ZS9        6
5 2020-03-31 3jtMiD          1
6 2020-04-01 0zOs6ZS9       11
7 2020-04-02 3jtMiD          2

---

我怎样才能做到这一点？

我们可以使用

complete

library(tidyr)
library(dplyr)
df2 %>%
  ungroup %>%
  complete(date, uid, fill = list(duration = 0))

-输出

# A tibble: 10 x 3
#   date       uid      duration
#   <date>     <chr>       <dbl>
# 1 2020-03-29 0zOs6ZS9       11
# 2 2020-03-29 3jtMiD          2
# 3 2020-03-30 0zOs6ZS9        0
# 4 2020-03-30 3jtMiD         12
# 5 2020-03-31 0zOs6ZS9        6
# 6 2020-03-31 3jtMiD          1
# 7 2020-04-01 0zOs6ZS9       11
# 8 2020-04-01 3jtMiD          0
# 9 2020-04-02 0zOs6ZS9        0
#10 2020-04-02 3jtMiD          2

#一个tible:10 x 3
#日期uid持续时间
#               
#1 2020-03-29 0zOs6ZS9 11
#2 2020-03-29第三季度中期报告2
#3 2020-03-30 0zOs6ZS9 0
#4 2020-03-30 3 JTMID 12
#5 2020-03-31 0zOs6ZS9 6
#6 2020-03-31第三季度中期报告1
#7 2020-04-01 0zOs6ZS9 11
#8 2020-04-01第三季度中期报告
#9 2020-04-02 0zOs6ZS9 0
#10 2020-04-02 3jtMiD 2

数据

df2
# A tibble: 10 x 3
#   date       uid      duration
#   <date>     <chr>       <dbl>
# 1 2020-03-29 0zOs6ZS9       11
# 2 2020-03-29 3jtMiD          2
# 3 2020-03-30 0zOs6ZS9        0
# 4 2020-03-30 3jtMiD         12
# 5 2020-03-31 0zOs6ZS9        6
# 6 2020-03-31 3jtMiD          1
# 7 2020-04-01 0zOs6ZS9       11
# 8 2020-04-01 3jtMiD          0
# 9 2020-04-02 0zOs6ZS9        0
#10 2020-04-02 3jtMiD          2

df2 <- structure(list(date = structure(c(18350, 18350, 18351, 18352, 
18352, 18353, 18354), class = "Date"), uid = c("0zOs6ZS9", "3jtMiD", 
"3jtMiD", "0zOs6ZS9", "3jtMiD", "0zOs6ZS9", "3jtMiD"), duration = c(11L, 
2L, 12L, 6L, 1L, 11L, 2L)), row.names = c("1", "2", "3", "4", 
"5", "6", "7"), class = "data.frame")