Question

我在下面给出了这个数据帧输出：

lf3 = structure(list(session_id = c(1L, 1L, 1L, 2L, 3L, 5L, 5L, 6L, 
6L, 7L), userId = c(1, 1, 1, 2, 2, 4, 4, 5, 5, 5), datetime = 
structure(c(1457029336, 
1457029337, 1457029340, 1457029596, 1457313569, 1457030783, 1457030784, 
1457030918, 1457030920, 1457370365), class = c("POSIXct", "POSIXt"
), tzone = "UTC"), referer = c(22, 2, 7, 5, 23, 20, 7, 24, 18, 
22), request = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 5)), .Names = c("session_id", 
"userId", "datetime", "referer", "request"), row.names = c(NA, 
10L), class = "data.frame")

现在我想退出那些具有最低指定标准/值的会话。我试试这段代码：

lf3 %>% group_by(session_id) %>% tally(sort = TRUE) %>% filter(n>2)

但是我希望返回相同的数据帧，只有会话通过这个条件，如下所示：

  session_id userId            datetime referer request
1          1      1 2016-03-03 18:22:16      22       1
2          1      1 2016-03-03 18:22:17       2       2
3          1      1 2016-03-03 18:22:20       7       3

如何使用

Answer 1

您可能需要group_by %>% filter：

lf3 %>% group_by(session_id) %>% filter(n() > 2)

# A tibble: 3 x 5
# Groups:   session_id [1]
#  session_id userId            datetime referer request
#       <int>  <dbl>              <dttm>   <dbl>   <dbl>
#1          1      1 2016-03-03 18:22:16      22       1
#2          1      1 2016-03-03 18:22:17       2       2
#3          1      1 2016-03-03 18:22:20       7       3

Answer 2

我们可以使用data.table

library(data.table)
setDT(lf3)[, if(.N >2) .SD, session_id]
#      session_id userId            datetime referer request
#1:          1      1 2016-03-03 18:22:16      22       1
#2:          1      1 2016-03-03 18:22:17       2       2
#3:          1      1 2016-03-03 18:22:20       7       3

删除不满足给定条件的行

2 个答案: