辍学行我有此数据帧dput下面给出:其不会满足给定条件
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")
现在我想退出具有最小指定的标准/值的那些会话。 我试试这个代码:
lf3 %>% group_by(session_id) %>% tally(sort = TRUE) %>% filter(n>2)
但我想退货同一数据框,只有会议通过此条件下,象下面这样:
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
如何去与
用预期输出更新您的问题。 –
因此,它只会给出频率大于2的session_id = 1行。所需输出将如下所示:'structure(list(session_id = c(1L,1L,1L),userId = c(1,1,1 (22,2,7),请求= 0(),datetime =结构(c(1457029336,1457029337,1457029340),class = c(“POSIXct”, “POSIXt”),tzone =“UTC” c(1, 2,3)),.Names = c(“session_id”,“userId”,“datetime”,“referer”, “request”),row.names = c(NA,3L) =“data.frame”)' – SumitArya
我更喜欢base R,'ave','lf3 [ave(lf3 $ userId,lf3 $ session_id,FUN = length)> 2,]' –