我有一个很大的数据集，看起来像这样简化:
row. member_id entry_id comment_count timestamp
1 1 a 4 2008-06-09 12:41:00
2 1 b 1 2008-07-14 18:41:00
3 1 c 3 2008-07-17 15:40:00
4 2 d 12 2008-06-09 12:41:00
5 2 e 50 2008-09-18 10:22:00
6 3 f 0 2008-10-03 13:36:00
我可以使用以下代码汇总计数:
transform(df, aggregated_count = ave(comment_count, member_id, FUN = cumsum))
但是我希望累积数据中的滞后时间为1，或者我希望
cumsum忽略当前行。结果应为:row. member_id entry_id comment_count timestamp previous_comments
1 1 a 4 2008-06-09 12:41:00 0
2 1 b 1 2008-07-14 18:41:00 4
3 1 c 3 2008-07-17 15:40:00 5
4 2 d 12 2008-06-09 12:41:00 0
5 2 e 50 2008-09-18 10:22:00 12
6 3 f 0 2008-10-03 13:36:00 0
知道如何在R中执行此操作吗？甚至滞后时间大于1？
再现性数据:
# dput(df)
structure(list(member_id = c(1L, 1L, 1L, 2L, 2L, 3L), entry_id = c("a",
"b", "c", "d", "e", "f"), comment_count = c(4L, 1L, 3L, 12L,
50L, 0L), timestamp = c("2008-06-09 12:41:00", "2008-07-14 18:41:00",
"2008-07-17 15:40:00", "2008-06-09 12:41:00", "2008-09-18 10:22:00",
"2008-10-03 13:36:00")), .Names = c("member_id", "entry_id",
"comment_count", "timestamp"), row.names = c("1", "2", "3", "4",
"5", "6"), class = "data.frame")
最佳答案
您可以将0用作第一个元素，并使用head(, -1)删除最后一个元素
transform(df, previous_comments=ave(comment_count, member_id,
FUN = function(x) cumsum(c(0, head(x, -1)))))
# member_id entry_id comment_count timestamp previous_comments
#1 1 a 4 2008-06-09 12:41:00 0
#2 1 b 1 2008-07-14 18:41:00 4
#3 1 c 3 2008-07-17 15:40:00 5
#4 2 d 12 2008-06-09 12:41:00 0
#5 2 e 50 2008-09-18 10:22:00 12
#6 3 f 0 2008-10-03 13:36:00 0
关于r - 累计和滞后，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/27649206/