当我浏览答案here时，我发现this solution与 data.frame 完全一样 .
library(dplyr) # dplyr_0.4.3
library(data.table) # data.table_1.9.5
df <- structure(list(id = c(1L, 1L, 1L, 2L, 2L, 3L, 3L, 3L, 3L),
a = c("AA",
"AB", "AA", "AB", "AB", "AB", "AB", "AA", "AA"), b = c(2L, 5L,
1L, 2L, 4L, 4L, 3L, 1L, 4L)), .Names = c("id", "a", "b"),
class = "data.frame", row.names = c(NA, -9L))
df %>%
group_by(id) %>%
mutate(relevance=+(a!='AA')) %>%
mutate(mean=cumsum(relevance * b) / cumsum(relevance))
Source: local data frame [9 x 5]
Groups: id [3]
id a b relevance mean
(int) (chr) (int) (int) (dbl)
1 1 AA 2 0 NaN
2 1 AB 5 1 5.0
3 1 AA 1 0 5.0
4 2 AB 2 1 2.0
5 2 AB 4 1 3.0
6 3 AB 4 1 4.0
7 3 AB 3 1 3.5
8 3 AA 1 0 3.5
9 3 AA 4 0 3.5
但是当使用 data.table 运行时，它会导致我无法理解的东西 .
setDT(df) %>%
group_by(id) %>%
mutate(relevance=+(a!='AA')) %>%
mutate(mean=cumsum(relevance * b) / cumsum(relevance))
Source: local data table [9 x 5]
id a b relevance mean
(int) (chr) (int) (int) (dbl)
1 1 AA 2 0 NaN
2 1 AB 5 1 5.000000
3 1 AA 1 0 5.000000
4 2 AB 2 1 3.500000
5 2 AB 4 1 3.666667
6 3 AB 4 1 3.750000
7 3 AB 3 1 3.600000
8 3 AA 1 0 3.600000
9 3 AA 4 0 3.600000
这是预期的行为吗？如果是的话，是否有任何关于何时不使用 data.table 后端与 dplyr 的指南？
1 回答
导致在
mutate之后在data.table was resolved in 0.5.0上删除分组的错误 .