ℕʘʘ*_*ḆḽḘ 0 r dplyr data.table tidyr
我有以下数据帧
one <- c('one',NA,NA,NA,NA,'two',NA,NA)
group1 <- c('A','A','A','A','B','B','B','B')
group2 <- c('C','C','C','D','E','E','F','F')
df = data.frame(one, group1,group2)
> df
one group1 group2
1 one A C
2 <NA> A C
3 <NA> A C
4 <NA> A D
5 <NA> B E
6 two B E
7 <NA> B F
8 <NA> B F
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我想获得的非缺失观测值的计数one的每个组合group1和group2.
在Pandas,我会用groupby(['group1','group2']).transform,但我怎么能在R?原始数据帧很大.
预期产出是:
> df
one group1 group2 count
1 one A C 1
2 <NA> A C 1
3 <NA> A C 1
4 <NA> A D 0
5 <NA> B E 1
6 two B E 1
7 <NA> B F 0
8 <NA> B F 0
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非常感谢!
library(dplyr)
df %>% group_by(group1, group2) %>% mutate(count = sum(!is.na(one)))
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Run Code Online (Sandbox Code Playgroud)Source: local data frame [8 x 4] Groups: group1, group2 [4] one group1 group2 count <fctr> <fctr> <fctr> <int> 1 one A C 1 2 NA A C 1 3 NA A C 1 4 NA A D 0 5 NA B E 1 6 two B E 1 7 NA B F 0 8 NA B F 0
用data.table:
setDT(df)
df[,count_B:=sum(!is.na(one)),by=c("group1","group2")]
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得到:
one group1 group2 count_B
1: one A C 1
2: NA A C 1
3: NA A C 1
4: NA A D 0
5: NA B E 1
6: two B E 1
7: NA B F 0
8: NA B F 0
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我们的想法是将真值(1转换为整数)相加,其中B不是NA在分组时group1和group2.