通过按多个分组变量a:c对数据帧进行分组，我想将其虚拟划分为子集。之后我想添加两列，一列包含子集大小的计数，另一列是子集的ID。
set.seed(67)
n <- 1000
df1 <- data.frame(
a=rbinom(n, 1, .5),
b=sample(20:40, n, replace = TRUE),
c=sample(seq(3000, 4000, 100), n, replace = TRUE),
d=rbinom(n, 1, .13),
k=rbinom(n, 1, .88),
l=rbinom(n, 1, .075),
m=rbinom(n, 1, .05),
n=rbinom(n, 1, .3)
)
> head(df1)
a b c d k l m n
1 1 21 3900 0 1 0 0 0
2 0 26 3600 0 1 0 0 0
3 0 23 3900 0 1 0 0 0
4 1 23 3900 0 1 0 0 0
5 0 32 4000 1 1 0 0 0
6 1 23 3200 0 0 0 0 0
我已经让小组计数正确，但我需要保留其他变量。
> with(df1, aggregate(d, list(a, b, c), length))
Group.1 Group.2 Group.3 x
1 0 20 3000 2
2 1 20 3000 3
3 0 21 3000 2
4 1 21 3000 3
5 0 22 3000 3
6 1 22 3000 1
...
当我将整个数据框定义为对象时，它还会显示计数但值会被覆盖：
> with(df1, aggregate(df1, list(a, b, c), length))
Group.1 Group.2 Group.3 a b c d k l m n
1 0 20 3000 2 2 2 2 2 2 2 2
2 1 20 3000 3 3 3 3 3 3 3 3
3 0 21 3000 2 2 2 2 2 2 2 2
4 1 21 3000 3 3 3 3 3 3 3 3
5 0 22 3000 3 3 3 3 3 3 3 3
6 1 22 3000 1 1 1 1 1 1 1 1
...
其实我想要这样的东西：
a b c d k l m n count id
847 0 20 3000 1 1 0 0 1 2 1
939 0 20 3000 0 0 0 0 0 2 1
264 1 21 3000 0 1 0 0 0 3 2
569 1 21 3000 0 1 0 0 0 3 2
876 1 21 3000 0 1 0 0 1 3 2
346 0 22 3000 0 1 0 0 1 3 3
846 0 22 3000 0 1 0 0 0 3 3
929 0 22 3000 0 1 0 0 1 3 3
...
我该怎么做？
答案 0 :(得分：2)
在基地R中，您可以使用ave ...
df1 <- df1[order(df1$c,df1$b,df1$a),]
df1$id <- cumsum(!duplicated(df1[,c("a","b","c")]))
df1$count <- ave(df1$a,df1$id,FUN=length)
head(df1)
a b c d k l m n id count
847 0 20 3000 1 1 0 0 1 1 2
939 0 20 3000 0 0 0 0 0 1 2
217 1 20 3000 0 1 0 0 0 2 3
458 1 20 3000 0 1 0 0 0 2 3
631 1 20 3000 0 1 0 0 0 2 3
360 0 21 3000 0 1 1 0 0 3 2
df的顺序会影响你得到的id值，但希望这并不重要。
答案 1 :(得分：1)
这是你之后的事吗？
library(tidyverse);
df1 %>%
group_by(a, b, c) %>%
mutate(count = n()) %>%
ungroup() %>%
arrange(b, c, a) %>%
mutate(id = cumsum(!duplicated(paste0(a, b, c))));
## A tibble: 1,000 x 10
# a b c d k l m n count id
# <int> <int> <dbl> <int> <int> <int> <int> <int> <int> <int>
# 1 0 20 3000. 1 1 0 0 1 2 1
# 2 0 20 3000. 0 0 0 0 0 2 1
# 3 1 20 3000. 0 1 0 0 0 3 2
# 4 1 20 3000. 0 1 0 0 0 3 2
# 5 1 20 3000. 0 1 0 0 0 3 2
# 6 0 20 3100. 0 1 0 0 0 2 3
# 7 0 20 3100. 0 1 1 0 0 2 3
# 8 1 20 3100. 0 1 0 0 0 1 4
# 9 0 20 3200. 1 1 0 0 0 3 5
#10 0 20 3200. 0 1 0 0 0 3 5
## ... with 990 more rows
答案 2 :(得分：1)
以下是使用group_indices
library(tidyverse)
df1 %>%
mutate(id = group_indices(., a,b,c)) %>% #extract the group indices when grouped by a, b and c
group_by(a, b, c) %>% #group by a, b and c
mutate(count = n()) %>% #get the number of elements in each group
arrange(a, b, c) #arrange by a, b, c or how ever you preffer
#output
# A tibble: 1,000 x 10
# Groups: a, b, c [414]
a b c d k l m n id count
<int> <int> <dbl> <int> <int> <int> <int> <int> <int> <int>
1 0 20 3000 1 1 0 0 1 1 2
2 0 20 3000 0 0 0 0 0 1 2
3 0 20 3100 0 1 0 0 0 2 2
4 0 20 3100 0 1 1 0 0 2 2
5 0 20 3200 1 1 0 0 0 3 3
6 0 20 3200 0 1 0 0 0 3 3
7 0 20 3200 0 1 0 0 0 3 3
8 0 20 3300 1 1 0 0 1 4 2
9 0 20 3300 0 1 0 0 0 4 2
10 0 20 3400 0 1 0 0 1 5 1
# ... with 990 more rows
答案 3 :(得分：1)
使用data.table，可以使用内置的.GRP和.N变量在几行中完成。
setDT(df1)
df1 <- df1[order(c,b,a)]
df1[,':='(count = .N, id = .GRP),.(a,b,c)]
print(head(df1))
a b c d k l m n count id
1: 0 20 3000 1 1 0 0 1 2 1
2: 0 20 3000 0 0 0 0 0 2 1
3: 1 20 3000 0 1 0 0 0 3 2
4: 1 20 3000 0 1 0 0 0 3 2
5: 1 20 3000 0 1 0 0 0 3 2
6: 0 21 3000 0 1 1 0 0 2 3