累积R中每个可能组合的值

Question

假设我有数据test(给定dput(，其中列表列为items:

test <- structure(list(items = list('a', c('b', 'c'), c('d', 'e'), 'f', c('g', 'h')),
ID = c(1,1,1,2,2)), row.names = c(NA, 5L), class = "data.frame")
library(tidyverse)
test %>% group_by(ID) %>%
mutate(dummy = accumulate(items, ~paste(.x, .y)))

我得到了一个类似的列表列输出

items ID        dummy
1     a  1            a
2  b, c  1     a b, a c
3  d, e  1 a b d, a c e
4     f  2            f
5  g, h  2     f g, f h

我希望第3行中有四个项目，每个项目都有可能的组合，即c("a b d", "a b e", "a c d", "a c e")。然而，这些是否是列表中的单独项目并不重要。换句话说，dummy的输出可以是多级列表类型，其中row3将包含列表中的四个项目。我试过使用expand.grid，但我在某个地方出错了！

所以我想要的输出看起来像

items ID                      dummy
1     a  1                          a
2  b, c  1                   a b, a c
3  d, e  1 a b d, a c d, a b e, a c e
4     f  2                          f
5  g, h  2                   f g, f h

Answer 1

使用Reduce+outer的data.table选项

setDT(test)[
,
dummy := .(Reduce(function(x, y) outer(x, y, paste),
items,
accumulate = TRUE
)),
ID
]

给出

> test
items ID                   dummy
1:     a  1                       a
2:   b,c  1                 a b,a c
3:   d,e  1 a b d,a c d,a b e,a c e
4:     f  2                       f
5:   g,h  2                 f g,f h

Answer 2

expand.grid()的另一种方法，

test %>% group_by(ID) %>%
mutate(dummy = accumulate(items, ~do.call("paste",expand.grid(.x, .y)))) %>% 
data.frame()

给予，

items ID                      dummy
1     a  1                          a
2  b, c  1                   a b, a c
3  d, e  1 a b d, a c d, a b e, a c e
4     f  2                          f
5  g, h  2                   f g, f h

Answer 3

如果您想要所有可能的组合，请使用sapply而不是.x

library(dplyr)
library(purrr)
test %>% 
group_by(ID) %>%
mutate(dummy = accumulate(items, ~c(sapply(.x, paste, .y)))) %>%
pull(dummy)
#[[1]]
#[1] "a"
#[[2]]
#[1] "a b" "a c"
#[[3]]
#[1] "a b d" "a b e" "a c d" "a c e"
#[[4]]
#[1] "f"
#[[5]]
#[1] "f g" "f h"

Answer 4

您可以使用外积来粘贴两个向量。。。

test2 <- test %>% group_by(ID) %>%
mutate(dummy = accumulate(items, ~outer(.x, .y, paste)))
str(test2)
grouped_df[,3] [5 x 3] (S3: grouped_df/tbl_df/tbl/data.frame)
$ items:List of 5
..$ : chr "a"
..$ : chr [1:2] "b" "c"
..$ : chr [1:2] "d" "e"
..$ : chr "f"
..$ : chr [1:2] "g" "h"
$ ID   : num [1:5] 1 1 1 2 2
$ dummy:List of 5
..$ : chr "a"
..$ : chr [1, 1:2] "a b" "a c"
..$ : chr [1, 1:2, 1:2] "a b d" "a c d" "a b e" "a c e"
..$ : chr "f"
..$ : chr [1, 1:2] "f g" "f h"

Answer 5

purrr包中还有cross和cross2：

library(tidyverse)
test %>% 
group_by(ID) %>% 
mutate(
dummy = accumulate(items, cross2) %>% map_depth(unlist, .depth = 2)
) %>% 
pull(dummy) %>% 
str()

#> List of 5
#>  $ :List of 1
#>   ..$ : chr "a"
#>  $ :List of 2
#>   ..$ : chr [1:2] "a" "b"
#>   ..$ : chr [1:2] "a" "c"
#>  $ :List of 4
#>   ..$ : chr [1:3] "a" "b" "d"
#>   ..$ : chr [1:3] "a" "c" "d"
#>   ..$ : chr [1:3] "a" "b" "e"
#>   ..$ : chr [1:3] "a" "c" "e"
#>  $ :List of 1
#>   ..$ : chr "f"
#>  $ :List of 2
#>   ..$ : chr [1:2] "f" "g"
#>   ..$ : chr [1:2] "f" "h"

^{创建于2021-05-18由reprex包(v1.0.0(}

Answer 6

也可以使用此解决方案：

library(dplyr)
library(purrr)
test %>%
group_by(ID) %>%
mutate(comb = accumulate(items[-1], .init = unlist(items[1]), 
~ expand.grid(.x, .y) %>% 
{map2(.$Var1, .$Var2, ~ paste(.x, .y, sep = " "))} %>%
unlist())) %>%
as.data.frame()
items ID                       comb
1     a  1                          a
2  b, c  1                   a b, a c
3  d, e  1 a b d, a c d, a b e, a c e
4     f  2                          f
5  g, h  2                   f g, f h