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我有 6 列的数据框，我想将它们分成 3 部分，每部分有 2 列。我想以所有可能的组合重新组合分区以创建 7 个新数据帧

part1,part2,part3
part1,par2
part1,part3
part2,part3
part1
part2
part3

我稍微修改了这个解决方案以重新组合它们将数据帧拆分为所有可能的数据帧组合，在 R 中按 3 列

>frame <- data.frame(id = letters[seq( from = 1, to = 10 )], a = rnorm(10, 4), b = rnorm(10, 6), c=rnorm(10, 5), d = rnorm(10, 2),e=rnorm(10, 5), f = rnorm(10, 2))

> frame
   id        a        b        c         d        e          f
1   a 6.322845 5.828619 5.465636 2.7658092 6.522706  1.4896078
2   b 2.352437 5.521230 6.555715 0.6612871 5.288508  2.4837969
3   c 2.790967 9.253197 3.724231 2.9954273 4.887744  1.3020424
4   d 2.017975 6.038846 4.540511 1.7989492 6.059974 -0.2463154
5   e 4.004463 4.384898 5.341084 1.9528288 4.186449  1.0823939
6   f 2.600336 6.562758 5.708489 2.1142707 6.769220  1.7942291
7   g 3.850400 7.231973 4.918542 3.3562489 6.090841  1.4202527
8   h 2.932744 6.377516 5.518261 1.7423230 4.422915  1.8789437
9   i 5.135185 5.218992 4.710196 1.1878825 5.421876  0.8455756
10  j 5.188278 7.233590 6.303500 0.3868047 4.390973  1.6997801 

>m <- seq(3) 
>j <-function(m){lapply(as.data.frame(combn(ncol(frame) - 1, m)), function(idx) frame[, c(1, idx + 1)])}

>lapply(m, function(m) j(m))

这将通过改组所有列来创建所有组合。我不想要列的组合，而是分区的组合。我怎样才能做到这一点？

4

2 回答 2

1

这是一个尝试：

library(dplyr)
library(purrr)

# Assign a partition to be used here
# (Updated from OP's clarification about pttns & @bouncyball's comment)
pttn <- split(names(frame)[-1], rep(1:3, each = 2))

# Create combinations of partitioned columns
do.call(c, lapply(seq_along(pttn), combn, x = pttn, simplify = FALSE)) %>% 
   map(~ frame %>% select(reduce(.x, c)))

do.call 的第一行创建了“分区”或分区列名的所有组合。如果要保留ID列，可以使用id, reduce(.x, c)而不是reduce(.x, c)

于 2021-03-29T17:48:26.083 回答
0

一种可能的解决方案，使用purrr::map()和一些数据争吵到长/宽。可能不是最有效或最优雅的解决方案，但它确实发挥了作用。

library(tidyverse)

# sample data
frame <- data.frame(
  id = letters[seq( from = 1, to = 10 )], a = rnorm(10, 4), b = rnorm(10, 6), c=rnorm(10, 5), 
  d = rnorm(10, 2),e=rnorm(10, 5), f = rnorm(10, 2))

# list of possible combinations 
list_of_combinations <- list(
  c(1), 
  c(2),
  c(3),
  c(1,2),
  c(1,3),
  c(2,3),
  c(1,2,3)
)

# data in long format and a category variable (for each "chunk")
df_long <- frame %>% pivot_longer(-id) %>% 
  mutate(
    cat = case_when(
      (name %in% c("a", "b")) ~ 1L, 
      (name %in% c("c", "d")) ~ 2L, 
      (name %in% c("e", "f")) ~ 3L)
  )
df_long
#> # A tibble: 60 x 4
#>    id    name   value   cat
#>    <chr> <chr>  <dbl> <int>
#>  1 a     a      3.93      1
#>  2 a     b      4.66      1
#>  3 a     c      2.78      2
#>  4 a     d      2.35      2
#>  5 a     e      5.93      3
#>  6 a     f     -0.500     3
#>  7 b     a      5.11      1
#>  8 b     b      5.37      1
#>  9 b     c      4.61      2
#> 10 b     d      3.58      2
#> # … with 50 more rows

# map list to generate a list of each combination and then map it back into wide format 
final_list_of_dfs <- list_of_combinations %>% map( ~ df_long %>% filter(cat %in% .x)) %>% 
  map(~ .x %>% select(-cat) %>% pivot_wider(names_from = "name", values_from = "value"))
glimpse(final_list_of_dfs)
#> List of 7
#>  $ : tibble [10 × 3] (S3: tbl_df/tbl/data.frame)
#>   ..$ id: chr [1:10] "a" "b" "c" "d" ...
#>   ..$ a : num [1:10] 3.93 5.11 4.16 2.59 2.69 ...
#>   ..$ b : num [1:10] 4.66 5.37 3.26 5.52 6.29 ...
#>  $ : tibble [10 × 3] (S3: tbl_df/tbl/data.frame)
#>   ..$ id: chr [1:10] "a" "b" "c" "d" ...
#>   ..$ c : num [1:10] 2.78 4.61 3.8 3.06 4.68 ...
#>   ..$ d : num [1:10] 2.353 3.579 0.744 1.582 3.377 ...
#>  $ : tibble [10 × 3] (S3: tbl_df/tbl/data.frame)
#>   ..$ id: chr [1:10] "a" "b" "c" "d" ...
#>   ..$ e : num [1:10] 5.93 3.89 5.43 3.88 5.51 ...
#>   ..$ f : num [1:10] -0.5 0.941 3.703 2.035 0.611 ...
#>  $ : tibble [10 × 5] (S3: tbl_df/tbl/data.frame)
#>   ..$ id: chr [1:10] "a" "b" "c" "d" ...
#>   ..$ a : num [1:10] 3.93 5.11 4.16 2.59 2.69 ...
#>   ..$ b : num [1:10] 4.66 5.37 3.26 5.52 6.29 ...
#>   ..$ c : num [1:10] 2.78 4.61 3.8 3.06 4.68 ...
#>   ..$ d : num [1:10] 2.353 3.579 0.744 1.582 3.377 ...
#>  $ : tibble [10 × 5] (S3: tbl_df/tbl/data.frame)
#>   ..$ id: chr [1:10] "a" "b" "c" "d" ...
#>   ..$ a : num [1:10] 3.93 5.11 4.16 2.59 2.69 ...
#>   ..$ b : num [1:10] 4.66 5.37 3.26 5.52 6.29 ...
#>   ..$ e : num [1:10] 5.93 3.89 5.43 3.88 5.51 ...
#>   ..$ f : num [1:10] -0.5 0.941 3.703 2.035 0.611 ...
#>  $ : tibble [10 × 5] (S3: tbl_df/tbl/data.frame)
#>   ..$ id: chr [1:10] "a" "b" "c" "d" ...
#>   ..$ c : num [1:10] 2.78 4.61 3.8 3.06 4.68 ...
#>   ..$ d : num [1:10] 2.353 3.579 0.744 1.582 3.377 ...
#>   ..$ e : num [1:10] 5.93 3.89 5.43 3.88 5.51 ...
#>   ..$ f : num [1:10] -0.5 0.941 3.703 2.035 0.611 ...
#>  $ : tibble [10 × 7] (S3: tbl_df/tbl/data.frame)
#>   ..$ id: chr [1:10] "a" "b" "c" "d" ...
#>   ..$ a : num [1:10] 3.93 5.11 4.16 2.59 2.69 ...
#>   ..$ b : num [1:10] 4.66 5.37 3.26 5.52 6.29 ...
#>   ..$ c : num [1:10] 2.78 4.61 3.8 3.06 4.68 ...
#>   ..$ d : num [1:10] 2.353 3.579 0.744 1.582 3.377 ...
#>   ..$ e : num [1:10] 5.93 3.89 5.43 3.88 5.51 ...
#>   ..$ f : num [1:10] -0.5 0.941 3.703 2.035 0.611 ...

由reprex 包于 2021-03-29 创建(v1.0.0)

于 2021-03-29T18:10:03.470 回答