在 R 中找到 vector1 的 n1 个元素和 vector2 的 n2 个元素的所有组合？
[英]Find all combinations of n1 elements from vector1 and n2 elements from vector2 in R?
问题描述
I have two vectors and I am trying to find all unique combinations of 3 elements from vector1 and 2 elements from vector2.
我有两个向量，我试图找到 vector1 中的 3 个元素和 vector2 中的 2 个元素的所有唯一组合。 I have tried the following code.我尝试了以下代码。
V1 = combn(1:5, 3) # 10 combinations in total
V2 = combn(6:11, 2) # 15 combinations in total
How to combine V1 and V2 so that there are 10 * 15 = 150 combinations in total?如何组合V1和V2，使得总共有10 * 15 = 150种组合？ Thanks.谢谢。
5 个解决方案
解决方案1
6 2022-12-02 21:47:25
The function comboGrid from RcppAlgos (I am the author) does just the trick:来自RcppAlgos （我是作者）的 function comboGrid就可以解决问题：
library(RcppAlgos)
grid <- comboGrid(c(rep(list(1:5), 3), rep(list(6:11), 2)),
repetition = FALSE)
head(grid)
#> Var1 Var2 Var3 Var4 Var5
#> [1,] 1 2 3 6 7
#> [2,] 1 2 3 6 8
#> [3,] 1 2 3 6 9
#> [4,] 1 2 3 6 10
#> [5,] 1 2 3 6 11
#> [6,] 1 2 3 7 8
tail(grid)
#> Var1 Var2 Var3 Var4 Var5
#> [145,] 3 4 5 8 9
#> [146,] 3 4 5 8 10
#> [147,] 3 4 5 8 11
#> [148,] 3 4 5 9 10
#> [149,] 3 4 5 9 11
#> [150,] 3 4 5 10 11
It is quite efficient as well.它也非常有效。 It is written in C++ and pulls together many ideas from the excellent question: Picking unordered combinations from pools with overlap .它写在C++中，汇集了来自优秀问题的许多想法： Picking unordered combinations from pools with overlap 。 The underlying algorithm avoids generating duplicates that would need to be filtered out.底层算法避免生成需要过滤掉的重复项。
Consider the following example where generating the Cartesian product contains more than 10 billion results:考虑以下示例，其中生成的笛卡尔积包含超过 100 亿个结果：
system.time(huge <- comboGrid(c(rep(list(1:20), 5), rep(list(21:35), 3)),
repetition = FALSE))
#> user system elapsed
#> 0.990 0.087 1.077
dim(huge)
#> [1] 7054320 8
解决方案2
6 2022-12-02 22:13:56
You can try expand.grid along with asplit , eg,您可以尝试expand.grid与asplit一起使用，例如，
expand.grid(asplit(V1,2), asplit(V2,2))
or要么
with(
expand.grid(asplit(V1, 2), asplit(V2, 2)),
t(mapply(c, Var1, Var2))
)
解决方案3
3 已采纳 2022-12-02 21:28:48
You can use expand.grid() :您可以使用expand.grid() ：
g <- expand.grid(seq_len(ncol(V1)), seq_len(ncol(V2)))
V3 <- rbind(V1[, g[, 1]], V2[, g[, 2]])
The result is in a similar format as V1 and V2 , ie a 5 × 150 matrix (here printed transposed):结果的格式与V1和V2类似，即 5 × 150 矩阵（此处打印转置）：
head(t(V3))
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 2 3 6 7
# [2,] 1 2 4 6 7
# [3,] 1 2 5 6 7
# [4,] 1 3 4 6 7
# [5,] 1 3 5 6 7
# [6,] 1 4 5 6 7
dim(unique(t(V3)))
# [1] 150 5
And a generalized approach that can handle more than two initial matrices of combinations, stored in a list V :还有一种通用方法，可以处理两个以上的初始组合矩阵，存储在列表V中：
V <- list(V1, V2)
g <- do.call(expand.grid, lapply(V, \(x) seq_len(ncol(x))))
V.comb <- do.call(rbind, mapply('[', V, T, g))
identical(V.comb, V3)
[1] TRUE
解决方案4
3 2022-12-02 21:35:07
After some helpful refactoring guidance from @onyambu, here is a shorter solution based on base::merge() :在@onyambu 提供了一些有用的重构指导之后，这里有一个基于base::merge()的更短的解决方案：
merge(t(combn(1:5, 3)),t(combn(6:11, 2)),by.x=NULL,by.y = NULL)
...and the first 20 rows of output: ...以及 output 的前 20 行：
> merge(t(combn(1:5, 3)),t(combn(6:11, 2)),by.x=NULL,by.y = NULL)
V1.x V2.x V3 V1.y V2.y
1 1 2 3 6 7
2 1 2 4 6 7
3 1 2 5 6 7
4 1 3 4 6 7
5 1 3 5 6 7
6 1 4 5 6 7
7 2 3 4 6 7
8 2 3 5 6 7
9 2 4 5 6 7
10 3 4 5 6 7
11 1 2 3 6 8
12 1 2 4 6 8
13 1 2 5 6 8
14 1 3 4 6 8
15 1 3 5 6 8
16 1 4 5 6 8
17 2 3 4 6 8
18 2 3 5 6 8
19 2 4 5 6 8
20 3 4 5 6 8
original solution原液
A base R solution to create a Cartesian product with merge() looks like this:使用merge()创建笛卡尔积的基本 R 解决方案如下所示：
df1 <- data.frame(t(combn(1:5, 3)))
df2 <- data.frame(t(combn(6:11, 2)))
colnames(df2) <- paste("y",1:2,sep=""))
merge(df1,df2,by.x=NULL,by.y = NULL)
...and the first 25 rows of output: ...以及 output 的前 25 行：
> merge(df1,df2,by.x=NULL,by.y = NULL)
X1 X2 X3 y1 y2
1 1 2 3 6 7
2 1 2 4 6 7
3 1 2 5 6 7
4 1 3 4 6 7
5 1 3 5 6 7
6 1 4 5 6 7
7 2 3 4 6 7
8 2 3 5 6 7
9 2 4 5 6 7
10 3 4 5 6 7
11 1 2 3 6 8
12 1 2 4 6 8
13 1 2 5 6 8
14 1 3 4 6 8
15 1 3 5 6 8
16 1 4 5 6 8
17 2 3 4 6 8
18 2 3 5 6 8
19 2 4 5 6 8
20 3 4 5 6 8
21 1 2 3 6 9
22 1 2 4 6 9
23 1 2 5 6 9
24 1 3 4 6 9
25 1 3 5 6 9
解决方案5
2 2022-12-02 21:33:10
Similar idea, using apply类似的想法，使用apply
apply(expand.grid(seq(ncol(V1)), seq(ncol(V2))), 1, function(i) {
c(V1[,i[1]], V2[,i[2]])})
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
#> [1,] 1 1 1 1 1 1 2 2 2 3 1 1 1 1
#> [2,] 2 2 2 3 3 4 3 3 4 4 2 2 2 3
#> [3,] 3 4 5 4 5 5 4 5 5 5 3 4 5 4
#> [4,] 6 6 6 6 6 6 6 6 6 6 6 6 6 6
#> [5,] 7 7 7 7 7 7 7 7 7 7 8 8 8 8
#> [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26]
#> [1,] 1 1 2 2 2 3 1 1 1 1 1 1
#> [2,] 3 4 3 3 4 4 2 2 2 3 3 4
#> [3,] 5 5 4 5 5 5 3 4 5 4 5 5
#> [4,] 6 6 6 6 6 6 6 6 6 6 6 6
#> [5,] 8 8 8 8 8 8 9 9 9 9 9 9
#> [,27] [,28] [,29] [,30] [,31] [,32] [,33] [,34] [,35] [,36] [,37] [,38]
#> [1,] 2 2 2 3 1 1 1 1 1 1 2 2
#> [2,] 3 3 4 4 2 2 2 3 3 4 3 3
#> [3,] 4 5 5 5 3 4 5 4 5 5 4 5
#> [4,] 6 6 6 6 6 6 6 6 6 6 6 6
#> [5,] 9 9 9 9 10 10 10 10 10 10 10 10
#> [,39] [,40] [,41] [,42] [,43] [,44] [,45] [,46] [,47] [,48] [,49] [,50]
#> [1,] 2 3 1 1 1 1 1 1 2 2 2 3
#> [2,] 4 4 2 2 2 3 3 4 3 3 4 4
#> [3,] 5 5 3 4 5 4 5 5 4 5 5 5
#> [4,] 6 6 6 6 6 6 6 6 6 6 6 6
#> [5,] 10 10 11 11 11 11 11 11 11 11 11 11
#> [,51] [,52] [,53] [,54] [,55] [,56] [,57] [,58] [,59] [,60] [,61] [,62]
#> [1,] 1 1 1 1 1 1 2 2 2 3 1 1
#> [2,] 2 2 2 3 3 4 3 3 4 4 2 2
#> [3,] 3 4 5 4 5 5 4 5 5 5 3 4
#> [4,] 7 7 7 7 7 7 7 7 7 7 7 7
#> [5,] 8 8 8 8 8 8 8 8 8 8 9 9
#> [,63] [,64] [,65] [,66] [,67] [,68] [,69] [,70] [,71] [,72] [,73] [,74]
#> [1,] 1 1 1 1 2 2 2 3 1 1 1 1
#> [2,] 2 3 3 4 3 3 4 4 2 2 2 3
#> [3,] 5 4 5 5 4 5 5 5 3 4 5 4
#> [4,] 7 7 7 7 7 7 7 7 7 7 7 7
#> [5,] 9 9 9 9 9 9 9 9 10 10 10 10
#> [,75] [,76] [,77] [,78] [,79] [,80] [,81] [,82] [,83] [,84] [,85] [,86]
#> [1,] 1 1 2 2 2 3 1 1 1 1 1 1
#> [2,] 3 4 3 3 4 4 2 2 2 3 3 4
#> [3,] 5 5 4 5 5 5 3 4 5 4 5 5
#> [4,] 7 7 7 7 7 7 7 7 7 7 7 7
#> [5,] 10 10 10 10 10 10 11 11 11 11 11 11
#> [,87] [,88] [,89] [,90] [,91] [,92] [,93] [,94] [,95] [,96] [,97] [,98]
#> [1,] 2 2 2 3 1 1 1 1 1 1 2 2
#> [2,] 3 3 4 4 2 2 2 3 3 4 3 3
#> [3,] 4 5 5 5 3 4 5 4 5 5 4 5
#> [4,] 7 7 7 7 8 8 8 8 8 8 8 8
#> [5,] 11 11 11 11 9 9 9 9 9 9 9 9
#> [,99] [,100] [,101] [,102] [,103] [,104] [,105] [,106] [,107] [,108]
#> [1,] 2 3 1 1 1 1 1 1 2 2
#> [2,] 4 4 2 2 2 3 3 4 3 3
#> [3,] 5 5 3 4 5 4 5 5 4 5
#> [4,] 8 8 8 8 8 8 8 8 8 8
#> [5,] 9 9 10 10 10 10 10 10 10 10
#> [,109] [,110] [,111] [,112] [,113] [,114] [,115] [,116] [,117] [,118]
#> [1,] 2 3 1 1 1 1 1 1 2 2
#> [2,] 4 4 2 2 2 3 3 4 3 3
#> [3,] 5 5 3 4 5 4 5 5 4 5
#> [4,] 8 8 8 8 8 8 8 8 8 8
#> [5,] 10 10 11 11 11 11 11 11 11 11
#> [,119] [,120] [,121] [,122] [,123] [,124] [,125] [,126] [,127] [,128]
#> [1,] 2 3 1 1 1 1 1 1 2 2
#> [2,] 4 4 2 2 2 3 3 4 3 3
#> [3,] 5 5 3 4 5 4 5 5 4 5
#> [4,] 8 8 9 9 9 9 9 9 9 9
#> [5,] 11 11 10 10 10 10 10 10 10 10
#> [,129] [,130] [,131] [,132] [,133] [,134] [,135] [,136] [,137] [,138]
#> [1,] 2 3 1 1 1 1 1 1 2 2
#> [2,] 4 4 2 2 2 3 3 4 3 3
#> [3,] 5 5 3 4 5 4 5 5 4 5
#> [4,] 9 9 9 9 9 9 9 9 9 9
#> [5,] 10 10 11 11 11 11 11 11 11 11
#> [,139] [,140] [,141] [,142] [,143] [,144] [,145] [,146] [,147] [,148]
#> [1,] 2 3 1 1 1 1 1 1 2 2
#> [2,] 4 4 2 2 2 3 3 4 3 3
#> [3,] 5 5 3 4 5 4 5 5 4 5
#> [4,] 9 9 10 10 10 10 10 10 10 10
#> [5,] 11 11 11 11 11 11 11 11 11 11
#> [,149] [,150]
#> [1,] 2 3
#> [2,] 4 4
#> [3,] 5 5
#> [4,] 10 10
#> [5,] 11 11
Created on 2022-12-02 with reprex v2.0.2创建于 2022-12-02，使用reprex v2.0.2
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