向量列表的所有可能的组合

I have a list of vectors and I would like to get a list of all possible combinations between the elements of every vector, i.e., combinations of n elements (from a vector) taken two and more than two at a time.

我有一个向量的列表，我想要一个列表，列出每个向量的元素之间所有可能的组合，例如。，n个元素(来自一个向量)的组合一次取2次，一次超过2次。

For instance, I have the following list:

例如，我有以下清单:

> DF
$`1`
   A B   C
1 11 2 432

$`2`
   A B   C
2 11 3 432

$`3`
   A B   C
3 13 4 241

Here's my code:

这是我的代码:

> d=list()
>   for (j in 1:length(DF)){
+     for (i in 2:length(DF)){
+       d[[j]]=combn(DF[[j]],i,simplify=F)
+     }
+   }
> d
[[1]]
[[1]][[1]]
   A B   C
1 11 2 432

[[2]]
[[2]][[1]]
   A B   C
2 11 3 432

[[3]]
[[3]][[1]]
   A B   C
3 13 4 241

It is wrong, because I just get combinations of three elements taken three at a time. I would have to add combinations of three elements taken two at a time. I just get the last loop value. It is a problem of dimensions inside the loop.

这是错误的，因为我每次只取三个元素的组合。我需要添加三个元素的组合一次取两个。得到最后一个循环值。这是循环内部的维数问题。

If I run the loop just for i=2, then I get:

如果我只对I =2进行循环，就得到:

> d
[[1]]
[[1]][[1]]
   A B
1 11 2

[[1]][[2]]
   A   C
1 11 432

[[1]][[3]]
  B   C
1 2 432


[[2]]
[[2]][[1]]
   A B
2 11 3

[[2]][[2]]
   A   C
2 11 432

[[2]][[3]]
  B   C
2 3 432

....

1 个解决方案

#1

you could try

你可以试试

lapply(2:3, function(k) { lapply(1:length(DF),function(x){ combn(DF[[x]],k, 
                          simplify = FALSE)})})