生成包含n个元素的所有k个置换并重复的矩阵的好算法吗?

问题描述

所以，假设我设置了S = {A1，A2}，并且我想以3为一组计算这两个元素的所有可能排列.

So, let's say I have set S = {A1, A2} and I want to calculate all possible permutations of these two elements in groups of 3.

我想生成一个像这样的矩阵:

I would like to generate a matrix such as this:

(A1 A1 A2)
(A1 A2 A1)
(A2 A1 A1)
(A2 A2 A1)
(A2 A1 A2)
(A1 A2 A2)

我正在使用R语言.我一直在尝试寻找某种算法来生成像这样的矩阵，但是到目前为止还没有成功.

I'm using R language. I've been trying to find some algorithm to generate a matrix like this, but haven't been successful so far.

感谢您的帮助.

推荐答案

获得所有排列的一种方法是使用expand.grid:

One way to obtain all permutations would be to use expand.grid:

a <- c("A", "B")

expand.grid(a, a, a)
#   Var1 Var2 Var3
# 1    A    A    A
# 2    B    A    A
# 3    A    B    A
# 4    B    B    A
# 5    A    A    B
# 6    B    A    B
# 7    A    B    B
# 8    B    B    B

如@ Forest1所建议，您可能要排除第一行和最后一行，因为它们包含AAA和BBB.这可以通过

As suggested by @Forest1, you may want to exclude the first and the last row, since they contain AAA and BBB. This can be achieved by

expand.grid(a, a, a)[2:7]

这篇关于生成包含n个元素的所有k个置换并重复的矩阵的好算法吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！