通过重复生成排列 [英] Generating permutations with repetitions

问题描述

我了解itertools，但似乎只能生成排列而不能重复.

I know about itertools, but it seems it can only generate permutations without repetitions.

例如，我想为2个骰子生成所有可能的骰子骰.因此，我需要大小为2的[1、2、3、4、5、6]的所有排列，包括重复:(1、1)，(1、2)，(2、1)...等等

For example, I'd like to generate all possible dice rolls for 2 dice. So I need all permutations of size 2 of [1, 2, 3, 4, 5, 6] including repetitions: (1, 1), (1, 2), (2, 1)... etc

如果可能的话，我不想从头开始实现

If possible I don't want to implement this from scratch

推荐答案

您正在寻找笛卡尔积.

在数学中，笛卡尔乘积(或乘积集)是两组的直接乘积.

In mathematics, a Cartesian product (or product set) is the direct product of two sets.

在您的情况下，这将是{1, 2, 3, 4, 5, 6} x {1, 2, 3, 4, 5, 6}. itertools 可以为您提供帮助:

In your case, this would be {1, 2, 3, 4, 5, 6} x {1, 2, 3, 4, 5, 6}. itertools can help you there:

import itertools
x = [1, 2, 3, 4, 5, 6]
[p for p in itertools.product(x, repeat=2)]
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), 
 (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), 
 (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), 
 (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]

要随机掷骰子(以完全无效的方式):

To get a random dice roll (in a totally inefficient way):

import random
random.choice([p for p in itertools.product(x, repeat=2)])
(6, 3)

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