如何获取两个列表之间的所有映射? [英] How to get all mappings between two lists?

问题描述

我们有两个列表，A和B:

We have two lists, A and B:

A = ['a','b','c']
B = [1, 2]

在A和B之间包含2 ^ n(此处为2 ^ 3 = 8)的所有映射的集合中，是否存在pythonic方法?那就是:

Is there a pythonic way to build the set of all maps between A and B containing 2^n (here 2^3=8)? That is:

[(a,1), (b,1), (c,1)]
[(a,1), (b,1), (c,2)]
[(a,1), (b,2), (c,1)]
[(a,1), (b,2), (c,2)]
[(a,2), (b,1), (c,1)]
[(a,2), (b,1), (c,2)]
[(a,2), (b,2), (c,1)]
[(a,2), (b,2), (c,2)]

使用itertools.product，可以获取所有元组:

Using itertools.product, it's possible to get all the tuples:

import itertools as it
P = it.product(A, B)
[p for p in P]

哪个给:

Out[3]: [('a', 1), ('a', 2), ('b', 1), ('b', 2), ('c', 1), ('c', 2)]

推荐答案

您可以使用 itertools.product 和 zip

You can do this with itertools.product and zip

from itertools import product
print [zip(A, item) for item in product(B, repeat=len(A))]

输出

[[('a', 1), ('b', 1), ('c', 1)],
 [('a', 1), ('b', 1), ('c', 2)],
 [('a', 1), ('b', 2), ('c', 1)],
 [('a', 1), ('b', 2), ('c', 2)],
 [('a', 2), ('b', 1), ('c', 1)],
 [('a', 2), ('b', 1), ('c', 2)],
 [('a', 2), ('b', 2), ('c', 1)],
 [('a', 2), ('b', 2), ('c', 2)]]

product(B, repeat=len(A))产生

[(1, 1, 1),
 (1, 1, 2),
 (1, 2, 1),
 (1, 2, 2),
 (2, 1, 1),
 (2, 1, 2),
 (2, 2, 1),
 (2, 2, 2)]

然后，我们从产品中选择每个元素，并用A压缩，以获得所需的输出.

Then we pick each element from the product and zip it with A, to get your desired output.

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