python生成一列有0有1的矩阵,如何在Python中生成0-1矩阵的所有可能组合？

最新推荐文章于 2022-11-12 00:57:23 发布

转载最新推荐文章于 2022-11-12 00:57:23 发布 · 1.1k 阅读

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#python生成一列有0有1的矩阵

本文介绍如何使用Python的numpy和itertools库生成一个KxN大小的矩阵的所有可能0-1组合，通过一个简单的one-liner解决方案，并解释了其背后的原理。

How can generate all possible combinations of a 0-1 matrix of size K by N?

For example, if I take K=2 and N=2, I get the following combinations.

combination 1

[0, 0;

0, 0];

combination 2

[1, 0;

0, 0];

combination 3

[0, 1;

0, 0];

combination 4

[0, 0;

1, 0];

combination 5

[0, 0;

0, 1];

combination 6

[1, 1;

0, 0];

combination 7

[1, 0;

1, 0];

combination 8

[1, 0;

0, 1];

combination 9

[0, 1;

1, 0];

combination 10

[0, 1;

0, 1];

combination 11

[0, 0;

1, 1];

combination 12

[1, 1;

1, 0];

combination 13

[0, 1;

1, 1];

combination 14

[1, 0;

1, 1];

combination 15

[1, 1;

0, 1];

combination 16

[1, 1;

1, 1];

解决方案

A one-liner solution with numpy and itertools:

[np.reshape(np.array(i), (K, N)) for i in itertools.product([0, 1], repeat = K*N)]

Explanation: the product function returns a Cartesian product of its input. For instance, product([0, 1], [0, 1]) returns an iterator that comprises all possible permutations of [0, 1] and [0, 1]. In other words, drawing from a product iterator:

for i, j in product([0, 1], [0, 1]):

is actually equivalent to running two nested for-loops:

for i in [0, 1]:

for j in [0, 1]:

The for-loops above already solve the problem at hand for a specific case of K, N = (1, 0). Continuing the above line of thought, to generate all possible zero/one states of a vector i, we need to draw samples from an iterator that is equivalent to a nested for-loop of depth l, where l = len(i). Luckily, itertools provides the framework to do just that with its repeat keyword argument. In the case of OP's problem this permutation depth should be K*N, so that it can be reshaped into a numpy array of proper sizes during each step of the list comprehension.