如何在Python中生成0-1矩阵的所有可能组合? [英] How to generate all possible combinations of 0-1 matrix in Python?
问题描述
如何生成大小为K的N的0-1矩阵的所有可能组合?
例如,如果我取K = 2和N = 2,我得到以下组合。
组合1
[0,0;
0,0];
组合2
[1,0;
0,0];
组合3
[0,1;
0,0];
组合4
[0,0;
1,0];
组合5
[0,0;
0,1];
组合6
[1,1;
0,0];
组合7
[1,0;
1,0];
组合8
[1,0;
0,1];
组合9
[0,1;
1,0];
组合10
[0,1;
0,1];
组合11
[0,0;
1,1];
组合12
[1,1;
1,0];
组合13
[0,1;
1,1];
组合14
[1,0;
1,1];
组合15
[1,1;
0,1];
组合16
[1,1;
1,1];
具有 numpy
和 itertools
:
[ np.reshape(np.array(i),(K,N))for it in itertools.product([0,1],repeat = K * N)]
说明: 积
函数返回其输入的笛卡尔积。例如, product([0,1],[0,1])
返回一个迭代器,该迭代器包含 [0,1]的所有可能排列
和 [0,1]
。换句话说,从乘积迭代器中提取:
对于乘积中的i,j([0,1],[0, 1]):
实际上等效于运行两个嵌套的for循环:
对于[0,1]中的i:
for j在[0,1]中:
上面的for循环已经解决了特定情况下 K的手头问题,N =(1,0)
。继续上述思路,要生成向量 i
的所有可能的零/一状态,我们需要从迭代器中提取样本,该迭代器等效于嵌套的for循环深度 l
,其中 l = len(i)
。幸运的是, itertools
提供了使用 repeat
关键字参数来实现此目的的框架。对于OP的问题,此排列深度应为 K * N
,以便可以在列表理解的每个步骤中将其重新排列为适当大小的Numpy数组。 / p>
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];
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.
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