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所有可能的矩阵组合

[英]All possible matrix combinations

How can I get all possible matrix of 0 and 1?如何获得所有可能的 0 和 1 矩阵? For example:例如:

My function is defined like我的 function 定义为

def matrix_combinations(m):

Where m is a matrix with only zeros其中 m 是只有零的矩阵

What I want is all possible matrix with 0 and 1 with the same dimensions as m我想要的是所有可能的矩阵,其中 0 和 1 的尺寸与 m 相同

For example: m = [[0, 0, 0], [0, 0, 0]] The output should be a list of matrix like this:例如: m = [[0, 0, 0], [0, 0, 0]] output 应该是这样的矩阵列表:

[[[0, 0, 0], [0, 0, 0]],
 [[0, 0, 0], [0, 0, 1]],
 [[0, 0, 0], [0, 1, 0]],
 [[0, 0, 0], [0, 1, 1]],
           ...
 [[0, 1, 1], [1, 1, 1]],
 [[1, 1, 1], [1, 1, 1]]]

I don't want to use package like itertools or numpy我不想像 itertools 或 numpy 那样使用 package

The idea behind my implementation is that the shape of the array does not matter, you just need all the combinations of 1s and 0s, and to do that you just have to count in binary.我的实现背后的想法是数组的形状无关紧要,您只需要 1 和 0 的所有组合,而要做到这一点,您只需要以二进制计数即可。

What we are doing is creating lists of binary numbers with the same size as the input matrix and then changing its shape to match the input matrix.我们正在做的是创建与输入矩阵大小相同的二进制数列表,然后更改其形状以匹配输入矩阵。

import numpy as np

def matrix_combinations(m):
    bits = m.shape[0] * m.shape[1]
    amount_numbers = 2**bits
    ret = []
    for i in range(amount_numbers):
        num = [int(x) for x in bin(i)[2:]]
        num = [0] * (bits - len(num)) + num
        ret.append(np.reshape(num, m.shape).tolist())
    return ret

I am using numpy because is really convinient to change shapes of arrays.我正在使用 numpy 因为真的很方便改变 arrays 的形状。 If you don't want to use numpy, assuming your matrix si bidimensional, you can create your own reshape function quite easyly.如果你不想使用 numpy,假设你的矩阵是二维的,你可以很容易地创建自己的重塑 function。 The code would look like this:代码如下所示:

def reshape(list, shape):
    ret = []
    for i in range(shape[0]):
        row = [] 
        for j in range(shape[1]):
            row.append(list[j*i+i])
        ret.append(row)

    return ret

def matrix_combinations(m):
    shape = (len(m),len(m[0]))
    bits = shape[0] * shape[1]
    amount_numbers = 2**bits
    ret = []
    for i in range(0,amount_numbers):
        num = [int(x) for x in bin(i)[2:]]
        num = [0] * (bits - len(num)) + num
        ret.append(reshape(num, shape))
    return ret

You first need the shape of your matrix - it would actually be best if you passed width and height rather than m .您首先需要矩阵的形状 - 实际上最好传递widthheight而不是m

Numpy answer below, since it's an excellent tool and you should definitely learn it if your working with matrices, it'll be worth your while. Numpy 在下面回答,因为它是一个很好的工具,如果你使用矩阵,你绝对应该学习它,这将是值得的。 Non-numpy answer first.首先是非numpy的答案。

I'm going to create numbers in the range of 0 to 2**(width*height) .我将创建02**(width*height)范围内的数字。

def matrix_combinations(m):
    height = len(m)
    width = len(m[0])
    size = height * width

    output_matrices = []

    for matrix_values in range(2**(width*height)):
        matrix = []
        for y in range(height):
            row = []
            for x in range(width):
                 last_bit = matrix_values & 1
                 matrix_values >>= 1
                 row.append(last_bit)
            matrix.append(row)
        output_matrices.append(matrix)

    return output_matrices

Now a numpy version.现在是 numpy 版本。

def matrix_combinations(m):
    indices = np.arange(m.size)
    output_matrices = []
    for value in range(2**m.size):
        matrix_values = np.bitwise_and(value >> indices, 1)
        matrix = matrix_values.reshape(m.shape)
        output_matrices.append(matrix)
    return output_matrices

More than the shape of the matrix is the number of elements that matters;比矩阵的形状更重要的是元素的数量; if it has n = h*w elements, then you'll have 2 ** n output matrices, just all numbers from 0 to 2**n-1 written in base 2. Below is an implementation:如果它有n = h*w元素,那么你将有2 ** n output 矩阵,只是从 0 到2**n-1的所有数字都写在基数 2 中。下面是一个实现:

def resize(x, height, width):
    return [
        x[i * width : (i+1) * width]
        for i in range(height)
    ]

def all_combinations(m):
    flattened = [x for y in m for x in y]
    n = len(flattened)
    width = len(m[0])
    height = len(m)
    result = []
    form = '{:0' + str(n) + 'b}'
    for i in range(2 ** n):
        s = form.format(i)
        x = [int(c) for c in s]
        result.append(resize(x, height, width))
    return result

This is really just every binary mask in the source digits of m , so I present a wildly-evil, string-based solution这实际上只是m的源数字中的每个二进制掩码,所以我提出了一个非常邪恶的基于字符串的解决方案

def matrix_combinations(m):
    target_length   = str(m).count("0")
    result_template = str(m).replace("0", "{}")
    for numeric in range(1, 2**target_length):
        yield eval(result_template.format(*list(bin(numeric)[2:].zfill(target_length))))

Example use示例使用

>>> _it = matrix_combinations([[0],[0]])
>>> list(_it)
[[[0], [1]], [[1], [0]], [[1], [1]]]
>>> _it = matrix_combinations([[0, 0],[0, 0]])
>>> list(_it)
[[[0, 0], [0, 1]], [[0, 0], [1, 0]], [[0, 0], [1, 1]], [[0, 1], [0, 0]], [[0, 1], [0, 1]], [[0, 1], [1, 0]], [[0, 1], [1, 1]], [[1, 0], [0, 0]], [[1, 0], [0, 1]], [[1, 0], [1, 0]], [[1, 0], [1, 1]], [[1, 1], [0, 0]], [[1, 1], [0, 1]], [[1, 1], [1, 0]], [[1, 1], [1, 1]]]

This works by这通过

  • discovering how many permutations there could be 2**(the count of 0)发现可能有多少排列2**(the count of 0)
  • creating a string template of the target matrix shape创建目标矩阵形状的字符串模板
    >>> str([[0, 0],[0, 0]]).replace("0", "{}") '[[{}, {}], [{}, {}]]'
  • representing each number in binary and packing it with leading 0s用二进制表示每个数字并用前导 0 包装它
    >>> bin(4)[2:].zfill(4) '0100'
  • filling the template with the values用值填充模板
    >>> '[[{}, {}], [{}, {}]]'.format(*list(bin(4)[2:].zfill(4))) '[[0, 1], [0, 0]]'
  • evaluating the template to create the final structure评估模板以创建最终结构

Do note that if you try to keep all of these in memory (for example with a list), your collection will fail (with any solution) around 25 zeros (5-10G of system memory and exponentially increasing).. however, if you process each individual list (which you can do with a generator like this one), it'll happily work with almost any number of zeros (my system started to get slow generating the initial lists at about 10**7 zeros)请注意,如果您尝试将所有这些保留在 memory 中(例如使用列表),您的收集将失败(使用任何解决方案)大约 25 个零(系统 memory 的 5-10G 并且呈指数增长)。但是,如果您处理每个单独的列表(您可以使用像这样的生成器来完成),它几乎可以处理任意数量的零(我的系统在大约 10**7 个零处生成初始列表时开始变得缓慢)

>>> a = next(matrix_combinations([[0]*(10**7)]))
>>>

This solution can take an input matrix of any depth and dimensions.该解决方案可以采用任何深度和维度的输入矩阵。 By using a recursive generator function, possible value ranges (in this case [0, 1] ) can be more easily traversed as part of the combinations building process:通过使用递归生成器 function,作为组合构建过程的一部分,可以更容易地遍历可能的值范围(在本例中为[0, 1] ):

def prod(d, c = []):
   yield from ([c] if not d else [j for k in d[0] for j in prod(d[1:], c+[k])])

def matrix_combinations(m):
   for i in prod([[0, 1] if not i else [*matrix_combinations(i)] for i in m]):
      yield list(i)

print(list(matrix_combinations([[0, 0, 0], [0, 0, 0]])))
print(list(matrix_combinations([[0, [0, [0, 0]], 0], [0, 0, [0]]])))

Output: Output:

[[[0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 1]], [[0, 0, 0], [0, 1, 0]], [[0, 0, 0], [0, 1, 1]], [[0, 0, 0], [1, 0, 0]], [[0, 0, 0], [1, 0, 1]], [[0, 0, 0], [1, 1, 0]], [[0, 0, 0], [1, 1, 1]], [[0, 0, 1], [0, 0, 0]], [[0, 0, 1], [0, 0, 1]], [[0, 0, 1], [0, 1, 0]], [[0, 0, 1], [0, 1, 1]], [[0, 0, 1], [1, 0, 0]], [[0, 0, 1], [1, 0, 1]], [[0, 0, 1], [1, 1, 0]], [[0, 0, 1], [1, 1, 1]], [[0, 1, 0], [0, 0, 0]], [[0, 1, 0], [0, 0, 1]], [[0, 1, 0], [0, 1, 0]], [[0, 1, 0], [0, 1, 1]], [[0, 1, 0], [1, 0, 0]], [[0, 1, 0], [1, 0, 1]], [[0, 1, 0], [1, 1, 0]], [[0, 1, 0], [1, 1, 1]], [[0, 1, 1], [0, 0, 0]], [[0, 1, 1], [0, 0, 1]], [[0, 1, 1], [0, 1, 0]], [[0, 1, 1], [0, 1, 1]], [[0, 1, 1], [1, 0, 0]], [[0, 1, 1], [1, 0, 1]], [[0, 1, 1], [1, 1, 0]], [[0, 1, 1], [1, 1, 1]], [[1, 0, 0], [0, 0, 0]], [[1, 0, 0], [0, 0, 1]], [[1, 0, 0], [0, 1, 0]], [[1, 0, 0], [0, 1, 1]], [[1, 0, 0], [1, 0, 0]], [[1, 0, 0], [1, 0, 1]], [[1, 0, 0], [1, 1, 0]], [[1, 0, 0], [1, 1, 1]], [[1, 0, 1], [0, 0, 0]], [[1, 0, 1], [0, 0, 1]], [[1, 0, 1], [0, 1, 0]], [[1, 0, 1], [0, 1, 1]], [[1, 0, 1], [1, 0, 0]], [[1, 0, 1], [1, 0, 1]], [[1, 0, 1], [1, 1, 0]], [[1, 0, 1], [1, 1, 1]], [[1, 1, 0], [0, 0, 0]], [[1, 1, 0], [0, 0, 1]], [[1, 1, 0], [0, 1, 0]], [[1, 1, 0], [0, 1, 1]], [[1, 1, 0], [1, 0, 0]], [[1, 1, 0], [1, 0, 1]], [[1, 1, 0], [1, 1, 0]], [[1, 1, 0], [1, 1, 1]], [[1, 1, 1], [0, 0, 0]], [[1, 1, 1], [0, 0, 1]], [[1, 1, 1], [0, 1, 0]], [[1, 1, 1], [0, 1, 1]], [[1, 1, 1], [1, 0, 0]], [[1, 1, 1], [1, 0, 1]], [[1, 1, 1], [1, 1, 0]], [[1, 1, 1], [1, 1, 1]]]
[[[0, [0, [0, 0]], 0], [0, 0, [0]]], [[0, [0, [0, 0]], 0], [0, 0, [1]]], [[0, [0, [0, 0]], 0], [0, 1, [0]]], [[0, [0, [0, 0]], 0], [0, 1, [1]]], [[0, [0, [0, 0]], 0], [1, 0, [0]]], [[0, [0, [0, 0]], 0], [1, 0, [1]]], [[0, [0, [0, 0]], 0], [1, 1, [0]]], [[0, [0, [0, 0]], 0], [1, 1, [1]]], [[0, [0, [0, 0]], 1], [0, 0, [0]]], [[0, [0, [0, 0]], 1], [0, 0, [1]]], [[0, [0, [0, 0]], 1], [0, 1, [0]]], [[0, [0, [0, 0]], 1], [0, 1, [1]]], [[0, [0, [0, 0]], 1], [1, 0, [0]]], [[0, [0, [0, 0]], 1], [1, 0, [1]]], [[0, [0, [0, 0]], 1], [1, 1, [0]]], [[0, [0, [0, 0]], 1], [1, 1, [1]]], [[0, [0, [0, 1]], 0], [0, 0, [0]]], [[0, [0, [0, 1]], 0], [0, 0, [1]]], [[0, [0, [0, 1]], 0], [0, 1, [0]]], [[0, [0, [0, 1]], 0], [0, 1, [1]]], [[0, [0, [0, 1]], 0], [1, 0, [0]]], [[0, [0, [0, 1]], 0], [1, 0, [1]]], [[0, [0, [0, 1]], 0], [1, 1, [0]]], [[0, [0, [0, 1]], 0], [1, 1, [1]]], [[0, [0, [0, 1]], 1], [0, 0, [0]]], [[0, [0, [0, 1]], 1], [0, 0, [1]]], [[0, [0, [0, 1]], 1], [0, 1, [0]]], [[0, [0, [0, 1]], 1], [0, 1, [1]]], [[0, [0, [0, 1]], 1], [1, 0, [0]]], [[0, [0, [0, 1]], 1], [1, 0, [1]]], [[0, [0, [0, 1]], 1], [1, 1, [0]]], [[0, [0, [0, 1]], 1], [1, 1, [1]]], [[0, [0, [1, 0]], 0], [0, 0, [0]]], [[0, [0, [1, 0]], 0], [0, 0, [1]]], [[0, [0, [1, 0]], 0], [0, 1, [0]]], [[0, [0, [1, 0]], 0], [0, 1, [1]]], [[0, [0, [1, 0]], 0], [1, 0, [0]]], [[0, [0, [1, 0]], 0], [1, 0, [1]]], [[0, [0, [1, 0]], 0], [1, 1, [0]]], [[0, [0, [1, 0]], 0], [1, 1, [1]]], [[0, [0, [1, 0]], 1], [0, 0, [0]]], [[0, [0, [1, 0]], 1], [0, 0, [1]]], [[0, [0, [1, 0]], 1], [0, 1, [0]]], [[0, [0, [1, 0]], 1], [0, 1, [1]]], [[0, [0, [1, 0]], 1], [1, 0, [0]]], [[0, [0, [1, 0]], 1], [1, 0, [1]]], [[0, [0, [1, 0]], 1], [1, 1, [0]]], [[0, [0, [1, 0]], 1], [1, 1, [1]]], [[0, [0, [1, 1]], 0], [0, 0, [0]]], [[0, [0, [1, 1]], 0], [0, 0, [1]]], [[0, [0, [1, 1]], 0], [0, 1, [0]]], [[0, [0, [1, 1]], 0], [0, 1, [1]]], [[0, [0, [1, 1]], 0], [1, 0, [0]]], [[0, [0, [1, 1]], 0], [1, 0, [1]]], [[0, [0, [1, 1]], 0], [1, 1, [0]]], [[0, [0, [1, 1]], 0], [1, 1, [1]]], [[0, [0, [1, 1]], 1], [0, 0, [0]]], [[0, [0, [1, 1]], 1], [0, 0, [1]]], [[0, [0, [1, 1]], 1], [0, 1, [0]]], [[0, [0, [1, 1]], 1], [0, 1, [1]]], [[0, [0, [1, 1]], 1], [1, 0, [0]]], [[0, [0, [1, 1]], 1], [1, 0, [1]]], [[0, [0, [1, 1]], 1], [1, 1, [0]]], [[0, [0, [1, 1]], 1], [1, 1, [1]]], [[0, [1, [0, 0]], 0], [0, 0, [0]]], [[0, [1, [0, 0]], 0], [0, 0, [1]]], [[0, [1, [0, 0]], 0], [0, 1, [0]]], [[0, [1, [0, 0]], 0], [0, 1, [1]]], [[0, [1, [0, 0]], 0], [1, 0, [0]]], [[0, [1, [0, 0]], 0], [1, 0, [1]]], [[0, [1, [0, 0]], 0], [1, 1, [0]]], [[0, [1, [0, 0]], 0], [1, 1, [1]]], [[0, [1, [0, 0]], 1], [0, 0, [0]]], [[0, [1, [0, 0]], 1], [0, 0, [1]]], [[0, [1, [0, 0]], 1], [0, 1, [0]]], [[0, [1, [0, 0]], 1], [0, 1, [1]]], [[0, [1, [0, 0]], 1], [1, 0, [0]]], [[0, [1, [0, 0]], 1], [1, 0, [1]]], [[0, [1, [0, 0]], 1], [1, 1, [0]]], [[0, [1, [0, 0]], 1], [1, 1, [1]]], [[0, [1, [0, 1]], 0], [0, 0, [0]]], [[0, [1, [0, 1]], 0], [0, 0, [1]]], [[0, [1, [0, 1]], 0], [0, 1, [0]]], [[0, [1, [0, 1]], 0], [0, 1, [1]]], [[0, [1, [0, 1]], 0], [1, 0, [0]]], [[0, [1, [0, 1]], 0], [1, 0, [1]]], [[0, [1, [0, 1]], 0], [1, 1, [0]]], [[0, [1, [0, 1]], 0], [1, 1, [1]]], [[0, [1, [0, 1]], 1], [0, 0, [0]]], [[0, [1, [0, 1]], 1], [0, 0, [1]]], [[0, [1, [0, 1]], 1], [0, 1, [0]]], [[0, [1, [0, 1]], 1], [0, 1, [1]]], [[0, [1, [0, 1]], 1], [1, 0, [0]]], [[0, [1, [0, 1]], 1], [1, 0, [1]]], [[0, [1, [0, 1]], 1], [1, 1, [0]]], [[0, [1, [0, 1]], 1], [1, 1, [1]]], [[0, [1, [1, 0]], 0], [0, 0, [0]]], [[0, [1, [1, 0]], 0], [0, 0, [1]]], [[0, [1, [1, 0]], 0], [0, 1, [0]]], [[0, [1, [1, 0]], 0], [0, 1, [1]]], [[0, [1, [1, 0]], 0], [1, 0, [0]]], [[0, [1, [1, 0]], 0], [1, 0, [1]]], [[0, [1, [1, 0]], 0], [1, 1, [0]]], [[0, [1, [1, 0]], 0], [1, 1, [1]]], [[0, [1, [1, 0]], 1], [0, 0, [0]]], [[0, [1, [1, 0]], 1], [0, 0, [1]]], [[0, [1, [1, 0]], 1], [0, 1, [0]]], [[0, [1, [1, 0]], 1], [0, 1, [1]]], [[0, [1, [1, 0]], 1], [1, 0, [0]]], [[0, [1, [1, 0]], 1], [1, 0, [1]]], [[0, [1, [1, 0]], 1], [1, 1, [0]]], 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