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问迭代2置换集
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Stack Overflow用户

提问于 2022-09-01 03:49:19

回答 2查看 56关注 0票数 -2

假设我从这两个假设的列表开始:颜色和形状。

我有15种颜色，假设有一个递归的无限多个形状(就像一个正方形的边数)。在最基本的情况下，它开始于

color_list, shape_list = [1], [1]

就我提到的复杂性而言，长度3的一个例子可能是

color_list, shape_list = [15,9,7], [1,3,7]

这些数字是任意的，以比较将颜色15与形状1、颜色9与形状3、颜色7与形状7相结合的特定排列。

我将如何在每一个可能的排列上创建一个迭代，从[1], [1]返回一些东西，比如[15,15,15,15,15], [n,n,n,n,n]。

python

list

loops

iteration

permutation

回答 2

Stack Overflow用户

发布于 2022-09-01 04:17:30

IIUC，你想做的是-

为每个(颜色= 3，形状= 5)，(颜色= 1，形状= 99)创建颜色和形状组合，等等。然后，您将尝试对从1到m(在您的示例中为m=5)的

的每个排列顺序对这些颜色组合进行置换。

下面是一段代码，可以为您提供类似的信息。请注意，这将是成倍的爆炸，因此避免更大的限制。

import itertools

color_i = 1
color_n = 5

shape_i = 1
shape_n = 6

permutes = 3

# color, shape combos
combos = list(itertools.product(range(color_i, color_n+1), range(shape_i, shape_n+1)))

# create permutations from the color, shape combos
permutations = []
for p in range(permutes):
    for i in itertools.permutations(combos,r=p+1):
        permutations.append(list(zip(*i)))

permutations

[[(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,)],
 [(1, 1), (1, 2)],
 [(1, 1), (1, 3)],
 ...,
 [(1, 4), (1, 2)],
 [(1, 4), (1, 3)],
 [(1, 4), (1, 4)],
 [(1, 4), (1, 5)],
 [(1, 4), (1, 6)],
 [(1, 5), (1, 1)],
 [(1, 5), (1, 2)],
 [(1, 5), (1, 3)],
 [(1, 5), (1, 4)],
 [(1, 5), (1, 5)],
 [(1, 5), (1, 6)],
 [(1, 1), (2, 1)],
 [(1, 1), (2, 3)],
 [(1, 1), (2, 4)],
 [(1, 1), (2, 5)],
 [(1, 1), (2, 6)],
 [(1, 2), (2, 1)],
 [(1, 2), (2, 2)],
 ...,
 [(4, 4), (3, 6)],
 [(4, 5), (3, 1)],
 [(4, 5), (3, 2)],
 [(4, 5), (3, 3)],
 [(4, 5), (3, 4)],
 [(4, 5), (3, 5)],
 [(4, 5), (3, 6)],
 [(4, 1), (4, 1)],
 [(4, 1), (4, 2)],
 [(4, 1), (4, 3)],
 [(4, 1), (4, 4)],
 [(4, 1), (4, 5)],
 ...,
 [(5, 5), (6, 2)],
 [(5, 5), (6, 3)],
 [(5, 5), (6, 4)],
 [(5, 5), (6, 5)],
 [(1, 1, 1), (1, 2, 3)],
 [(1, 1, 1), (1, 2, 4)],
 [(1, 1, 1), (1, 2, 5)],
 [(1, 1, 1), (1, 2, 6)],
 [(1, 1, 2), (1, 2, 1)],
 [(1, 1, 2), (1, 2, 2)],
 [(1, 1, 2), (1, 2, 3)],
 ...,
 [(1, 1, 4), (1, 2, 6)],
 [(1, 1, 5), (1, 2, 1)],
 [(1, 1, 5), (1, 2, 2)],
 [(1, 1, 5), (1, 2, 3)],
 [(1, 1, 5), (1, 2, 4)],
 [(1, 1, 5), (1, 2, 5)],
 [(1, 1, 5), (1, 2, 6)],
 [(1, 1, 1), (1, 3, 2)],
 [(1, 1, 1), (1, 3, 4)],
 [(1, 1, 1), (1, 3, 5)],
 [(1, 1, 1), (1, 3, 6)],
 [(1, 1, 2), (1, 3, 1)],
 [(1, 1, 2), (1, 3, 2)],
 [(1, 1, 2), (1, 3, 3)],
 [(1, 1, 2), (1, 3, 4)],
 [(1, 1, 2), (1, 3, 5)],
 [(1, 1, 2), (1, 3, 6)],
 [(1, 1, 3), (1, 3, 1)],
 [(1, 1, 3), (1, 3, 2)],
 [(1, 1, 3), (1, 3, 3)],
 [(1, 1, 3), (1, 3, 4)],
 [(1, 1, 3), (1, 3, 5)],
 [(1, 1, 3), (1, 3, 6)],
 [(1, 1, 4), (1, 3, 1)],
 [(1, 1, 4), (1, 3, 2)],
 [(1, 1, 4), (1, 3, 3)],
 [(1, 1, 4), (1, 3, 4)],
 [(1, 1, 4), (1, 3, 5)],
 [(1, 1, 4), (1, 3, 6)],
 [(1, 1, 5), (1, 3, 1)],
 [(1, 1, 5), (1, 3, 2)],
 [(1, 1, 5), (1, 3, 3)],
 [(1, 1, 5), (1, 3, 4)],
 [(1, 1, 5), (1, 3, 5)],
 [(1, 1, 5), (1, 3, 6)],
 [(1, 1, 1), (1, 4, 2)],
 [(1, 1, 1), (1, 4, 3)],
 [(1, 1, 1), (1, 4, 5)],
 [(1, 1, 1), (1, 4, 6)],
 [(1, 1, 2), (1, 4, 1)],
 [(1, 1, 2), (1, 4, 2)],
 [(1, 1, 2), (1, 4, 3)],
 [(1, 1, 2), (1, 4, 4)],
 [(1, 1, 2), (1, 4, 5)],
 [(1, 1, 2), (1, 4, 6)],
 [(1, 1, 3), (1, 4, 1)],
 [(1, 1, 3), (1, 4, 2)],
 [(1, 1, 3), (1, 4, 3)],
 [(1, 1, 3), (1, 4, 4)],
 [(1, 1, 3), (1, 4, 5)],
 [(1, 1, 3), (1, 4, 6)],
 ...]

它是如何运作的-

使用颜色和形状的范围创建itertools.product。这为您提供了所有可能的(颜色、形状)元组。

置换阶为1到m (1，2，3，.，m)的

在所有可能的元组k倍上置换，其中1>=k>=m

[(color1, color2, color3), (shape1, shape2, shape3)]

在每个生成的置换上使用list(zip(*i))将其从[(color1, shape1), (color2, shape2), (color3, shape3)]转到[(color1, shape1), (color2, shape2), (color3, shape3)]

票数 0

Stack Overflow用户

发布于 2022-09-01 04:17:44

您有许多参数：

可能的颜色数(例如，
)可能的形状数(例如，
)颜色/形状组合的数目(例如，

)

在Python中：

n_colours = 2
colours = range(1, n_colours+1)

n_shapes = 3
shapes = range(1, n_shapes+1)

# creating a list of all possible combinations of colours and shapes:
all_pairs = list(product(colours, shapes))

您要求在两个不同的颜色和相应形状的列表中获得这些组合。将它们作为成对的元组来生成似乎更明智：

from itertools import product

max_len = 4
tuples = [t for p in [product(all_pairs, repeat=n) for n in range(1, max_len+1)] for t in p]

但是，如果您想要将它们生成成一对独立的颜色和形状，下面是一个实现，它一个接一个地生成它们：

from itertools import product


def all_colour_shape_pairs(n_colours, n_shapes, max_len):
    ps = list(product(range(1, n_colours+1), range(1, n_shapes+1)))
    for t in [t for p in [product(ps, repeat=n) for n in range(1, max_len+1)] for t in p]:
        yield zip(*t)


result = all_colour_shape_pairs(2, 3, 4)
while input('Press enter for another, or enter "x" to stop:') != 'x':
    print(tuple(next(result)))

输出

((1,), (1,))
((1,), (2,))
((1,), (3,))
((2,), (1,))
...
((2,), (3,))
((1, 1), (1, 1))
((1, 1), (1, 2))
...
((2, 2), (3, 3))
((1, 1, 1), (1, 1, 1))
((1, 1, 1), (1, 1, 2))
...

如果希望生成器继续生成“不合理”的结果(或者更确切地说，在计算机耗尽资源并无法继续时停止)：

from itertools import product


def all_colour_shape_pairs(n_colours, n_shapes, max_len):
    ps = list(product(range(1, n_colours+1), range(1, n_shapes+1)))
    n = 1
    while True:
        for t in product(ps, repeat=n):
            yield zip(*t)
        n += 1


result = all_colour_shape_pairs(2, 3, 4)
while input('Press enter for another, or enter "x" to stop:') != 'x':
    print(tuple(next(result)))

票数 0

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/73564123

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