Question

我将如何以最快的方式计算元素数组的排列，每个排列在排列内重复一定次数？

例如：

elements = [0, 1]
repetitions = [2, 3]

我希望它返回10个唯一的排列：

[(0, 0, 1, 1, 1), (0, 1, 0, 1, 1), (1, 0, 1, 0, 1) ....]

Answer 1

这是一个解决方案：想法是找到所有可能的索引，我们可以将第一个元素放在这里，然后递归地找到其余索引的可能索引。这样，我们可以保证直接生成唯一的输出。

from itertools import combinations


def combs(elements, repetitions, index=0, indices_left=None, already_set=None):
    if indices_left is None:
        already_set = [None] * sum(repetitions)
        indices_left = set(range(sum(repetitions)))

    element = elements[index]
    number = repetitions[index]

    for indices_choice in combinations(indices_left, number):
        currently_set = already_set[:]
        for i in indices_choice:
            currently_set[i] = element
        remaining_indices = indices_left - set(indices_choice)
        if not remaining_indices:
            yield currently_set
        else:
            yield from combs(elements, repetitions, index+1, remaining_indices, currently_set)

在您输入示例的情况下，这将为我们提供：

elements = [0, 1]
repetitions = [2, 3]

for comb in combs(elements, repetitions):
    print(comb)

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

另一个具有更多元素的示例：

elements = [0, 1, 2]
repetitions = [2, 3, 2]

c = combs(elements, repetitions)

print(next(c))
# [0, 0, 1, 1, 1, 2, 2]

print(next(c))
# [0, 0, 1, 1, 2, 1, 2]

print(len(list(combs(elements, repetitions))))
# 210

一点速度测试：

elements = [0, 1]
repetitions = [10, 10]

a = list(combs(elements, repetitions))
print(len(a))
# 184756

%timeit list(combs(elements, repetitions))
# 1.15 s ± 5.52 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

Answer 2

这不是一个完整的答案，而是总体思路。我几乎可以肯定，没有递归和组合（itertools），没有简单的方法可以做到这一点。假设您已经定义了数据，元素和重复项。

n是元素的大小
s是所有重复项的总和，所以它是您每个排列的元素的最终数量寻找。

我猜您必须进行一次递归，这将转化为递归函数。 En是元素数组。 Rn是重复数组。因此，En-1是n-1个第一元素和Rn-1个第一元素的数组。如果您知道En-1和Rn-1的解决方案，则只需计算与itertools软件包see here的组合。以您的示例为例，您知道E2和R2对吗？我会尝试抽出时间来编写代码，不知道您对递归的了解是多少，这可能很难理解。

from itertools import combinations

def multipleElementsPermutations(elements, repetitions): #elements is En, repetitions is Rn
    if len(elements) != len(repetitions) or len(elements) <2 or len(repetitions) <2:
        return None 
    elif len(elements) == 2:
        #Write some code using itertools to solve the case n=2
        return list(permutations(elements[0]*repetitions[0] + elements[1]*repetitions[1], repetitions[0]+repetitions[1]))
    else:
        last_element = elements[-1]
        last_repetition = repetitions[-1]
        #Here thus function will call itself with the n-1 arguments
        previous_case_solution = multipleElementsPermutation(
            elements[:-1], #En-1
            repetitions[:-1] #Rn-1
        )
        #Write some code here to find different ways to add last element

很抱歉，没有时间编写完整的解决方案。我猜这更多是数学问题，而不是代码问题。

使用下面的Erwan的想法，也许这段代码可以工作：

from itertools import permutations

n = len(elements)
repeated_elements_array = [element[i]*repetition[i] for i in range(n)]
repeated_elements_list = []
for repeated_elements in repeated_elements_array:
    repeated_elements_list += repeated_elements
s = 0
for repetition in repetitions:
    s += repetition
solution = permutations(repeated_elements_list, s)

Answer 3

您可以从permutations模块使用itertools

>>> from itertools import permutations
>>> list(permutations([0]*2 + [1]*3, 5))
[(0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0)]

更自动的版本：

>>> e = [0, 1]
>>> r = [2, 3]
>>> b = sum([[i]*j for i, j in zip(e, r)], [])
>>> list(permutations(b, len(b)))
[(0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 0, 1, 1, 1), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 0, 1, 1), (0, 1, 0, 1, 1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 0, 1, 1), (1, 0, 0, 1, 1), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 0, 1, 0, 1), (1, 0, 1, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 0, 0, 1), (1, 1, 0, 1, 0), (1, 1, 1, 0, 0), (1, 1, 1, 0, 0)]

编辑：如@JosephWood所述，如果您不想重复（00111可能出现多次），请查看here

Edit2：如果元素中只有0.1，并且不想重复，则可以使用类似(j for j in ([i%2, i//2%2, i//4%2, i//8%2, i//16%2] for i in range(0, 2**5)) if sum(j)==3)的二进制表示形式。我不知道它是否更快。

Answer 4

类似于@Erwan，它也可以工作：

import itertools as it
elements = [0, 1]
repetitions = [2, 3]
arr = []
for j,i in enumerate(repetitions):
   arr = arr +[elements[j]]*i

set(list(it.permutations(arr)))

具有特定重复项的元素数组的Python排列

4 个答案: