python详尽的组合

Posted 2021-04-19

我正在尝试写一个函数有输入：

given_num=6, n=3

如何将所有可能组合的结果作为列表返回，其中n个元素和sum = given_num嵌套在列表中：

[[0,0,6],[0,1,5],[0,2,4],[0,3,3],[0,4,2],[0,5,1],[0,6,0],
[1,0,5],[1,1,4],[1,2,3],[1,3,2],[1, 4,1],[1,5,0],
...
[6,0,0]]

---

求解，可能的组合计算为

factorial(given_num+n-1)/(factorial(given_num) * factorial(n-1))

答案

在这里尝试使用itertools product：

from itertools import product

given_num = 6
n = 3
all_comb = product(range(given_num+1), repeat=n)
final_lst = [i for i in all_comb if sum(i) == given_num]
print(final_lst)

另一答案

from itertools import product
import numpy as np

tuples = []
given_num = 6
n = 3

for numbers in product(np.arange(given_num+1), repeat=n):

    if sum(numbers) == given_num:
        tuples.append(numbers)

print(tuples)

另一答案

如果你想通过递归函数获得一些乐趣

def summands(num, N):
    if N==1:
        return [[num]]
    else:
        return [[i] + j for i in range(num+1) for j in summands(num - i, N-1)]