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]]
已解决,可能的组合计算为
factorial(given_num+n-1)/(factorial(given_num) * factorial(n-1))
答案 0 :(得分:1)
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)
答案 1 :(得分:0)
pathname
答案 2 :(得分:0)
如果您想要使用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)]