用for循环用n个值解包元组？

Question

我正在进行排列，我想以特定方式打印我的结果。

我的代码：

    from itertools import permutations as p

    n = 3      #This can be change

    permu_lst = [i for i in p(range(1, n+1)]

    for a, b, c in permu_lst:
       print(a, b, c)

    Output: 
    1 2 3
    1 3 2
    2 1 3
    2 3 1
    3 1 2
    3 2 1

所以我的问题是如何在 n 更改时自动执行 for 循环以打印结果。

Answer 1

您可以通过定义一个函数来概括这一点 n :

from itertools import permutations as p
def print_permutations(n):
    permu_lst = [i for i in p(range(1, n+1))]

    for per in permu_lst:
       print(*per)

per 前的 * 解包元组。

Answer 2

使用iterable unpacking operator(*)

from itertools import permutations as p

n = 4      #This can be change
permu_lst = [i for i in p(range(1, n+1))]

for tup in permu_lst:
    print(*tup)

输出：

参考：https://www.python.org/dev/peps/pep-0448/

用for循环用n个值解包元组？

Tuple unpacking with n values with for loop?

python

iteration

loops

tuples

permutation