问题描述
我正在做排列,我想以一种特定的方式打印结果。
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,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,n
更改时,如何自动执行for循环以打印结果。
解决方法
您可以通过定义一个以n
作为参数的函数来对此进行概括:
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*将元组解包。
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:
1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 2 3
1 4 3 2
2 1 3 4
2 1 4 3
2 3 1 4
2 3 4 1
2 4 1 3
2 4 3 1
3 1 2 4
3 1 4 2
3 2 1 4
3 2 4 1
3 4 1 2
3 4 2 1
4 1 2 3
4 1 3 2
4 2 1 3
4 2 3 1
4 3 1 2
4 3 2 1