Question

假设我有一个包含5个元素x=[a,b,c,d,e]的列表，我想运行一个for循环，打印所有列表，其中两个条目比原始列表中的相应条目少1个。

在Python中执行此操作的简单方法是什么？提前谢谢。

编辑：如果我想要x=[4,5,6,7,8]：

[3,4,6,7,8], [3,5,5,7,8], [3,5,6,6,8] etc.

Answer 1

这样的事情：

>>> from itertools import combinations
>>> lis = [0,1,2,3,4]
>>> for x,y in combinations(range(len(lis)),2):
    l = lis[:]
    l[x] -= 1
    l[y] -= 1
    print l
...     
[-1, 0, 2, 3, 4]
[-1, 1, 1, 3, 4]
[-1, 1, 2, 2, 4]
[-1, 1, 2, 3, 3]
[0, 0, 1, 3, 4]
[0, 0, 2, 2, 4]
[0, 0, 2, 3, 3]
[0, 1, 1, 2, 4]
[0, 1, 1, 3, 3]
[0, 1, 2, 2, 3]

更短的版本：

for x,y in combinations(range(len(lis)),2):
    print [item - 1 if i in (x,y) else item  for i,item in enumerate(lis)]
...     
[-1, 0, 2, 3, 4]
[-1, 1, 1, 3, 4]
[-1, 1, 2, 2, 4]
[-1, 1, 2, 3, 3]
[0, 0, 1, 3, 4]
[0, 0, 2, 2, 4]
[0, 0, 2, 3, 3]
[0, 1, 1, 2, 4]
[0, 1, 1, 3, 3]
[0, 1, 2, 2, 3]

Answer 2

from itertools import combinations
a = [1,2,3,4]
for combination in combinations(range(len(a)),r=2):
    print [c-(1 if i in combination else 0) for i,c in enumerate(a)]

Python在for循环中更改列表元素

2 个答案: