这可能是一个愚蠢的问题，但鉴于以下 dict：
combination_dict = {"one": [1, 2, 3], "two": [2, 3, 4], "three": [3, 4, 5]}
我将如何实现此列表：
result_list = [{"one": [1, 2, 3], "two": [2, 3, 4]}, {"one": [1, 2, 3], "three": [3, 4, 5]}, {"two": [2, 3, 4], "three": [3, 4, 5]}]
换句话说，我希望字典中两个键/值对的所有组合都没有替换，而不管顺序如何。
一种解决方案是使用itertools.combinations()：
result_list = map(dict, itertools.combinations(
combination_dict.iteritems(), 2))
编辑：由于大众需求，这里有一个 Python 3.x 版本：
result_list = list(map(dict, itertools.combinations(
combination_dict.items(), 2)))
我更喜欢@JollyJumper 的解决方案以提高可读性，尽管这个解决方案执行得更快
>>> from itertools import combinations
>>> d = {"one": [1, 2, 3], "two": [2, 3, 4], "three": [3, 4, 5]}
>>> [{j: d[j] for j in i} for i in combinations(d, 2)]
[{'three': [3, 4, 5], 'two': [2, 3, 4]}, {'three': [3, 4, 5], 'one': [1, 2, 3]}, {'two': [2, 3, 4], 'one': [1, 2, 3]}]
时间：
>python -m timeit -s "d = {'three': [3, 4, 5], 'two': [2, 3, 4], 'one': [1, 2, 3]}; from itertools import combinations" "map(dict, combinations(d.iteritems(), 2))"
100000 loops, best of 3: 3.27 usec per loop
>python -m timeit -s "d = {'three': [3, 4, 5], 'two': [2, 3, 4], 'one': [1, 2, 3]}; from itertools import combinations" "[{j: d[j] for j in i} for i in combinations(d, 2)]"
1000000 loops, best of 3: 1.92 usec per loop
from itertools import combinations
combination_dict = {"one": [1, 2, 3], "two": [2, 3, 4], "three": [3, 4, 5]}
lis=[]
for i in range(1,len(combination_dict)):
for x in combinations(combination_dict,i):
dic={z:combination_dict[z] for z in x}
lis.append(dic)
print lis
输出：
[{'three': [3, 4, 5]}, {'two': [2, 3, 4]}, {'one': [1, 2, 3]}, {'three': [3, 4, 5], 'two': [2, 3, 4]}, {'three': [3, 4, 5], 'one': [1, 2, 3]}, {'two': [2, 3, 4], 'one': [1, 2, 3]}]
我相信这会给你你所需要的。
result list = [{combination_dict['one','two'],combination_dict['one','three']}]
我发现本教程非常有帮助：
http://bdhacker.wordpress.com/2010/02/27/python-tutorial-dictionaries-key-value-pair-maps-basics/