我需要做的是像这样转换
{'key1': [1, 2, 3], 'key2': [4, 5, 6]}
进入
[{'key1': 1, 'key2': 4}, {'key1': 2, 'key2': 5}, {'key1': 3, 'key2': 6}]
值列表的长度可以变化! 执行此操作的最快方法是什么(最好没有 for 循环)？
最佳答案
适用于任意数量的键
>>> map(dict, zip(*[[(k, v) for v in value] for k, value in d.items()]))
[{'key2': 4, 'key1': 1}, {'key2': 5, 'key1': 2}, {'key2': 6, 'key1': 3}]
例如:
d = {'key3': [7, 8, 9], 'key2': [4, 5, 6], 'key1': [1, 2, 3]}
>>> map(dict, zip(*[[(k, v) for v in value] for k, value in d.items()]))
[{'key3': 7, 'key2': 4, 'key1': 1}, {'key3': 8, 'key2': 5, 'key1': 2}, {'key3': 9, 'key2': 6, 'key1': 3}]
适用于任意数量的值或键的通用解决方案:(python2.6)
>>> from itertools import izip_longest
>>> d = {'key2': [3, 4, 5, 6], 'key1': [1, 2]}
>>> map(lambda a: dict(filter(None, a)), izip_longest(*[[(k, v) for v in value] for k, value in d.items()]))
[{'key2': 3, 'key1': 1}, {'key2': 4, 'key1': 2}, {'key2': 5}, {'key2': 6}]
如果你没有python2.6:
>>> d = {'key2': [3, 4, 5, 6], 'key1': [1, 2]}
>>> map(lambda a: dict(filter(None, a)), map(None, *[[(k, v) for v in value] for k, value in d.items()]))
[{'key2': 3, 'key1': 1}, {'key2': 4, 'key1': 2}, {'key2': 5}, {'key2': 6}]
关于python - 将列表字典拆分为字典列表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/1780174/