如何用列表中的键和Python中的空值初始化字典？

我想从这里得到：

keys = [1,2,3] 

对此：

 {1: None, 2: None, 3: None} 

有没有pythonic的做法呢？

这是一个丑陋的做法：

 >>> keys = [1,2,3] >>> dict([(1,2)]) {1: 2} >>> dict(zip(keys, [None]*len(keys))) {1: None, 2: None, 3: None} 

Answer 1

dict.fromkeys([1, 2, 3, 4])

这实际上是一个classmethod，所以它也适用于dict-subclasses（比如collections.defaultdict ）。 可选的第二个参数指定用于键的值（默认为None 。

Answer 2

没有人关心提供一个字典的理解解决scheme？

 >>> keys = [1,2,3,5,6,7] >>> {key: None for key in keys} {1: None, 2: None, 3: None, 5: None, 6: None, 7: None} 

Answer 3

 dict.fromkeys(keys, None) 

Answer 4

 >>> keyDict = {"a","b","c","d"} >>> dict([(key, []) for key in keyDict]) 

输出：

 {'a': [], 'c': [], 'b': [], 'd': []} 

Answer 5

 d = {} for i in keys: d[i] = None 

Answer 6

 d = {} for i in range(len(keys)): d.update({i+1:[]})