问题：如何在Python中使用列表中的键和空值初始化字典？
我想从中得到：
keys = [1,2,3]
对此：
{1: None, 2: None, 3: None}
有pythonic的方法吗？
这是一个丑陋的方法：
>>> keys = [1,2,3]
>>> dict([(1,2)])
{1: 2}
>>> dict(zip(keys, [None]*len(keys)))
{1: None, 2: None, 3: None}
I’d like to get from this:
keys = [1,2,3]
to this:
{1: None, 2: None, 3: None}
Is there a pythonic way of doing it?
This is an ugly way to do it:
>>> keys = [1,2,3]
>>> dict([(1,2)])
{1: 2}
>>> dict(zip(keys, [None]*len(keys)))
{1: None, 2: None, 3: None}
回答 0
dict.fromkeys([1, 2, 3, 4])
这实际上是一个类方法，因此它也适用于dict子类（如collections.defaultdict）。可选的第二个参数指定用于键的值（默认为）None。
dict.fromkeys([1, 2, 3, 4])
This is actually a classmethod, so it works for dict-subclasses (like collections.defaultdict) as well. The optional second argument specifies the value to use for the keys (defaults to None.)
回答 1
没有人愿意提供字典理解的解决方案吗？
>>> 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}
nobody cared to give a dict-comprehension solution ?
>>> 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}
回答 2
dict.fromkeys(keys, None)
dict.fromkeys(keys, None)
回答 3
>>> keyDict = {"a","b","c","d"}
>>> dict([(key, []) for key in keyDict])
输出：
{'a': [], 'c': [], 'b': [], 'd': []}
>>> keyDict = {"a","b","c","d"}
>>> dict([(key, []) for key in keyDict])
Output:
{'a': [], 'c': [], 'b': [], 'd': []}
回答 4
d = {}
for i in keys:
d[i] = None
d = {}
for i in keys:
d[i] = None
回答 5
在许多地方要附加任意键默认/初始值的工作流程，你没有需要到哈希每个键单独的时间提前。您可以使用collections.defaultdict。例如：
from collections import defaultdict
d = defaultdict(lambda: None)
print(d[1]) # None
print(d[2]) # None
print(d[3]) # None
这样效率更高，它省去了在实例化时散列所有键的麻烦。此外，defaultdict是的子类dict，因此通常无需转换回常规字典。
对于需要对许可键进行控制的工作流，可以dict.fromkeys按照已接受的答案使用：
d = dict.fromkeys([1, 2, 3, 4])
In many workflows where you want to attach a default / initial value for arbitrary keys, you don’t need to hash each key individually ahead of time. You can use collections.defaultdict. For example:
from collections import defaultdict
d = defaultdict(lambda: None)
print(d[1]) # None
print(d[2]) # None
print(d[3]) # None
This is more efficient, it saves having to hash all your keys at instantiation. Moreover, defaultdict is a subclass of dict, so there’s usually no need to convert back to a regular dictionary.
For workflows where you require controls on permissible keys, you can use dict.fromkeys as per the accepted answer:
d = dict.fromkeys([1, 2, 3, 4])
