How to Initialize a Dict with Keys from a List and Empty Value in Python?

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}

dict.fromkeys(keys, None)

>>> keyDict = {"a","b","c","d"}

>>> dict([(key, []) for key in keyDict])

Output:

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

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

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])

Just because it's fun how the dict constructor works nicely with zip, you can repeat the default value and zip it to the keys:

from itertools import repeat

keys = [1, 2, 3]
default_value = None

d = dict(zip(keys, repeat(default_value)))
print(d)

Will give:

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

repeat creates an infinite iterator of the element passed to it but as zip stops on the shortest iterable all works well.