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How to initialize a dict with keys from a list and empty value in Python?


Question

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}
2016/07/22
1
261
7/22/2016 7:30:37 PM

Accepted Answer

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.)

2010/02/11
403
2/11/2010 6:36:40 AM


dict.fromkeys(keys, None)
2010/02/11

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

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

Output:

{'a': [], 'c': [], 'b': [], 'd': []}
2015/08/24

d = {}
for i in keys:
    d[i] = None
2016/05/24

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])
2018/12/17

Source: https://stackoverflow.com/questions/2241891
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