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

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}

default_keys = [1, "name"]

To get dictionary with None as values:

dict.fromkeys(default_keys)  

Output :

{1: None, 'name': None}

To get dictionary with default values:

dict.fromkeys(default_keys, [])  

Output :

{1: [], 'name': []}

dict.fromkeys(keys, None)

You could use dict.fromkeys as follows:

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

This will create a list object for each key. If you change value for any specific key it won't affect other keys (as most people would want, I presume).

d = {}
for i in keys:
    d[i] = 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}

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

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

Output:

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

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])