How to initialize a dict with keys from a list and empty value in Python?
DictionaryPythonDictionary Problem Overview
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}
Dictionary Solutions
Solution 1 - Dictionary
You can use fromkeys
:
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
.)
Solution 2 - Dictionary
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}
Solution 3 - Dictionary
dict.fromkeys(keys, None)
Solution 4 - Dictionary
>>> keyDict = {"a","b","c","d"}
>>> dict([(key, []) for key in keyDict])
Output:
{'a': [], 'c': [], 'b': [], 'd': []}
Solution 5 - Dictionary
d = {}
for i in keys:
d[i] = None
Solution 6 - Dictionary
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])
Solution 7 - Dictionary
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.