How do I create a dictionary with keys from a list and values defaulting to (say) zero?


I have a = [1,2,3,4] and I want d = {1:0, 2:0, 3:0, 4:0}

d = dict(zip(q,[0 for x in range(0,len(q))]))

works but is ugly. What’s a cleaner way?

Asked By: blahster



d = dict([(x,0) for x in a])

**edit Tim’s solution is better because it uses generators see the comment to his answer.

Answered By: GWW

dict((el,0) for el in a) will work well.

Python 2.7 and above also support dict comprehensions. That syntax is {el:0 for el in a}.

Answered By: Tim McNamara

In addition to Tim’s answer, which is very appropriate to your specific example, it’s worth mentioning collections.defaultdict, which lets you do stuff like this:

>>> d = defaultdict(int)
>>> d[0] += 1
>>> d
{0: 1}
>>> d[4] += 1
>>> d
{0: 1, 4: 1}

For mapping [1, 2, 3, 4] as in your example, it’s a fish out of water. But depending on the reason you asked the question, this may end up being a more appropriate technique.

Answered By: intuited
d = dict.fromkeys(a, 0)

a is the list, 0 is the default value. Pay attention not to set the default value to some mutable object (i.e. list or dict), because it will be one object used as value for every key in the dictionary (check here for a solution for this case). Numbers/strings are safe.

Answered By: eumiro

In python version >= 2.7 and in python 3:

d = {el:0 for el in a}
Answered By: Andrey
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