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

This question already has answers here: How to initialize a dict with keys from a list and empty value in Python? (7 answers) 开发者_JAVA技巧 Closed 6 years ago.

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?

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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}.

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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.

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In python version >= 2.7 and in python 3:

d = {el:0 for el in a}

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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.

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

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