Creating a dictionary from an iterable

Question

What is the easiest way to create a dictionary from an iterable and assigning it some default value? I tried:

>>> x = dict(zip(range(0, 10), range(0))) 

But that doesn't work since range(0) is not an iterable as I thought it would not be (but I tried anyways!)

So how do I go about it? If I do:

>>> x = dict(zip(range(0, 10), 0)) Traceback (most recent call last):   File "<stdin>", line 1, in <module> TypeError: zip argument #2 must support iteration 

This doesn't work either. Any suggestions?

Answer 1

In python 3, You can use a dict comprehension.

>>> {i:0 for i in range(0,10)} {0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0, 6: 0, 7: 0, 8: 0, 9: 0} 

Fortunately, this has been backported in python 2.7 so that's also available there.

Answer 2

You need the dict.fromkeys method, which does exactly what you want.

From the docs:

fromkeys(...)     dict.fromkeys(S[,v]) -> New dict with keys from S and values equal to v.     v defaults to None. 

So what you need is:

>>> x = dict.fromkeys(range(0, 10), 0) >>> x {0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0, 6: 0, 7: 0, 8: 0, 9: 0}