How does collections.defaultdict work?

Question

I\'ve read the examples in python docs, but still can\'t figure out what this method means. Can somebody help? Here are two examples from the python docs

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Answer 1

Since the question is about "how it works", some readers may want to see more nuts and bolts. Specifically, the method in question is the __missing__(key) method. See: https://docs.python.org/2/library/collections.html#defaultdict-objects .

More concretely, this answer shows how to make use of __missing__(key) in a practical way: https://stackoverflow.com/a/17956989/1593924

To clarify what 'callable' means, here's an interactive session (from 2.7.6 but should work in v3 too):

>>> x = int
>>> x

>>> y = int(5)
>>> y
5
>>> z = x(5)
>>> z
5

>>> from collections import defaultdict
>>> dd = defaultdict(int)
>>> dd
defaultdict(, {})
>>> dd = defaultdict(x)
>>> dd
defaultdict(, {})
>>> dd['a']
0
>>> dd
defaultdict(, {'a': 0})

That was the most typical use of defaultdict (except for the pointless use of the x variable). You can do the same thing with 0 as the explicit default value, but not with a simple value:

>>> dd2 = defaultdict(0)

Traceback (most recent call last):
  File "", line 1, in 
    dd2 = defaultdict(0)
TypeError: first argument must be callable

Instead, the following works because it passes in a simple function (it creates on the fly a nameless function which takes no arguments and always returns 0):

>>> dd2 = defaultdict(lambda: 0)
>>> dd2
defaultdict( at 0x02C4C130>, {})
>>> dd2['a']
0
>>> dd2
defaultdict( at 0x02C4C130>, {'a': 0})
>>> 

And with a different default value:

>>> dd3 = defaultdict(lambda: 1)
>>> dd3
defaultdict( at 0x02C4C170>, {})
>>> dd3['a']
1
>>> dd3
defaultdict( at 0x02C4C170>, {'a': 1})
>>>