How to natively increment a dictionary element's value?

Question

When working with Python 3 dictionaries, I keep having to do something like this:

d=dict()
if 'k' in d:
    d['k']+=1
else:
    d['k']=0

I seem to remember there being a native way to do this, but was looking through the documentation and couldn't find it. Do you know what this is?

Answer 1

This is the use case for collections.defaultdict, here simply using the int callable for the default factory.

>>> from collections import defaultdict
>>> d = defaultdict(int)
>>> d
defaultdict(<class 'int'>, {})
>>> d['k'] +=1
>>> d
defaultdict(<class 'int'>, {'k': 1})

A defaultdict is configured to create items whenever a missing key is searched. You provide it with a callable (here int()) which it uses to produce a default value whenever the lookup with __getitem__ is passed a key that does not exist. This callable is stored in an instance attribute called default_factory.

If you don't provide a default_factory, you'll get a KeyError as per usual for missing keys.

Then suppose you wanted a different default value, perhaps 1 instead of 0. You would simply have to pass a callable that provides your desired starting value, in this case very trivially

>>> d = defaultdict(lambda: 1)

This could obviously also be any regular named function.

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It's worth noting however that if in your case you are attempting to just use a dictionary to store the count of particular values, a collections.Counter is more suitable for the job.

>>> from collections import Counter
>>> Counter('kangaroo')
Counter({'a': 2, 'o': 2, 'n': 1, 'r': 1, 'k': 1, 'g': 1})