Python how to create a dict of dict of list with defaultdict

问题

How do I create a dict of dict of lists using defaultdict? I am getting the following error.

>>> from collections import defaultdict
>>> a=defaultdict()
>>> a["testkey"]=None
>>> a
defaultdict(None, {'testkey': None})
>>> a["testkey"]["list"]=[]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'NoneType' object does not support item assignment

回答1:

It's a little tricky. You make a defaultdict of defaultdicts, like so:

defaultdict(lambda: defaultdict(list))

回答2:

Slightly faster than using a lambda:

defaultdict(defaultdict(list).copy)

This has the same observable behavior as wim's answer, but avoids a lambda in favor of a (in CPython) bound built-in method implemented in C, which means default value generation doesn't have to execute any Python byte code or look up any names, and it runs a small amount faster. In microbenchmarks, it looks like the cost paid when a key did not exist at access time is about 5-10% lower this way than with the otherwise equivalent lambda.

Really, the reason I prefer it is because I hate lambda due to people overusing it when it's a bad idea (e.g. map/filter with lambda is always more verbose and slower than an equivalent listcomp/genexpr, but people keep doing it anyway for no discernible reason), even though in this case it hardly matters.

回答3:

You may have to do like this.

>>> from collections import defaultdict
>>> a=defaultdict()
>>> a["testkey"]=None
>>> a["testkey"]=defaultdict(list)
>>> a["testkey"]["list"]=["a","b","c"]
>>> a
defaultdict(None, {'testkey': defaultdict(<type 'list'>, {'list': ['a', 'b', 'c']})})

来源：https://stackoverflow.com/questions/33080869/python-how-to-create-a-dict-of-dict-of-list-with-defaultdict