Why is reduce() is not working with a defaultdict object in the following case:
>>> from collections import defaultdict
>>> d = defaultdict(lambda: defaultdict(int))
>>> reduce(dict.get, [1,2], d)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: descriptor 'get' requires a 'dict' object but received a 'NoneType'
The above reduce is equivalent to d[1][2], so I expected 0, i.e. the default return value as output but I got a NoneType exception instead.
If I use d[1][2] it works fine as shown below:
>>> d[1][2]
0
Am I doing anything wrong?
dict.get() returns None if a key is not present, even for a defaultdict object.
That's because the job of dict.get() is to return a default value instead when a key is missing. If it didn't you'd never be able to return a different default (the second argument to dict.get()):
>>> from collections import defaultdict
>>> d = defaultdict(lambda: defaultdict(int))
>>> d.get(1, 'default')
'default'
In other words, your reduce(dict.get, ..) function is not the equivalent of the d[1][2] expression. It is the equivalent of d.get(1).get(2), which fails the exact same way:
>>> d = defaultdict(lambda: defaultdict(int))
>>> d.get(1).get(2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'NoneType' object has no attribute 'get'
Use dict.__getitem__ instead if you want to rely on the auto-insertion behaviour:
>>> reduce(dict.__getitem__, [1,2], d)
0
The expression d[1] translates directly to d.__getitem__(1).