Python defaultdict and lambda

Question

In someone else\'s code I read the following two lines:

x = defaultdict(lambda: 0)
y = defaultdict(lambda: defaultdict(lambda: 0))

As the a

Answer 1

You are correct for what the first one does. As for y, it will create a defaultdict with default 0 when a key doesn't exist in y, so you can think of this as a nested dictionary. Consider the following example:

y = defaultdict(lambda: defaultdict(lambda: 0))
print y['k1']['k2']   # 0
print dict(y['k1'])   # {'k2': 0}

To create an equivalent nested dictionary structure without defaultdict you would need to create an inner dict for y['k1'] and then set y['k1']['k2'] to 0, but defaultdict does all of this behind the scenes when it encounters keys it hasn't seen:

y = {}
y['k1'] = {}
y['k1']['k2'] = 0

The following function may help for playing around with this on an interpreter to better your understanding:

def to_dict(d):
    if isinstance(d, defaultdict):
        return dict((k, to_dict(v)) for k, v in d.items())
    return d

This will return the dict equivalent of a nested defaultdict, which is a lot easier to read, for example:

>>> y = defaultdict(lambda: defaultdict(lambda: 0))
>>> y['a']['b'] = 5
>>> y
defaultdict( at 0xb7ea93e4>, {'a': defaultdict( at 0xb7ea9374>, {'b': 5})})
>>> to_dict(y)
{'a': {'b': 5}}