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 argument of defaultdict is a default factory, I think the first line means that when I call x[k] for a nonexistent key k (such as a statement like v=x[k]), the key-value pair (k,0) will be automatically added to the dictionary, as if the statement x[k]=0 is first executed. Am I correct?

And what about y? It seems that the default factory will create a defaultdict with default 0. But what does that mean concretely? I tried to play around with it in Python shell, but couldn't figure out what it is exactly.

Answer 1

I think the first line means that when I call x[k] for a nonexistent key k (such as a statement like v=x[k]), the key-value pair (k,0) will be automatically added to the dictionary, as if the statement x[k]=0 is first executed.

That's right. This is more idiomatically written

x = defaultdict(int) 

In the case of y, when you do y["ham"]["spam"], the key "ham" is inserted in y if it does not exist. The value associated with it becomes a defaultdict in which "spam" is automatically inserted with a value of 0.

I.e., y is a kind of "two-tiered" defaultdict. If "ham" not in y, then evaluating y["ham"]["spam"] is like doing

y["ham"] = {} y["ham"]["spam"] = 0 

in terms of ordinary dict.

Answer 2

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(<function <lambda> at 0xb7ea93e4>, {'a': defaultdict(<function <lambda> at 0xb7ea9374>, {'b': 5})}) >>> to_dict(y) {'a': {'b': 5}}