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.