Why defaultdict constructor takes a function and not a constant

Question

This is how defaultdict works:

from collections import defaultdict
a=defaultdict(lambda:3)
a[200]==3 #True

Why is it that defaultdict was designed to take a function with no arguments, and not simply a constant value?

---

Here's the alternative definition.

class dd(dict):
    def __init__(self,x):
        self._default=x
    def __getitem__(self,key):
        if key in self: return dict.__getitem__(self, key)
        else:
            self[key]=self._default
            return self[key]

So that

a=dd(3)
a[200]==3 #True

Answer 1

Because if you want the default value to be a mutable object, you probably want it to be a different mutable object for each key.

If you passed a constant and did defaultdict([]), then every time a missing key was accessed, its value would be set to the same list. Then you'd get this:

>>> d = defaultdict([])
>>> d[1].append("Whoops")
>>> print d[2]
["Whoops"]

Having a mutable default value is in fact very common and useful, since it lets you do things like d[key].append("Blah") without having to first check that d[key] exists. For this case, you need some way to return a new value each time, and the simplest way to do that is to have a callable that returns a default value.