5.3.2.1 defaultdict Examples

Using list as the default_factory, it is easy to group a sequence of key-value pairs into a dictionary of lists:

>>> s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]
>>> d = defaultdict(list)
>>> for k, v in s:
        d[k].append(v)

>>> d.items()
[('blue', [2, 4]), ('red', [1]), ('yellow', [1, 3])]

When each key is encountered for the first time, it is not already in the mapping; so an entry is automatically created using the default_factory function which returns an empty list. The list.append() operation then attaches the value to the new list. When keys are encountered again, the look-up proceeds normally (returning the list for that key) and the list.append() operation adds another value to the list. This technique is simpler and faster than an equivalent technique using dict.setdefault():

>>> d = {}
>>> for k, v in s:
	d.setdefault(k, []).append(v)

>>> d.items()
[('blue', [2, 4]), ('red', [1]), ('yellow', [1, 3])]

Setting the default_factory to int makes the defaultdict useful for counting (like a bag or multiset in other languages):

>>> s = 'mississippi'
>>> d = defaultdict(int)
>>> for k in s:
        d[k] += 1

>>> d.items()
[('i', 4), ('p', 2), ('s', 4), ('m', 1)]

When a letter is first encountered, it is missing from the mapping, so the default_factory function calls int() to supply a default count of zero. The increment operation then builds up the count for each letter.

The function int() which always returns zero is just a special case of constant functions. A faster and more flexible way to create constant functions is to use itertools.repeat() which can supply any constant value (not just zero):

>>> def constant_factory(value):
...     return itertools.repeat(value).next
>>> d = defaultdict(constant_factory('<missing>'))
>>> d.update(name='John', action='ran')
>>> '%(name)s %(action)s to %(object)s' % d
'John ran to <missing>'

Setting the default_factory to set makes the defaultdict useful for building a dictionary of sets:

>>> s = [('red', 1), ('blue', 2), ('red', 3), ('blue', 4), ('red', 1), ('blue', 4)]
>>> d = defaultdict(set)
>>> for k, v in s:
        d[k].add(v)

>>> d.items()
[('blue', set([2, 4])), ('red', set([1, 3]))]