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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]))]