Python collections.defaultdict()

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Documentation

class collections.defaultdict([default_factory[, ...]])

Returns a new dictionary-like object. defaultdict is a subclass of the built-in dict class. It overrides one method and adds one writable instance variable. The remaining functionality is the same as for the dictclass and is not documented here.

The first argument provides the initial value for the default_factory attribute; it defaults to None. All remaining arguments are treated the same as if they were passed to the dict constructor, including keyword arguments.

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

A faster and more flexible way to create constant functions is to use a lambda function which can supply any constant value (not just zero):

>>> def constant_factory(value):
...     return lambda: value
>>> 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)
...
>>> sorted(d.items())
[('blue', {2, 4}), ('red', {1, 3})]