python defaultdict 简单了解

首先看官方文档doc

总结：

1. 使用一个factory函数作为defaultdict对象的参数

   In [1]: from collections import defaultdict
   
   In [2]: d = defaultdict(list)
   

2. 当给d中不存在的键k赋值时，d会使用初始化时的factory函数初始化键k的对应的值

   In [3]: d['a']
   Out[3]: []
   

   但是如果初始化时，不提供这样的factory函数会报异常
   使用int初始化时，默认值为0

   In [4]: d = defaultdict(int)
   In [5]: d['s']
   Out[5]: 0
   

3. 其他与普通dict使用方式相同

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 dict class 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.

New in version 2.5.

defaultdict objects support the following method in addition to the standard dict operations:

__missing__(key)
If the default_factory attribute is None, this raises a KeyError exception with the key as argument.

If default_factory is not None, it is called without arguments to provide a default value for the given key, this value is inserted in the dictionary for the key, and returned.

If calling default_factory raises an exception this exception is propagated unchanged.

This method is called by the __getitem__() method of the dict class when the requested key is not found; whatever it returns or raises is then returned or raised by __getitem__().

Note that __missing__() is not called for any operations besides __getitem__(). This means that get() will, like normal dictionaries, return None as a default rather than using default_factory.

defaultdict objects support the following instance variable:

default_factory
This attribute is used by the __missing__() method; it is initialized from the first argument to the constructor, if present, or to None, if absent.

8.3.3.1. 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]))]