How does collections.defaultdict work?

Question

I\'ve read the examples in python docs, but still can\'t figure out what this method means. Can somebody help? Here are two examples from the python docs

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Answer 1

defaultdict

"The standard dictionary includes the method setdefault() for retrieving a value and establishing a default if the value does not exist. By contrast, defaultdict lets the caller specify the default(value to be returned) up front when the container is initialized."

as defined by Doug Hellmann in The Python Standard Library by Example

How to use defaultdict

Import defaultdict

>>> from collections import defaultdict

Initialize defaultdict

Initialize it by passing

callable as its first argument(mandatory)

>>> d_int = defaultdict(int)
>>> d_list = defaultdict(list)
>>> def foo():
...     return 'default value'
... 
>>> d_foo = defaultdict(foo)
>>> d_int
defaultdict(, {})
>>> d_list
defaultdict(, {})
>>> d_foo
defaultdict(, {})

**kwargs as its second argument(optional)

>>> d_int = defaultdict(int, a=10, b=12, c=13)
>>> d_int
defaultdict(, {'a': 10, 'c': 13, 'b': 12})

or

>>> kwargs = {'a':10,'b':12,'c':13}
>>> d_int = defaultdict(int, **kwargs)
>>> d_int
defaultdict(, {'a': 10, 'c': 13, 'b': 12})

How does it works

As is a child class of standard dictionary, it can perform all the same functions.

But in case of passing an unknown key it returns the default value instead of error. For ex:

>>> d_int['a']
10
>>> d_int['d']
0
>>> d_int
defaultdict(, {'a': 10, 'c': 13, 'b': 12, 'd': 0})

In case you want to change default value overwrite default_factory:

>>> d_int.default_factory = lambda: 1
>>> d_int['e']
1
>>> d_int
defaultdict( at 0x7f34a0a91578>, {'a': 10, 'c': 13, 'b': 12, 'e': 1, 'd': 0})

or

>>> def foo():
...     return 2
>>> d_int.default_factory = foo
>>> d_int['f']
2
>>> d_int
defaultdict(, {'a': 10, 'c': 13, 'b': 12, 'e': 1, 'd': 0, 'f': 2})

Examples in the Question

Example 1

As int has been passed as default_factory, any unknown key will return 0 by default.

Now as the string is passed in the loop, it will increase the count of those alphabets in d.

>>> s = 'mississippi'
>>> d = defaultdict(int)
>>> d.default_factory

>>> for k in s:
...     d[k] += 1
>>> d.items()
[('i', 4), ('p', 2), ('s', 4), ('m', 1)]
>>> d
defaultdict(, {'i': 4, 'p': 2, 's': 4, 'm': 1})

Example 2

As a list has been passed as default_factory, any unknown(non-existent) key will return [ ](ie. list) by default.

Now as the list of tuples is passed in the loop, it will append the value in the d[color]

>>> s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]
>>> d = defaultdict(list)
>>> d.default_factory

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