I saw this example at pythontips. I do not understand the second line when defaultdict takes an argument "tree" and return a "tree".
import collections
tree = lambda: collections.defaultdict(tree)
some_dict = tree()
some_dict['color']['favor'] = "yellow"
# Works fine
After I run this code, I checked the type of some_dict
defaultdict(< function < lambda > at 0x7f19ae634048 >,
{'color': defaultdict(
< function < lambda > at 0x7f19ae634048 >, {'favor': 'yellow'})})
defaultdict takes a zero-argument callable to its constructor, which is called when the key is not found, as you correctly explained. lambda: 0 will of course always return zero, but the preferred method to do that is defaultdict(int) , which will do the same thing.
A defaultdict works exactly like a normal dict, but it is initialized with a function (“default factory”) that takes no arguments and provides the default value for a nonexistent key. A defaultdict will never raise a KeyError. Any key that does not exist gets the value returned by the default factory.
Just use the same syntax as you would on a regular dictionary. del mydict['stock2'] for example, or mydict. pop('stock2') if you want the value returned at the same time. Note that you don't need to be using a defaultdict at all in your example.
defaultdict objects Return 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.
This is a pretty clever way to create a recursive defaultdict
. It's a little tricky to understand at first but once you dig into what's happening, it's actually a pretty simple use of recursion.
In this example, we define a recursive lambda function, tree
, that returns a defaultdict
whose constructor is tree
. Let's rewrite this using regular functions for clarity.
from collections import defaultdict
from pprint import pprint
def get_recursive_dict():
return defaultdict(get_recursive_dict)
Note that we're returning defaultdict(get_recursive_dict)
and not defaultdict(get_recursive_dict())
. We want to pass defaultdict
a callable object (i.e. the function get_recursive_dict
). Actually calling get_recursive_dict()
would result in infinite recursion.
If we call get_recursive_dict
, we get an empty defaultdict
whose default value is the function get_recursive_dict
.
recursive_dict = get_recursive_dict()
print(recursive_dict)
# defaultdict(<function get_recursive_dict at 0x0000000004FFC4A8>, {})
Let's see this in action. Create the key 'alice'
and it's corresponding value defaults to an empty defaultdict
whose default value is the function get_recursive_dict
. Notice that this is the same default value as our recursive_dict
!
print(recursive_dict['alice'])
# defaultdict(<function get_recursive_dict at 0x0000000004AF46D8>, {})
print(recursive_dict)
# defaultdict(<function get_recursive_dict at 0x0000000004AF46D8>, {'alice': defaultdict(<function get_recursive_dict at 0x0000000004AF46D8>, {})})
So we can create as many nested dictionaries as we want.
recursive_dict['bob']['age'] = 2
recursive_dict['charlie']['food']['dessert'] = 'cake'
print(recursive_dict)
# defaultdict(<function get_recursive_dict at 0x00000000049BD4A8>, {'charlie': defaultdict(<function get_recursive_dict at 0x00000000049BD4A8>, {'food': defaultdict(<function get_recursive_dict at 0x00000000049BD4A8>, {'dessert': 'cake'})}), 'bob': defaultdict(<function get_recursive_dict at 0x00000000049BD4A8>, {'age': 2}), 'alice': defaultdict(<function get_recursive_dict at 0x00000000049BD4A8>, {})})
Once you overwrite the default value with a key, you can no longer create arbitrarily deep nested dictionaries.
recursive_dict['bob']['age']['year'] = 2016
# TypeError: 'int' object does not support item assignment
I hope this clears things up!
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With