initialize and add value to a list in dictionary at one step in Python 3.7

Question

I have a dictionary where key is string and value is list.

Now while adding a value associated with given key, I always have to check if there is any list yet, otherwise

Answer 1

There are at least three ways:

Use dict.setdefault

>>> data = {}
>>> data.setdefault('foo', []).append(42)
>>> data
{'foo': [42]}

Use defaultdict, which unlike .setdefault, takes a callable:

>>> from collections import defaultdict
>>> data = defaultdict(list)
>>> data
defaultdict(<class 'list'>, {})
>>> data['foo'].append(42)
>>> data
defaultdict(<class 'list'>, {'foo': [42]})

Finally, subclass dict and implement __missing__:

>>> class MyDict(dict):
...     def __missing__(self, key):
...         self[key] = value  = []
...         return value
...
>>> data = MyDict()
>>> data['foo'].append(42)
>>> data
{'foo': [42]}

Note, you can think of the last one as the most flexible, you have access to the actual key that's missing when you deal with it. defaultdict is a class factory, and it generates a subclass of dict as well. But, the callable is not passed any arguments, nevertheless, it is sufficient for most needs.

Answer 2

You can use dict.setdefault:

>>> d = {}
>>> d.setdefault('k', []).append(1)
>>> d
{'k': [1]}
>>> d.setdefault('k', []).append(2)
>>> d
{'k': [1, 2]}

Help on method_descriptor in dict:

dict.setdefault = setdefault(...) D.setdefault(k[,d]) -> D.get(k,d), also set D[k]=d if k not in D