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

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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.

Further, note that the defaultdict and __missing__ approaches will keep the default behavior, this may be undesirable after you create your data structure, you probably want a KeyError usually, or at least, you don’t want mydict[key] to add a key anymore.

In both cases, you can just create a regular dict from the dict subclasses, e.g. dict(data). This should generally be very fast, even for large dict objects, especially if it is a one-time cost. For defaultdict, you can also set the default_factory to None and the old behavior returns:

>>> data = defaultdict(list)
>>> data
defaultdict(<class 'list'>, {})
>>> data['foo']
[]
>>> data['bar']
[]
>>> data
defaultdict(<class 'list'>, {'foo': [], 'bar': []})
>>> data.default_factory = None
>>> data['baz']
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 'baz'
>>>

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