不可变字典的可变包装

Question

Here is the problem; I have an immutable dictionary with huge set of items. The key type and the value types contained within this dict are themselves immutable. I would like to be able to mutate this dict (adding/removing/replacing key-value pairs) without having to do a full copy of the dict.

I am imagining some wrapper class for the immutable dict which adheres to the dict contract, and defaults to the immutable dict for values that have not been updated. I see the post How to “perfectly” override a dict? which I plan to leverage to make this wrapper.

Before I embark on implementing this design I just wanted to ask- is this construct already provided by the language? Or how else can I achieve the desired effect? I am on the latest version of Python (3.7) so I can use all language features available. Thanks!

Answer 1

Take a look at collections.ChainMap . It's a wrapper around multiple dictionaries: all writes go to the first dictionary, and lookups are searched in order of the maps. So I think you could just do something like:

modified_map = {}
mutable_map = collections.ChainMap(modified_map, huge_immutable_map)

Answer 2

Let's say you used a frozendict implementation like this one :

class frozendict(collections.Mapping):
    """
    An immutable wrapper around dictionaries that implements the complete :py:class:`collections.Mapping`
    interface. It can be used as a drop-in replacement for dictionaries where immutability is desired.
    """

    dict_cls = dict

    def __init__(self, *args, **kwargs):
        self._dict = self.dict_cls(*args, **kwargs)
        self._hash = None

    def __getitem__(self, key):
        return self._dict[key]

    def __contains__(self, key):
        return key in self._dict

    def copy(self, **add_or_replace):
        return self.__class__(self, **add_or_replace)

    def __iter__(self):
        return iter(self._dict)

    def __len__(self):
        return len(self._dict)

    def __repr__(self):
        return '<%s %r>' % (self.__class__.__name__, self._dict)

    def __hash__(self):
        if self._hash is None:
            h = 0
            for key, value in self._dict.items():
                h ^= hash((key, value))
            self._hash = h
        return self._hash

If you wanted to mutate it, you could just reach in and mutate self._dict :

d = frozendict({'a': 1, 'b': 2})
d['a'] = 3  # This fails
mutable_dict = d._dict
mutable_dict['a'] = 3  # This works
print(d['a'])

It's a little yucky reaching into a class's protected members, but I'd say that's okay because what you're trying to do is a little yucky. If you want a mutable dictionary (just a dict ) then use one. If you never want to mutate it, use a frozendict implementation like the one above. A hyrid of mutable and immutable makes no sense. All frozendict does is it doesn't implement the mutation dunder methods ( __setitem__ , __delitem__ , etc.). Under the hood a frozendict is represented by a regular, mutable dict .

The above approach is superior in my view to the one you linked. Composability (having frozendict have a _dict property) is much easier to reason about than inheritance (subclassing dict ) in many cases.