[Python-Dev] Add a frozendict builtin type (original) (raw)

Victor Stinner victor.stinner at haypocalc.com
Tue Feb 28 13:14:15 CET 2012

Updated patch and more justifications.

New patch:

 * frozendict.hash computes hash(frozenset(self.items())) and caches the result is its private hash attribute

hash(frozenset(self.items())) is preferred over hash(sorted(self.items())) because keys and values may be unorderable. frozenset() is faster than sorted(): O(n) vs O(n*log(n)).

frozendict hash doesn't care of the item order creation:

a=frozendict.fromkeys('ai') a frozendict({'a': None, 'i': None}) b=frozendict.fromkeys('ia') b frozendict({'i': None, 'a': None}) hash(a) == hash(b) True a == b True tuple(a.items()) == tuple(b.items()) False

frozendict supports unorderable keys and values:

hash(frozendict({b'abc': 1, 'abc': 2})) 935669091 hash(frozendict({1: b'abc', 2: 'abc'})) 1319859033

 * Add a frozendict abstract base class to collections?

I realized that Mapping already exists and so the following patch is enough:

+Mapping.register(frozendict)

See also the PEP 351.

I read the PEP and the email explaining why it was rejected.

Just to be clear: the PEP 351 tries to freeze an object, try to convert a mutable or immutable object to an immutable object. Whereas my frozendict proposition doesn't convert anything: it just raises a TypeError if you use a mutable key or value.

For example, frozendict({'list': ['a', 'b', 'c']}) doesn't create frozendict({'list': ('a', 'b', 'c')}) but raises a TypeError.

Victor -------------- next part -------------- A non-text attachment was scrubbed... Name: frozendict-2.patch Type: text/x-patch Size: 29099 bytes Desc: not available URL: <http://mail.python.org/pipermail/python-dev/attachments/20120228/c2ed03a7/attachment-0001.bin>

More information about the Python-Dev mailing list