Why does `frozenset.copy()` exist? even though it doesn't copy the frozenset

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1

frozenset.copy() doesn’t copy the frozenset to avoid wasting memory according to the post as shown below:

import copy

A = frozenset([0, 1, 2])

B = A.copy()

print(A is B)
# True

Now, why does frozenset.copy() exist? even though it doesn’t copy the frozenset.

2

It’s so that you can write functions that for instance take both a set and frozenset, then just call .copy() and have it still work. There’s a couple of these “useless” functions here and there. Another example is int.is_integer(), which matches float.is_integer().

3

While there’s no reasonable use case to create a copy of a frozenset, the existence of frozenset.copy provides an interface for a frozenset subclass, where creating a copy may actually make sense.

Note that this outputs False as intended:

class S(frozenset):
    pass

A = S([0, 1, 2])
B = A.copy()
print(A is B)
2 Likes
4

So, people already told you about subclasses.

Correct, for bultin type (frozenset) this just make a new reference. But what’s the difference? Remember, frozenset entries must be hashable.