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I was doing some coding exercises and I ended up using a set of frozensets. Here is the code:

cities = 4
roads = [[0, 1], [1, 2], [2, 0]]
roads = set([frozenset(road) for road in roads])
    
output = []
    
for i in range(cities-1):
    for j in range(i+1, cities):
        if set([i,j]) not in roads:
            output.append([i,j])

As you can see, the if in the nested for tests for the presence of the set in the set of sets. However, it was my understanding that in this case, hashables need to be used with the in operator.

If I replace set([i,j]) with [i,j], I do get the following error:

TypeError: unhashable type: 'list'

So, here is my question: why does it work with the set, which is not (as far as I know) hashable and not with the list? Should it not also throw an error, what am I missing?

Ben
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2 Answers2

5

From my reading of the CPython source it appears that the test for contains checks if the key is found in the set; if not, and if the key is a set object, an attempt is made to convert the key to a frozenset, and then that key is tested. The same behavior exists for operations like remove, as seen here:

>>> s = set([frozenset([1,2])])
>>> s
{frozenset({1, 2})}
>>> s.remove(set([1,2]))
>>> s
set()

The code in question in the interpreter is the set_contains() function in Objects/setobject.c.

sj95126
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2

Nevermind, found the answer in the documentation, for anyone wondering:

Note, the elem argument to the __contains__(), remove(), and discard() methods may be a set. To support searching for an equivalent frozenset, a temporary one is created from elem.

Ben
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    I do not think this answers your question. You essentially wonder why `set().__contains__()` accepts an unhashable `{0,1}` but does not accept another unhashable `[0,1]`. It is not about comparison. – DYZ Jul 17 '21 at 22:33
  • Could you provide the *URL*? – CristiFati Jul 17 '21 at 22:34
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    The actual answer is just a little further down the page: "Note, the _elem_ argument to the `__contains__()`, `remove()`, and `discard()` methods may be a set. To support searching for an equivalent frozenset, a temporary one is created from _elem_.". – Davis Herring Jul 17 '21 at 22:58
  • @DYZ My bad, you are right, updated with the correct answer from DavisHerring – Ben Jul 17 '21 at 23:06