I'm answering the question in the title of this post. :)
Because lists are mutable, dict keys need to be hashable, and hashing mutable objects is a bad idea because hash values should be computed on the basis of instance attributes.
Example 1: hashing a mutable object where the hash value is based on a mutable characteristic of the object.
>>> class stupidlist(list):
... def __hash__(self):
... return len(self)
...
>>> stupid = stupidlist([1, 2, 3])
>>> d = {stupid: 0}
>>> stupid.append(4)
>>> stupid
[1, 2, 3, 4]
>>> d
{[1, 2, 3, 4]: 0}
>>> stupid in d
False
>>> stupid in d.keys()
False
>>> stupid in list(d.keys())
True
After mutating stupid, it cannot be found in the dict any longer because the hash changed. Only a linear scan over the list of the dict's keys finds stupid.
Example 2: ... but why not just a constant hash value?
>>> class stupidlist2(list):
... def __hash__(self):
... return id(self)
...
>>> stupidA = stupidlist2([1, 2, 3])
>>> stupidB = stupidlist2([1, 2, 3])
>>>
>>> stupidA == stupidB
True
>>> stupidA in {stupidB: 0}
False
That's not a good idea as well because equal objects should hash identically such that you can find them in a dict or set.
Example 3: ... ok, what about constant hashes across all instances?!
>>> class stupidlist3(list):
... def __hash__(self):
... return 1
...
>>> stupidC = stupidlist3([1, 2, 3])
>>> stupidD = stupidlist3([1, 2, 3])
>>> stupidE = stupidlist3([1, 2, 3, 4])
>>>
>>> stupidC in {stupidD: 0}
True
>>> stupidC in {stupidE: 0}
False
>>> d = {stupidC: 0}
>>> stupidC.append(5)
>>> stupidC in d
True
Things seem to work as expected, but think about what's happening: when all instances of your class produce the same hash value, you will have a hash collision whenever there are more than two instances as keys in a dict or present in a set.
Finding the right instance with d[key] or key in d needs to perform as many equality checks as there are instances of stupidlist3 in the dict's keys. At this point, the purpose of the dictionary - O(1) lookup - is completely defeated. This is demonstrated in the following timings (done with IPython).
Some Timings
>>> lists_list = [[i] for i in range(1000)]
>>> stupidlists_set = {stupidlist3([i]) for i in range(1000)}
>>> tuples_set = {(i,) for i in range(1000)}
>>> l = [999]
>>> s = stupidlist3([999])
>>> t = (999,)
>>>
>>> %timeit l in lists_list
25.5 µs ± 442 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
>>> %timeit s in stupidlists_set
38.5 µs ± 61.2 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
>>> %timeit t in tuples_set
77.6 ns ± 1.5 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
As you can see, the membership test in our stupidlists_set is even slower than a linear scan over the whole lists_list, while you have the expected super fast lookup time (factor 500) in a set without loads of hash collisions.
TL; DR: you can use tuple(yourlist) as dict keys, because tuples are immutable and hashable.
