Hashing a tuple in Python where order matters?

Question

I have:

tuple1 = token1, token2
tuple2 = token2, token1
for tuple in [tuple1, tuple2]:
    if tuple in dict:
        dict[tuple] += 1
    else:
        dict[tuple] = 1

However, both tuple 1 and tuple2 are getting the same counts. What is a way to hash a group of 2 things such that order matters?

Answer 1

Order is taken into account when hashing:

>>> hash((1,2))
1299869600
>>> hash((2,1))
1499606158

This assumes that the objects themselves have unique hashes. Even if they don't, you could still be OK when using it in a dictionary (as long as the objects themselves aren't equal as defined by their __eq__ method):

>>> t1 = 'a',hash('a') 
>>> [hash(x) for x in t1]  #both elements in the tuple have same hash value since `int` hash to themselves in cpython
[-468864544, -468864544]
>>> t2 = hash('a'),'a'
>>> hash(t1)
1486610051
>>> hash(t2)
1486610051
>>> d = {t1:1,t2:2}  #This is OK.  dict's don't fail when there is a hash collision
>>> d
{('a', -468864544): 1, (-468864544, 'a'): 2}
>>> d[t1]+=7
>>> d[t1]
8
>>> d[t1]+=7
>>> d[t1]
15
>>> d[t2]   #didn't touch d[t2] as expected.
2

Note that due to hash collisions, this dict is likely to be less efficient than another dict where there aren't hash collisions :)