hash – Time complexity of accessing a Python dict

hash – Time complexity of accessing a Python dict

See Time Complexity. The python dict is a hashmap, its worst case is therefore O(n) if the hash function is bad and results in a lot of collisions. However that is a very rare case where every item added has the same hash and so is added to the same chain which for a major Python implementation would be extremely unlikely. The average time complexity is of course O(1).

The best method would be to check and take a look at the hashs of the objects you are using. The CPython Dict uses int PyObject_Hash (PyObject *o) which is the equivalent of hash(o).

After a quick check, I have not yet managed to find two tuples that hash to the same value, which would indicate that the lookup is O(1)

l = []
for x in range(0, 50):
    for y in range(0, 50):
        if hash((x,y)) in l:
            print Fail: , (x,y)
print Test Finished

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You are not correct. dict access is unlikely to be your problem here. It is almost certainly O(1), unless you have some very weird inputs or a very bad hashing function. Paste some sample code from your application for a better diagnosis.

hash – Time complexity of accessing a Python dict

It would be easier to make suggestions if you provided example code and data.

Accessing the dictionary is unlikely to be a problem as that operation is O(1) on average, and O(N) amortized worst case. Its possible that the built-in hashing functions are experiencing collisions for your data. If youre having problems with has the built-in hashing function, you can provide your own.

Pythons dictionary implementation
reduces the average complexity of
dictionary lookups to O(1) by
requiring that key objects provide a
hash function. Such a hash function
takes the information in a key object
and uses it to produce an integer,
called a hash value. This hash value
is then used to determine which
bucket this (key, value) pair should
be placed into.

You can overwrite the __hash__ method in your class to implement a custom hash function like this:

def __hash__(self):    
    return hash(str(self))

Depending on what your data actually looks like, you might be able to come up with a faster hash function that has fewer collisions than the standard function. However, this is unlikely. See the Python Wiki page on Dictionary Keys for more information.

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