Is a Python dictionary an example of a hash table?

Question:

One of the basic data structures in Python is the dictionary, which allows one to record “keys” for looking up “values” of any type. Is this implemented internally as a hash table? If not, what is it?

Asked By: Tommy Herbert

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Answers:

Yes, it is a hash mapping or hash table. You can read a description of python’s dict implementation, as written by Tim Peters, here.

That’s why you can’t use something ‘not hashable’ as a dict key, like a list:

>>> a = {}
>>> b = ['some', 'list']
>>> hash(b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: list objects are unhashable
>>> a[b] = 'some'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: list objects are unhashable

You can read more about hash tables or check how it has been implemented in python and why it is implemented that way.

Answered By: nosklo

Yes. Internally it is implemented as open hashing based on a primitive polynomial over Z/2 (source).

Answered By: Ben Hoffstein

To expand upon nosklo’s explanation:

a = {}
b = ['some', 'list']
a[b] = 'some' # this won't work
a[tuple(b)] = 'some' # this will, same as a['some', 'list']
Answered By: Jeremy Cantrell

There must be more to a Python dictionary than a table lookup on hash(). By brute experimentation I found this hash collision:

>>> hash(1.1)
2040142438
>>> hash(4504.1)
2040142438

Yet it doesn’t break the dictionary:

>>> d = { 1.1: 'a', 4504.1: 'b' }
>>> d[1.1]
'a'
>>> d[4504.1]
'b'

Sanity check:

>>> for k,v in d.items(): print(hash(k))
2040142438
2040142438

Possibly there’s another lookup level beyond hash() that avoids collisions between dictionary keys. Or maybe dict() uses a different hash.

(By the way, this in Python 2.7.10. Same story in Python 3.4.3 and 3.5.0 with a collision at hash(1.1) == hash(214748749.8).)

(I haven’t found any collisions in Python 3.9.6. Since the hashes are bigger — hash(1.1) == 230584300921369601 — I estimate it would take my desktop a thousand years to find one. So I’ll get back to you on this.)

Answered By: Bob Stein