Why can a Python dict have multiple keys with the same hash?

Question

I am trying to understand the Python hash function under the hood. I created a custom class where all instances return the same hash value.

class C:
    def __hash__(self):
        return 42

I just assumed that only one instance of the above class can be in a dict at any time, but in fact a dict can have multiple elements with the same hash.

c, d = C(), C()
x = {c: 'c', d: 'd'}
print(x)
# {<__main__.C object at 0x7f0824087b80>: 'c', <__main__.C object at 0x7f0823ae2d60>: 'd'}
# note that the dict has 2 elements

I experimented a little more and found that if I override the __eq__ method such that all the instances of the class compare equal, then the dict only allows one instance.

class D:
    def __hash__(self):
        return 42
    def __eq__(self, other):
        return True

p, q = D(), D()
y = {p: 'p', q: 'q'}
print(y)
# {<__main__.D object at 0x7f0823a9af40>: 'q'}
# note that the dict only has 1 element

So I am curious to know how a dict can have multiple elements with the same hash.

Answer 1

Here is everything about Python dicts that I was able to put together (probably more than anyone would like to know; but the answer is comprehensive). A shout out to Duncan for pointing out that Python dicts use slots and leading me down this rabbit hole.

• Python dictionaries are implemented as hash tables.
• Hash tables must allow for hash collisions i.e. even if two keys have same hash value, the implementation of the table must have a strategy to insert and retrieve the key and value pairs unambiguously.
• Python dict uses open addressing to resolve hash collisions (explained below) (see dictobject.c:296-297).
• Python hash table is just a continguous block of memory (sort of like an array, so you can do O(1) lookup by index).
• Each slot in the table can store one and only one entry. This is important
• Each entry in the table actually a combination of the three values - . This is implemented as a C struct (see dictobject.h:51-56)

• The figure below is a logical representation of a python hash table. In the figure below, 0, 1, ..., i, ... on the left are indices of the slots in the hash table (they are just for illustrative purposes and are not stored along with the table obviously!).

  # Logical model of Python Hash table
  -+-----------------+
  0| <hash|key|value>|
  -+-----------------+
  1|      ...        |
  -+-----------------+
  .|      ...        |
  -+-----------------+
  i|      ...        |
  -+-----------------+
  .|      ...        |
  -+-----------------+
  n|      ...        |
  -+-----------------+
  

• When a new dict is initialized it starts with 8 slots. (see dictobject.h:49)

• When adding entries to the table, we start with some slot, i that is based on the hash of the key. CPython uses initial i = hash(key) & mask. Where mask = PyDictMINSIZE - 1, but that's not really important). Just note that the initial slot, i, that is checked depends on the hash of the key.
• If that slot is empty, the entry is added to the slot (by entry, I mean, <hash|key|value>). But what if that slot is occupied!? Most likely because another entry has the same hash (hash collision!)
• If the slot is occupied, CPython (and even PyPy) compares the the hash AND the key (by compare I mean == comparison not the is comparison) of the entry in the slot against the key of the current entry to be inserted (dictobject.c:337,344-345). If both match, then it thinks the entry already exists, gives up and moves on to the next entry to be inserted. If either hash or the key don't match, it starts probing.
• Probing just means it searches the slots by slot to find an empty slot. Technically we could just go one by one, i+1, i+2, ... and use the first available one (that's linear probing). But for reasons explained beautifully in the comments (see dictobject.c:33-126), CPython uses random probing. In random probing, the next slot is picked in a pseudo random order. The entry is added to the first empty slot. For this discussion, the actual algorithm used to pick the next slot is not really important (see dictobject.c:33-126 for the algorithm for probing). What is important is that the slots are probed until first empty slot is found.
• The same thing happens for lookups, just starts with the initial slot i (where i depends on the hash of the key). If the hash and the key both don't match the entry in the slot, it starts probing, until it finds a slot with a match. If all slots are exhausted, it reports a fail.
• BTW, the dict will be resized if it is two-thirds full. This avoids slowing down lookups. (see dictobject.h:64-65)

There you go! The Python implementation of dict checks for both hash equality of two keys and the normal equality (==) of the keys when inserting items. So in summary, if there are two keys, a and b and hash(a)==hash(b), but a!=b, then both can exist harmoniously in a Python dict. But if hash(a)==hash(b) and a==b, then they cannot both be in the same dict.

Because we have to probe after every hash collision, one side effect of too many hash collisions is that the lookups and insertions will become very slow (as Duncan points out in the comments).

I guess the short answer to my question is, "Because that's how it's implemented in the source code ;)"

While this is good to know (for geek points?), I am not sure how it can be used in real life. Because unless you are trying to explicitly break something, why would two objects that are not equal, have same hash?

Answer 2

For a detailed description of how Python's hashing works see my answer to Why is early return slower than else?

Basically it uses the hash to pick a slot in the table. If there is a value in the slot and the hash matches, it compares the items to see if they are equal.

If the hash matches but the items aren't equal, then it tries another slot. There's a formula to pick this (which I describe in the referenced answer), and it gradually pulls in unused parts of the hash value; but once it has used them all up, it will eventually work its way through all slots in the hash table. That guarantees eventually we either find a matching item or an empty slot. When the search finds an empty slot, it inserts the value or gives up (depending whether we are adding or getting a value).

The important thing to note is that there are no lists or buckets: there is just a hash table with a particular number of slots, and each hash is used to generate a sequence of candidate slots.

Answer 3

Edit: the answer below is one of possible ways to deal with hash collisions, it is however not how Python does it. Python's wiki referenced below is also incorrect. The best source given by @Duncan below is the implementation itself: https://github.com/python/cpython/blob/master/Objects/dictobject.c I apologize for the mix-up.

---

It stores a list (or bucket) of elements at the hash then iterates through that list until it finds the actual key in that list. A picture says more than a thousand words:

Here you see John Smith and Sandra Dee both hash to 152. Bucket 152 contains both of them. When looking up Sandra Dee it first finds the list in bucket 152, then loops through that list until Sandra Dee is found and returns 521-6955.

The following is wrong it's only here for context: On Python's wiki you can find (pseudo?) code how Python performs the lookup.

There's actually several possible solutions to this problem, check out the wikipedia article for a nice overview: http://en.wikipedia.org/wiki/Hash_table#Collision_resolution

Answer 4

Hash tables, in general have to allow for hash collisions! You will get unlucky and two things will eventually hash to the same thing. Underneath, there is a set of objects in a list of items that has that same hash key. Usually, there is only one thing in that list, but in this case, it'll keep stacking them into the same one. The only way it knows they are different is through the equals operator.

When this happens, your performance will degrade over time, which is why you want your hash function to be as "random as possible".

Answer 5

In the thread I did not see what exactly python does with instances of a user-defined classes when we put it into a dictionary as a keys. Let's read some documentation: it declares that only hashable objects can be used as a keys. Hashable are all immutable built-in classes and all user-defined classes.

User-defined classes have __cmp__() and __hash__() methods by default; with them, all objects compare unequal (except with themselves) and x.__hash__() returns a result derived from id(x).

So if you have a constantly __hash__ in your class, but not providing any __cmp__ or __eq__ method, then all your instances are unequal for the dictionary. In the other hand, if you providing any __cmp__ or __eq__ method, but not providing __hash__, your instances are still unequal in terms of dictionary.

class A(object):
    def __hash__(self):
        return 42


class B(object):
    def __eq__(self, other):
        return True


class C(A, B):
    pass


dict_a = {A(): 1, A(): 2, A(): 3}
dict_b = {B(): 1, B(): 2, B(): 3}
dict_c = {C(): 1, C(): 2, C(): 3}

print(dict_a)
print(dict_b)
print(dict_c)

Output

{<__main__.A object at 0x7f9672f04850>: 1, <__main__.A object at 0x7f9672f04910>: 3, <__main__.A object at 0x7f9672f048d0>: 2}
{<__main__.B object at 0x7f9672f04990>: 2, <__main__.B object at 0x7f9672f04950>: 1, <__main__.B object at 0x7f9672f049d0>: 3}
{<__main__.C object at 0x7f9672f04a10>: 3}