Understanding Hash Tables with python as a reference

Question

I am going through online lectures on data structures and I want to confirm my understanding of the hash table.

I understand that a hash table will use a hashing function to reduce the universe of all possible keys down to a set m and use chaining to resolve collisions.

I can't seem to visualize the m part of it. Say I create an empty dict() in python. Does python create a table with some predefined number of empty slots?

Answer 1

Overview

An overview of how Python's dictionaries are implemented can be found in the 2017 Pycon talk, Modern Python Dictionaries A confluence of a dozen great ideas.

How to visualize reduction

I understand that a hash table will use a hashing function to reduce the universe of all possible keys down to a set m and use chaining to resolve collisions. ... I can't seem to visualize the m part of it.

The easiest visualization is with m == 2 so that hashing divides keys into two groups:

>>> from pprint import pprint
>>> def hash(n):
        'Hash a number into evens or odds'
        return n % 2

>>> table = [[], []]
>>> for x in [10, 15, 12, 41, 80, 13, 40, 9]:
        table[hash(x)].append(x)

>>> pprint(table, width=25)
[[10, 12, 80, 40],
 [15, 41, 13, 9]]

In the above example, the eight keys all get divided into two groups (the evens and the odds).

The example also works with bigger values of m such as m == 7:

>>> table = [[], [], [], [], [], [], []]
>>> for x in [10, 15, 12, 41, 80, 13, 40, 9]:
        table[x % 7].append(x)

>>> pprint(table, width=25)
[[],
 [15],
 [9],
 [10, 80],
 [],
 [12, 40],
 [41, 13]]

As you can see, the above example has two empty slots and slots with a collision.

Table for an empty dict

Say I create an empty dict() in python. Does python create a table with some predefined number of null entries?

Yes, Python creates eight slots for an empty table. In Python's source code, we see #define PyDict_MINSIZE 8 in cpython/Objects/dictobject.c.