Simple LRU cache implementation on Python 3

What is LRU Cache?

This is caching item replacement policy, so the least used items will be discarded first.

Problem

The LRU Cache algorithm requires keeping track of what was used when, which is expensive if one wants to make sure the algorithm always discards the least recently used item.

Solution

Approaching a problem I was thinking of two capabilities for data structures: a FIFO queue and Hash table.

FIFO queue will be responsible to evict least used items. Hash table will be responsible to get cached items. Using this data structures make both operations in O(1) time complexity.

Python collections.OrderedDict combine both of this capabilities:

  • Queue: dict items are ordered as FIFO queue, so inserts and evictions are done in O(1)
  • Hash table: dict keys provide access to data in O(1) time
import collections
LRUCache = collections.OrderedDict()

Insert operation:

LRUCache[key] = value

It will add a key to the end of the dict, so position of new items is always fixed.

Check for hit:

if key in LRUCache:
  LRUCache.move_to_end(key)

Here we are doing two things: 1) Checking if key exist 2)If key is exist we move they key to the end of dict, so the keys which got a hit always update it’s position to become like newest key.

Discard or evict operation:

if len(LRUCache) > CACHE_SIZE:
  evict_key, evict_val = LRUCache.popitem(last=False)

Evict operation pop item from the beginning of the dict, so removing the oldest key in the dict.

Python 3 implementation

"""Simple LRUCache implementation."""

import collections
import os
import random
from math import factorial

CACHE_SIZE = 100
if os.getenv('CACHE_SIZE'):
    CACHE_SIZE = int(os.getenv('CACHE_SIZE'))

SAMPLE_SIZE = 100
if os.getenv('SAMPLE_SIZE'):
    SAMPLE_SIZE = int(os.getenv('SAMPLE_SIZE'))

LRUCache = collections.OrderedDict()


def expensive_call(number):
    """Calculate factorial. Example of expensive call."""
    return factorial(number)


if __name__ == '__main__':

    test_cases = random.choices(
        [x for x in range(SAMPLE_SIZE*3)],
        [x for x in range(SAMPLE_SIZE*3)],
        k=SAMPLE_SIZE
    )

    for test in test_cases:
        if test in LRUCache:
            print("hit:", test, LRUCache[test])
            # Update position of the hit item to first. Optional.
            LRUCache.move_to_end(test, last=True)
        else:
            LRUCache[test] = expensive_call(test)
            print("miss:", test, LRUCache[test])
        if len(LRUCache) > CACHE_SIZE:
            evict_key, evict_val = LRUCache.popitem(last=False)
            print("evict:", evict_key, evict_val)

As a use case I have used LRU cache to cache the output of expensive function call like factorial.

Sample size and Cache size are controllable through environment variables. Try to run it on small numbers to see how it behave:

CACHE_SIZE=4 SAMPLE_SIZE=10 python lru.py

Next steps are

  • Encapsulate business logic into class
  • Add Python magic functions to provide Pythonic way of dealing with class objects
  • Add unit test

References

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