Simple LRU Cache

In-memory cache that deletes Least-Recently-Used items when limit is reached.

Prerequisites

Python >= 3.0

Usage and Behavior

Instantiate a cache that will hold up to a user-defined number of items:

from lruCache import LruCache

myCache = LruCache(itemLimit=3)

Add entries to cache:

myCache.set('first', 1) # None
# None<-{'first'=>1}->None

myCache.set('second', 2) # None
# None<-{'second'=>2}-><-{'first'=>1}->None

myCache.set('third', 3) # None
# None<-{'third'=>3}-><-{'second'=>2}-><-{'first'=>1}->None

Accessing a value via its key moves it, the most-recently used entry, to the front of the cache:

myCache.get('second') # 2
# None<-{'second'=>2}-><-{'third'=>3}-><-{'first'=>1}->None

Once the cache reaches its item limit, the least-recently-used entry is discarded:

myCache.set('fourth', 4) # None
# None<-{'fourth'=>4}-><-{'second'=>2}-><-{'third'=>3}->None

myCache.get('first') # None
# None<-{'fourth'=>4}-><-{'second'=>2}-><-{'third'=>3}->None

myCache.get('third') # 3
# None<-{'third'=>3}-><-{'fourth'=>4}-><-{'second'=>2}->None

Manually remove an entry from the cache:

myCache.remove('fourth') # 4
# None<-{'third'=>3}-><-{'second'=>2}->None

myCache.get('fourth') # None