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2.3.0 • Public • Published


Simpler, faster LRU cache algorithm

A Least Recently Used cache is used to speedup requests to a key-value oriented resource, while making a bounded memory commitment.

I've recently benchmarked the various lru implementations available on npm and found wildly varing performance. There where some that performed well overall, and others that performed extremely well in some cases, but poorly in others, due to compromises made to maintain correctness.

After writing the benchmark, of course I had to try my hand at my own LRU implementation. I soon found a few things, LRUs are quite difficult to implement, first of all contain a linked list. LRUs use a linked list to maintain the order that keys have been accessed, so that when the cache fills, the old values (which presumably are the least likely to be needed again) can be removed from the cache. Linked Lists are not easy to implement correctly!

Then I discovered why some of the fast algorithms where so slow - they used delete cache[key] which is much slower than cache[key] = value, much much slower.

So, why looking for a way to avoid delete I had an idea - have two cache objects, and when one fills - create a new one and start putting items in that, and then it's sufficiently full, throw it away. It avoids delete, at at max, only commits us to only N values and between N and 2N keys.

Then I realized with this pattern, you don't actually need the linked list anymore! This makes a N-2N least recently used cache very very simple. This both has performance benefits, and it's also very easy to verify it's correctness.

This algorithm does not give you an ordered list of the N most recently used items, but you do not really need that! The property of dropping the least recent items is still preserved.

see a benchmark of this against the other LRU implementations on npm.


var HLRU = require('hashlru')
var lru = HLRU(100)
lru.set(key, value)


create two caches - old_cache and new_cache, and a counter, size.

When an key, value pair is added, if key is already in new_cache update the value, not currently in new_cache, set new_cache[key] = value. If the key was not already in new_cache then size is incremented. If size > max, move the old_cache = new_cache, reset size = 0, and initialize a new new_cache={}

To get a key, check if new_cache contains key, and if so, return it. If not, check if it is in old_cache and if so, move that value to new_cache, and increment size. If size > max, move the old_cache = new_cache, reset size = 0, and initialize a new new_cache={}


Writes are O(1) on average, like a hash table.

When implemented in a garbage collected language, the old cache is thrown away when the new cache is full. To better manage memory usage, it could also be implemented as two fixes sized hash tables. In this case, instead of discarding the old cache, it would be zeroed. This means at most every N writes when the caches are rotated, that write will require N operations (to clear the old cache)

This still averages out to O(1) but it does cost O(N) but only every N writes (except for updates) so N/N is still 1.

HashLRU (max) => lru

initialize a lru object.

lru.get (key) => value | undefined

Returns the value in the cache, or undefined if the value is not in the cache.

lru.set(key, value)

update the value for key.

lru.has(key) => boolean

Checks if the key is in the cache.


Removes the key from the cache.


Empties the entire cache.






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  • davidmarkclements
  • dominictarr
  • kikobeats