LRU是Least Recent Used的简写,翻译成中文就是 最近最少被使用。
无论是对某个key的get,还是set都算做是对该key的一次使用。当set一个不存在的key,并且LRU Cache中key的数量超过cache size的时候,需要将使用时间距离现在最长的那个key从LRU Cache中清除。
在Java中,LRUCache是通过LinkedHashMap实现的。鄙人照猫画虎,实现一个Python版的LRU Cache(可能和其他大神的实现有所区别)。
首先,需要说明的是:
__hash__
)具体实现是:
class Entry: def __init__(self, hash_code, v, prev=None, next=None): self.hash_code = hash_code self.v = v self.prev = prev self.next = next def __str__(self): return "Entry{hash_code=%d, v=%s}" % ( self.hash_code, self.v) __repr__ = __str__ class LRUCache: def __init__(self, max_size): self._max_size = max_size self._dict = dict() self._head = Entry(None, None) self._head.prev = self._head self._head.next = self._head def __setitem__(self, k, v): try: hash_code = hash(k) except TypeError: raise old_entry = self._dict.get(hash_code) new_entry = Entry(hash_code, v) self._dict[hash_code] = new_entry if old_entry: prev = old_entry.prev next = old_entry.next prev.next = next next.prev = prev head = self._head head_prev = self._head.prev head_next = self._head.next head.next = new_entry if head_prev is head: head.prev = new_entry head_next.prev = new_entry new_entry.prev = head new_entry.next = head_next if not old_entry and len(self._dict) > self._max_size: last_one = head.prev last_one.prev.next = head head.prev = last_one.prev self._dict.pop(last_one.hash_code) def __getitem__(self, k): entry = self._dict[hash(k)] head = self._head head_next = head.next prev = entry.prev next = entry.next if entry.prev is not head: if head.prev is entry: head.prev = prev head.next = entry head_next.prev = entry entry.prev = head entry.next = head_next prev.next = next next.prev = prev return entry.v def get_dict(self): return self._dict if __name__ == "__main__": cache = LRUCache(2) inner_dict = cache.get_dict() cache[1] = 1 assert inner_dict.keys() == [1], "test 1" cache[2] = 2 assert sorted(inner_dict.keys()) == [1, 2], "test 2" cache[3] = 3 assert sorted(inner_dict.keys()) == [2, 3], "test 3" cache[2] assert sorted(inner_dict.keys()) == [2, 3], "test 4" assert inner_dict[hash(2)].next.v == 3 cache[4] = 4 assert sorted(inner_dict.keys()) == [2, 4], "test 5" assert inner_dict[hash(4)].v == 4, "test 6"