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"