Python-memcached是官方支持的Django memcached驱动程序。
是否支持
如果是,请问如何在Django中使用这些功能?我找不到任何文件。
答案 0 :(得分:2)
查看python-memcached v1.45上的_get_server
方法,它似乎不使用一致哈希,而是一个简单的hash % len(buckets)
。
对于二进制协议也是如此,就我在源代码中看到的而言,python-memcache只使用文本命令。
答案 1 :(得分:1)
您可以使用此功能:http://amix.dk/blog/post/19370
它封装了python-memcache的Client类,因此使用一致的散列来分发密钥。
编辑 - 我正在挖掘python-memcached
1.4.5源代码,看起来它实际上可能支持一致的散列。
相关代码:
from binascii import crc32 # zlib version is not cross-platform
def cmemcache_hash(key):
return((((crc32(key) & 0xffffffff) >> 16) & 0x7fff) or 1)
serverHashFunction = cmemcache_hash
-- SNIP --
def _get_server(self, key):
if isinstance(key, tuple):
serverhash, key = key
else:
serverhash = serverHashFunction(key)
for i in range(Client._SERVER_RETRIES):
server = self.buckets[serverhash % len(self.buckets)]
if server.connect():
#print "(using server %s)" % server,
return server, key
serverhash = serverHashFunction(str(serverhash) + str(i))
return None, None
基于此代码,看起来它确实实现了算法,除非cmemcache_hash
不是有意义的名称且不是真正的算法。 (现在已退休的cmemcache执行一致哈希)
但我认为OP指的是更具“弹性”的一致性散列,例如libketama。我不认为那里的解决方案有所下降,看起来你需要卷起袖子编译/安装一个更高级的memcached lib,如pylibmc,并编写一个使用它的自定义Django后端而不是python-memcached。
无论如何,在任何一种情况下,当您向池中添加/删除存储桶时,都会发生一些键的重新映射(即使使用libketama,只比其他算法少)
答案 2 :(得分:1)
请检查此示例python实现的一致哈希。 consistent-hashing python-memcached continnum-circle
实施主体:想象一个连续的圈子,其中有许多复制的服务器点。当我们添加新服务器时,总数的1 / n 缓存键将丢失
'''consistent_hashing.py is a simple demonstration of consistent
hashing.'''
import bisect
import hashlib
class ConsistentHash:
'''
To imagine it is like a continnum circle with a number of replicated
server points spread across it. When we add a new server, 1/n of the total
cache keys will be lost.
consistentHash(n,r) creates a consistent hash object for a
cluster of size n, using r replicas.
It has three attributes. num_machines and num_replics are
self-explanatory. hash_tuples is a list of tuples (j,k,hash),
where j ranges over machine numbers (0...n-1), k ranges over
replicas (0...r-1), and hash is the corresponding hash value,
in the range [0,1). The tuples are sorted by increasing hash
value.
The class has a single instance method, get_machine(key), which
returns the number of the machine to which key should be
mapped.'''
def __init__(self,replicas=1):
self.num_replicas = replicas
def setup_servers(self,servers=None):
hash_tuples = [(index,k,my_hash(str(index)+"_"+str(k))) \
for index,server in enumerate(servers)
for k in range(int(self.num_replicas) * int(server.weight)) ]
self.hash_tuples=self.sort(hash_tuples);
def sort(self,hash_tuples):
'''Sort the hash tuples based on just the hash values '''
hash_tuples.sort(lambda x,y: cmp(x[2],y[2]))
return hash_tuples
def add_machine(self,server,siz):
'''This mathod adds a new machine. Then it updates the server hash
in the continuum circle '''
newPoints=[(siz,k,my_hash(str(siz)+"_"+str(k))) \
for k in range(self.num_replicas*server.weight)]
self.hash_tuples.extend(newPoints)
self.hash_tuples=self.sort(self.hash_tuples);
def get_machine(self,key):
'''Returns the number of the machine which key gets sent to.'''
h = my_hash(key)
# edge case where we cycle past hash value of 1 and back to 0.
if h > self.hash_tuples[-1][2]: return self.hash_tuples[0][0]
hash_values = map(lambda x: x[2],self.hash_tuples)
index = bisect.bisect_left(hash_values,h)
return self.hash_tuples[index][0]
def my_hash(key):
'''my_hash(key) returns a hash in the range [0,1).'''
return (int(hashlib.md5(key).hexdigest(),16) % 1000000)/1000000.0
答案 3 :(得分:0)
现在vbucket即将解决一致哈希问题,对缓存未命中影响最小。
答案 4 :(得分:0)
如果您想要django的即插即用解决方案,请使用django-memcached-hashring
:https://github.com/jezdez/django-memcached-hashring。
它是django.core.cache.backends.memcached.MemcachedCache
和hash_ring
库附近的适配器。
答案 5 :(得分:0)
我使用了Consistent散列算法。丢失的密钥是密钥总数的1 / n。这意味着成功的密钥提取将是大约85%的6/7 * 100。 here