在Redis中的LRU算法文中说到，LRU有一个缺陷，在如下情况下：

~~~~~A~~~~~A~~~~~A~~~~A~~~~~A~~~~~A~~|

~~B~~B~~B~~B~~B~~B~~B~~B~~B~~B~~B~~B~|

~~~~~~~~~~C~~~~~~~~~C~~~~~~~~~C~~~~~~|

~~~~~D~~~~~~~~~~D~~~~~~~~~D~~~~~~~~~D|

会将数据D误认为将来最有可能被访问到的数据。

Redis作者曾想改进LRU算法，但发现Redis的LRU算法受制于随机采样数maxmemory_samples，在maxmemory_samples等于10的情况下已经很接近于理想的LRU算法性能，也就是说，LRU算法本身已经很难再进一步了。

于是，将思路回到原点，淘汰算法的本意是保留那些将来最有可能被再次访问的数据，而LRU算法只是预测最近被访问的数据将来最有可能被访问到。我们可以转变思路，采用一种LFU(Least Frequently Used)算法，也就是最频繁被访问的数据将来最有可能被访问到。在上面的情况中，根据访问频繁情况，可以确定保留优先级：B>A>C=D。

Redis中的LFU思路

在LFU算法中，可以为每个key维护一个计数器。每次key被访问的时候，计数器增大。计数器越大，可以约等于访问越频繁。

上述简单算法存在两个问题：

在LRU算法中可以维护一个双向链表，然后简单的把被访问的节点移至链表开头，但在LFU中是不可行的，节点要严格按照计数器进行排序，新增节点或者更新节点位置时，时间复杂度可能达到O(N)。

只是简单的增加计数器的方法并不完美。访问模式是会频繁变化的，一段时间内频繁访问的key一段时间之后可能会很少被访问到，只增加计数器并不能体现这种趋势。

第一个问题很好解决，可以借鉴LRU实现的经验，维护一个待淘汰key的pool。第二个问题的解决办法是，记录key最后一个被访问的时间，然后随着时间推移，降低计数器。

Redis对象的结构如下：

typedef struct redisObject {

unsigned type:4;

unsigned encoding:4;

unsigned lru:LRU_BITS; /* LRU time (relative to global lru_clock) or

* LFU data (least significant 8 bits frequency

* and most significant 16 bits access time). */

int refcount;

void *ptr;

} robj;

在LRU算法中，24 bits的lru是用来记录LRU time的，在LFU中也可以使用这个字段，不过是分成16 bits与8 bits使用：

16 bits 8 bits

+----------------+--------+

+ Last decr time | LOG_C |

+----------------+--------+

高16 bits用来记录最近一次计数器降低的时间ldt，单位是分钟，低8 bits记录计数器数值counter。

LFU配置

Redis4.0之后为maxmemory_policy淘汰策略添加了两个LFU模式：

volatile-lfu：对有过期时间的key采用LFU淘汰算法

allkeys-lfu：对全部key采用LFU淘汰算法

还有2个配置可以调整LFU算法：

lfu-log-factor 10

lfu-decay-time 1

lfu-log-factor可以调整计数器counter的增长速度，lfu-log-factor越大，counter增长的越慢。

lfu-decay-time是一个以分钟为单位的数值，可以调整counter的减少速度

源码实现

robj *lookupKey(redisDb *db, robj *key, int flags) {

dictEntry *de = dictFind(db->dict,key->ptr);

if (de) {

robj *val = dictGetVal(de);

/* Update the access time for the ageing algorithm.

* Don't do it if we have a saving child, as this will trigger

* a copy on write madness. */

if (server.rdb_child_pid == -1 &&

server.aof_child_pid == -1 &&

!(flags & LOOKUP_NOTOUCH))

{

if (server.maxmemory_policy & MAXMEMORY_FLAG_LFU) {

updateLFU(val);

} else {

val->lru = LRU_CLOCK();

}

}

return val;

} else {

return NULL;

}

}

当采用LFU策略时，updateLFU更新lru：

/* Update LFU when an object is accessed.

* Firstly, decrement the counter if the decrement time is reached.

* Then logarithmically increment the counter, and update the access time. */

void updateLFU(robj *val) {

unsigned long counter = LFUDecrAndReturn(val);

counter = LFULogIncr(counter);

val->lru = (LFUGetTimeInMinutes()<<8) | counter;

}

降低LFUDecrAndReturn

首先，LFUDecrAndReturn对counter进行减少操作：

/* If the object decrement time is reached decrement the LFU counter but

* do not update LFU fields of the object, we update the access time

* and counter in an explicit way when the object is really accessed.

* And we will times halve the counter according to the times of

* elapsed time than server.lfu_decay_time.

* Return the object frequency counter.

*

* This function is used in order to scan the dataset for the best object

* to fit: as we check for the candidate, we incrementally decrement the

* counter of the scanned objects if needed. */

unsigned long LFUDecrAndReturn(robj *o) {

unsigned long ldt = o->lru >> 8;

unsigned long counter = o->lru & 255;

unsigned long num_periods = server.lfu_decay_time ? LFUTimeElapsed(ldt) / server.lfu_decay_time : 0;

if (num_periods)

counter = (num_periods > counter) ? 0 : counter - num_periods;

return counter;

}

函数首先取得高16 bits的最近降低时间ldt与低8 bits的计数器counter，然后根据配置的lfu_decay_time计算应该降低多少。

LFUTimeElapsed用来计算当前时间与ldt的差值：

/* Return the current time in minutes, just taking the least significant

* 16 bits. The returned time is suitable to be stored as LDT (last decrement

* time) for the LFU implementation. */

unsigned long LFUGetTimeInMinutes(void) {

return (server.unixtime/60) & 65535;

}

/* Given an object last access time, compute the minimum number of minutes

* that elapsed since the last access. Handle overflow (ldt greater than

* the current 16 bits minutes time) considering the time as wrapping

* exactly once. */

unsigned long LFUTimeElapsed(unsigned long ldt) {

unsigned long now = LFUGetTimeInMinutes();

if (now >= ldt) return now-ldt;

return 65535-ldt+now;

}

具体是当前时间转化成分钟数后取低16 bits，然后计算与ldt的差值now-ldt。当ldt > now时，默认为过了一个周期(16 bits，最大65535)，取值65535-ldt+now。

然后用差值与配置lfu_decay_time相除，LFUTimeElapsed(ldt) / server.lfu_decay_time，已过去n个lfu_decay_time，则将counter减少n，counter - num_periods。

增长LFULogIncr

增长函数LFULogIncr如下：

/* Logarithmically increment a counter. The greater is the current counter value

* the less likely is that it gets really implemented. Saturate it at 255. */

uint8_t LFULogIncr(uint8_t counter) {

if (counter == 255) return 255;

double r = (double)rand()/RAND_MAX;

double baseval = counter - LFU_INIT_VAL;

if (baseval < 0) baseval = 0;

double p = 1.0/(baseval*server.lfu_log_factor+1);

if (r < p) counter++;

return counter;

}

counter并不是简单的访问一次就+1，而是采用了一个0-1之间的p因子控制增长。counter最大值为255。取一个0-1之间的随机数r与p比较，当r

+--------+------------+------------+------------+------------+------------+

| factor | 100 hits | 1000 hits | 100K hits | 1M hits | 10M hits |

+--------+------------+------------+------------+------------+------------+

| 0 | 104 | 255 | 255 | 255 | 255 |

+--------+------------+------------+------------+------------+------------+

| 1 | 18 | 49 | 255 | 255 | 255 |

+--------+------------+------------+------------+------------+------------+

| 10 | 10 | 18 | 142 | 255 | 255 |

+--------+------------+------------+------------+------------+------------+

| 100 | 8 | 11 | 49 | 143 | 255 |

+--------+------------+------------+------------+------------+------------+

可见counter增长与访问次数呈现对数增长的趋势，随着访问次数越来越大，counter增长的越来越慢。

新生key策略

另外一个问题是，当创建新对象的时候，对象的counter如果为0，很容易就会被淘汰掉，还需要为新生key设置一个初始counter，createObject:

robj *createObject(int type, void *ptr){

robj *o = zmalloc(sizeof(*o));

o->type= type;

o->encoding = OBJ_ENCODING_RAW;

o->ptr = ptr;

o->refcount = 1;

/* Set the LRU to the current lruclock (minutes resolution), or

* alternatively the LFU counter. */

if (server.maxmemory_policy & MAXMEMORY_FLAG_LFU) {

o->lru = (LFUGetTimeInMinutes()<<8) | LFU_INIT_VAL;

} else {

o->lru = LRU_CLOCK();

}

return o;

}

counter会被初始化为LFU_INIT_VAL，默认5。

pool

pool算法就与LRU算法一致了：

if (server.maxmemory_policy & (MAXMEMORY_FLAG_LRU|MAXMEMORY_FLAG_LFU) ||

server.maxmemory_policy == MAXMEMORY_VOLATILE_TTL)

计算idle时有所不同：

} else if (server.maxmemory_policy & MAXMEMORY_FLAG_LFU) {

/* When we use an LRU policy, we sort the keys by idle time

* so that we expire keys starting from greater idle time.

* However when the policy is an LFU one, we have a frequency

* estimation, and we want to evict keys with lower frequency

* first. So inside the pool we put objects using the inverted

* frequency subtracting the actual frequency to the maximum

* frequency of 255. */

idle = 255-LFUDecrAndReturn(o);

使用了255-LFUDecrAndReturn(o)当做排序的依据。

参考链接