| 分值 | 2.0 | 4.0 | 6.0 | 8.0 |
| 元素 | 西瓜 | 葡萄 | 芒果 | 香蕉 |
| 分值 | 6000 | 8000 | 10000 | 12000 |
| 元素 | lucy | tom | jim | jack |
| 分值 | 300 | 400 | 500 | 600 |
| 元素 | 核心编程 | 数据库 | django | redis |
# 在有序集合中添加一个成员 zadd key score member # 查看指定区间元素（升序) zrange key start stop [withscores] # 查看指定区间元素（降序） ZREVRANGE key start stop [withscores] # 查看指定元素的分值 ZSCORE key member # 返回指定区间元素 # offset : 跳过多少个元素 # count : 返回几个 # 小括号 : 开区间 zrangebyscore fruits (2.0 8.0 zrangebyscore key min max [withscores] [limit offset count] # 删除成员 zrem key member # 增加或者减少分值 zincrby key increment member 值 成员 # 返回元素排名 zrank key member # 返回元素逆序排名 zrevrank key member # 删除指定区间内的元素 zremrangebyscore key min max # 返回集合中元素个数 zcard key # 返回指定范围中元素的个数 zcount key min max zcount fruits 4 7 zcount fruits (4 7 4<fruits<=7#'('是开区间不包括 (4 (7 4<fruits<7 # 并集 zunionstore destination numkeys key [weights 权重值] [AGGREGATE SUM|MIN|MAX] # 有几个集合numkeys就是几 权重值也要相应的给几个 # 交集：和并集类似，只取相同的元素 ZINTERSTORE destination numkeys key1 key2 WEIGHTS weight AGGREGATE SUM|MIN|MAX
三
127.0.0.1:6379> ZRANGE salary 0 -1 1) "lucy" 2) "tom" 3) "jim" 4) "jack" 127.0.0.1:6379> ZRANGE salary 0 -1 withscores 1) "lucy" 2) "6000" 3) "tom" 4) "8000" 5) "jim" 6) "10000" 7) "jack" 8) "12000"
ZREVRANGE key start stop [withscores]
3.显示指定元素的分值
ZSCORE key member
127.0.0.1:6379> zscore salary jack "14000"
zrangebyscore key min max [withscores] [limit offset count]
offset : 跳过多少个元素
count : 返回几个
小括号 : 开区间 zrangebyscore fruits (2.0 8.0
127.0.0.1:6379> ZRANGEBYSCORE salary (8000 12000 1) "jim" 2) "jack" 127.0.0.1:6379> ZRANGE salary 0 -1 withscores 1) "lucy" 2) "6000" 3) "tom" 4) "8000" 5) "jim" 6) "10000" 7) "jack" 8) "12000"
5.删除
zrem key member
127.0.0.1:6379> ZREM salary jim (integer) 1 127.0.0.1:6379> ZRANGE salary 0 -1 withscores 1) "lucy" 2) "6000" 3) "tom" 4) "8000" 5) "jack" 6) "12000"
6.增加或者减少分值
zincrby key increment member
127.0.0.1:6379> ZINCRBY salary 2000 jack "14000" 127.0.0.1:6379> ZRANGE salary 0 -1 withscores 1) "lucy" 2) "6000" 3) "tom" 4) "8000" 5) "jack" 6) "14000"
7.返回元素的排名（索引）
zrank key member
127.0.0.1:6379> zrank salary jack
(integer) 2
8.返回元素逆序排名
zrevrank key member
127.0.0.1:6379> ZREVRANK salary jack (integer) 0 127.0.0.1:6379> ZREVRANK salary lucy (integer) 2
9.删除指定区间内的元素
zremrangebyscore key min max
127.0.0.1:6379> ZREMRANGEBYSCORE salary 4000 6000 (integer) 1 127.0.0.1:6379> ZRANGE salary 0 -1 withscores 1) "tom" 2) "8000" 3) "jack" 4) "14000"
10.返回集合中元素个数
zcard key
127.0.0.1:6379> ZCARD salary
(integer) 2
11.返回指定范围中元素的个数
zcount fruits 4 7
zcount fruits (4 7
127.0.0.1:6379> ZRANGE salary 0 -1 withscores 1) "tom" 2) "8000" 3) "jack" 4) "14000" 127.0.0.1:6379> zcount salary 8000 14000 (integer) 2 # 不包含8000，包含14000 127.0.0.1:6379> zcount salary (8000 14000 (integer) 1
12.并集
zunionstore destination numkeys key [weights ] [AGGREGATE SUM|MIN|MAX]
127.0.0.1:6379> zadd stu_score1 60 tom 70 jim (integer) 2 127.0.0.1:6379> zadd stu_score2 80 tom 90 lucy (integer) 2 # 默认为SUM 127.0.0.1:6379> ZUNIONSTORE stu_score3 2 stu_score1 stu_score2 (integer) 3 127.0.0.1:6379> ZRANGE stu_score3 0 -1 withscores 1) "jim" 2) "70" 3) "lucy" 4) "90" 5) "tom" 6) "140" 127.0.0.1:6379> # WEIGHTS 和 AGGREGATE 127.0.0.1:6379> ZRANGE stu_score1 0 -1 withscores 1) "tom" 2) "60" 3) "jim" 4) "70" 127.0.0.1:6379> ZRANGE stu_score2 0 -1 withscores 1) "tom" 2) "80" 3) "lucy" 4) "90" # 权重1给stu_score1，权重0.5给stu_score2，算完权重之后求和SUM 127.0.0.1:6379> ZUNIONSTORE stu_score8 2 stu_score1 stu_score2 weights 1 0.5 AGGREGATE SUM (integer) 3 127.0.0.1:6379> ZRANGE stu_score8 0 -1 withscores 1) "lucy" 2) "45" 3) "jim" 4) "70" 5) "tom" 6) "100" 127.0.0.1:6379>
12.交集
ZINTERSTORE destination numkeys key1 key2 WEIGHTS weight AGGREGATE SUM|MIN|MAX
和并集类似，只取相同的元素
python操作sorted set
import redis r = redis.Redis(host='192.168.43.49',port=6379,password='123456',db=0) # 注意第二个参数为字典 r.zadd('salary',{'tom':6000,'jim':8000,'jack':12000}) # 结果为列表中存放元组[(),(),()] print(r.zrange('salary',0,-1,withscores=True)) print(r.zrevrange('salary',0,-1,withscores=True)) # start:起始值,num:显示条数 print(r.zrangebyscore('salary',6000,12000,start=1,num=2,withscores=True)) # 删除 r.zrem('salary','tom') print(r.zrange('salary',0,-1,withscores=True)) # 增加分值 r.zincrby('salary',5000,'jack') print(r.zrange('salary',0,-1,withscores=True)) # 返回元素排名 print(r.zrank('salary','jack')) print(r.zrevrank('salary','jack')) # 删除指定区间内的元素 r.zremrangebyscore('salary',6000,8000) print(r.zrange('salary',0,-1,withscores=True)) # 统计元素个数 print(r.zcard('salary')) # 返回指定范围内元素个数 print(r.zcount('salary',6000,20000)) # 并集 r.zadd('salary2',{'jack':17000,'lucy':8000}) r.zunionstore('salary3',('salary','salary2'),aggregate='max') print(r.zrange('salary3',0,-1,withscores=True)) # 交集 r.zinterstore('salary4',('salary','salary2'),aggregate='max') print(r.zrange('salary4',0,-1,withscores=True))
案例1：网易音乐排行榜
1、每首歌的歌名作为元素（先不考虑重复）
2、每首歌的播放次数作为分值
3、使用ZREVRANGE来获取播放次数最多的歌曲
import redis r = redis.Redis(host='192.168.43.49',port=6379,password='123456',db=0) r.zadd('ranking',{'song1':1,'song2':1,'song3':1,'song4':1}) r.zadd('ranking',{'song5':1,'song6':1,'song7':1}) r.zadd('ranking',{'song8':1,'song9':1}) r.zincrby('ranking',50,'song3') r.zincrby('ranking',60,'song5') r.zincrby('ranking',80,'song7') # 获取前3名 rlist = r.zrevrange('ranking',0,2,withscores=True) i = 1 for r in rlist: print('第%d名:%s' % (i,r[0].decode())) #print('第{}名:{}播放次数:{}'.format(i,name[0].decode(),int(name[1]))) i += 1
案例2: 京东商品畅销榜
# 第1天
ZADD mobile-001 5000 'huawei' 4000 'oppo' 3000 'iphone'
# 第2天
ZADD mobile-002 5200 'huawei' 4300 'oppo' 3230 'iphone'
# 第3天
ZADD mobile-003 5500 'huawei' 4660 'oppo' 3580 'iphone'
问题：如何获取三款收集的销量排名？
ZUNIONSTORE mobile-001:003 mobile-001 mobile-002 mobile-003 # 可否？
# 正确
1、ZADD mobile-003 5500 'huawei' 4660 'oppo' 3580 'iphone'
2、ZUNIONSTORE mobile-001:003 mobile-001 mobile-002 mobile-003 AGGREGATE MAX
import redis r = redis.Redis(host='127.0.0.1', port=6379, db=0) # 先定义三个集合mobile-001,mobile-002,mobile-003 # 定义三个字典 day01_dict = { 'huawei': 5000, 'oppo': 4000, 'iphone': 3000 } day02_dict = { 'huawei': 5200, 'oppo': 4300, 'iphone': 3890 } day03_dict = { 'huawei': 5500, 'oppo': 4600, 'iphone': 3500 } r.zadd('mobile-001', day01_dict) r.zadd('mobile-002', day02_dict) r.zadd('mobile-003', day03_dict) r.zinterstore( 'mobile-001:003', ('mobile-001', 'mobile-002', 'mobile-003'), aggregate='max' ) # 逆序 res = r.zrevrange('mobile-001:003', 0, 2, withscores=True) for i in res: print('{}-{}'.format(i[0].decode(), i[1]))