声明

Question

上下文

我有一个应用程序从表中选择一个加权随机条目，其中前缀总和（权重）是关键部分。简化的表定义如下所示：

CREATE TABLE entries (
    id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
    weight DECIMAL(9, 3),
    fenwick DECIMAL(9, 3)
) ENGINE=MEMORY;

其中`fenwick`将值存储在`weights`的Fenwick树表示中。

让每个条目的“范围”跨越其前缀和与其前缀sum +其权重之间。应用程序必须在@r和0之间生成一个随机数SUM(weight)，并找到其范围包含@r的条目，如下所示：

Weighted random entry selection

Fenwick树结合MEMORY引擎和二分搜索，应该允许我在O(lg^2(n))时间内找到相应的条目，而不是O(n)时间与天真的查询：

SELECT a.id-1 FROM (SELECT *, (@x:=@x+weight) AS counter FROM entries 
    CROSS JOIN (SELECT @x:=0) a
    HAVING counter>@r LIMIT 1) a;

研究

由于多个查询的开销，我一直在尝试将前缀sum操作压缩为一个查询（而不是脚本语言中的几个数组访问）。在这个过程中，我意识到传统的求和方法，即涉及按降序键顺序访问元素，只会求和第一个元素。我怀疑MySQL在WHERE子句中存在变量时线性地运行表。这是查询：

SELECT
SUM(1) INTO @garbage
FROM entries 
CROSS JOIN (
    SELECT @sum:=0,
        @n:=@entryid
) a
WHERE id=@n AND @n>0 AND (@n:=@n-(@n&(-@n))) AND (@sum:=@sum+entries.fenwick);
/*SELECT @sum*/

其中@entryid是我们正在计算其前缀和的条目的ID。我确实创建了一个可以工作的查询（以及一个返回整数最左边位的函数lft）：

SET @n:=lft(@entryid);
SET @sum:=0;
SELECT
    SUM(1) INTO @garbage
    FROM entries
    WHERE id=@n 
      AND @n<=@entryid 
      AND (@n:=@n+lft(@entryid^@n)) 
      AND (@sum:=@sum+entries.fenwick);
/*SELECT @sum*/

但它只证实了我对线性搜索的怀疑。 EXPLAIN查询也是如此：

+------+-------------+---------+------+---------------+------+---------+------+--------+-------------+
| id   | select_type | table   | type | possible_keys | key  | key_len | ref  | rows   | Extra       |
+------+-------------+---------+------+---------------+------+---------+------+--------+-------------+
|    1 | SIMPLE      | entries | ALL  | NULL          | NULL | NULL    | NULL | 752544 | Using where |
+------+-------------+---------+------+---------------+------+---------+------+--------+-------------+
1 row in set (0.00 sec)

索引：

SHOW INDEXES FROM entries;
+---------+------------+----------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
| Table   | Non_unique | Key_name | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment | Index_comment |
+---------+------------+----------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
| entries |          0 | PRIMARY  |            1 | id          | NULL      |       752544 |     NULL | NULL   |      | HASH       |         |               |
+---------+------------+----------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
1 row in set (0.00 sec)

现在，我已经看到很多问题，询问如何在WHERE子句中消除变量，以便优化器可以处理查询。但是，我想不出这种查询在没有id=@n的情况下可以做到的方式。我已经考虑将我想要求的条目的关键值放入表中并使用连接，但我相信我会得到不良影响：要么过多的表，要么通过评估@entryid进行线性搜索无论如何。

问题

是否有任何方式强制MySQL使用此查询的索引？如果他们提供此功能，我甚至会尝试不同的DBMS。

Answer 1

声明

Fenwick树对我来说是新的，我只是在找到这篇文章时才发现它们。这里给出的结果是基于我的理解和一些研究，但我绝不是一个芬威克树专家，我可能错过了一些东西。

参考资料

说明fenwick树是如何工作的

https://stackoverflow.com/a/15444954/1157540转载自 https://cs.stackexchange.com/a/10541/38148

https://cs.stackexchange.com/a/42816/38148

fenwick树的用法

https://en.wikipedia.org/wiki/Fenwick_tree

https://en.wikipedia.org/wiki/Prefix_sum

问题1，找到给定条目的权重

给出下表

CREATE TABLE `entries` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `weight` decimal(9,3) DEFAULT NULL,
  `fenwick` decimal(9,3) NOT NULL DEFAULT '0.000',
  PRIMARY KEY (`id`)
) ENGINE=INNODB;

并且已经填充了给定数据（请参阅concat提供的http://sqlfiddle.com/#!9/be1f2/1），如何计算给定条目的权重@entryid？

这里要理解的关键概念是，fenwick索引的结构基于id值本身的数学和按位运算。

查询通常只应使用主键查找（WHERE ID = value）。

使用排序（ORDER BY）或范围（WHERE (value1 < ID) AND (ID < value2))的任何查询都会错过该点，并且不会按预期的顺序遍历树。

例如，使用密钥60：

SET @entryid := 60;

让我们用二进制

分解值60

mysql> SELECT (@entryid & 0x0080) as b8,
    ->        (@entryid & 0x0040) as b7,
    ->        (@entryid & 0x0020) as b6,
    ->        (@entryid & 0x0010) as b5,
    ->        (@entryid & 0x0008) as b4,
    ->        (@entryid & 0x0004) as b3,
    ->        (@entryid & 0x0002) as b2,
    ->        (@entryid & 0x0001) as b1;
+------+------+------+------+------+------+------+------+
| b8   | b7   | b6   | b5   | b4   | b3   | b2   | b1   |
+------+------+------+------+------+------+------+------+
|    0 |    0 |   32 |   16 |    8 |    4 |    0 |    0 |
+------+------+------+------+------+------+------+------+
1 row in set (0.00 sec)

换句话说，只保留设置的位，我们有

32 + 16 + 8 + 4 = 60

现在，逐个删除设置的最低位以导航树：

32 + 16 + 8 + 4 = 60
32 + 16 + 8 = 56
32 + 16 = 48
32

这给出了访问元素60的路径（32,48,56,60）。

请注意，将60转换为(32, 48, 56, 60)只需要对ID值本身进行位数学运算：不需要访问表或数据库，并且可以在发出查询的客户端中完成此计算

然后

元素60的分数权重

mysql> select sum(fenwick) from entries where id in (32, 48, 56, 60);
+--------------+
| sum(fenwick) |
+--------------+
|       32.434 |
+--------------+
1 row in set (0.00 sec)

验证

mysql> select sum(weight) from entries where id <= @entryid;
+-------------+
| sum(weight) |
+-------------+
|      32.434 |
+-------------+
1 row in set (0.00 sec)

现在，我们来比较这些查询的效率。

mysql> explain select sum(fenwick) from entries where id in (32, 48, 56, 60);
+----+-------------+---------+------------+-------+---------------+---------+---------+------+------+----------+-------------+
| id | select_type | table   | partitions | type  | possible_keys | key     | key_len | ref  | rows | filtered | Extra       |
+----+-------------+---------+------------+-------+---------------+---------+---------+------+------+----------+-------------+
|  1 | SIMPLE      | entries | NULL       | range | PRIMARY       | PRIMARY | 4       | NULL |    4 |   100.00 | Using where |
+----+-------------+---------+------------+-------+---------------+---------+---------+------+------+----------+-------------+

或以不同的方式呈现

explain format=json select sum(fenwick) from entries where id in (32, 48, 56, 60);
{
  "query_block": {
    "select_id": 1,
    "cost_info": {
      "query_cost": "5.61"
    },
    "table": {
      "table_name": "entries",
      "access_type": "range",
      "possible_keys": [
        "PRIMARY"
      ],
      "key": "PRIMARY",
      "used_key_parts": [
        "id"
      ],
      "key_length": "4",
      "rows_examined_per_scan": 4,
      "rows_produced_per_join": 4,
      "filtered": "100.00",
      "cost_info": {
        "read_cost": "4.81",
        "eval_cost": "0.80",
        "prefix_cost": "5.61",
        "data_read_per_join": "64"
      },
      "used_columns": [
        "id",
        "fenwick"
      ],
      "attached_condition": "(`test`.`entries`.`id` in (32,48,56,60))"
    }
  }

因此，优化器按主键获取了4行（IN子句中有4个值）。

当不使用fenwick索引时，我们有

mysql> explain select sum(weight) from entries where id <= @entryid;
+----+-------------+---------+------------+-------+---------------+---------+---------+------+------+----------+-------------+
| id | select_type | table   | partitions | type  | possible_keys | key     | key_len | ref  | rows | filtered | Extra       |
+----+-------------+---------+------------+-------+---------------+---------+---------+------+------+----------+-------------+
|  1 | SIMPLE      | entries | NULL       | range | PRIMARY       | PRIMARY | 4       | NULL |   60 |   100.00 | Using where |
+----+-------------+---------+------------+-------+---------------+---------+---------+------+------+----------+-------------+

或者，以不同的方式呈现

explain format=json select sum(weight) from entries where id <= @entryid;
{
  "query_block": {
    "select_id": 1,
    "cost_info": {
      "query_cost": "25.07"
    },
    "table": {
      "table_name": "entries",
      "access_type": "range",
      "possible_keys": [
        "PRIMARY"
      ],
      "key": "PRIMARY",
      "used_key_parts": [
        "id"
      ],
      "key_length": "4",
      "rows_examined_per_scan": 60,
      "rows_produced_per_join": 60,
      "filtered": "100.00",
      "cost_info": {
        "read_cost": "13.07",
        "eval_cost": "12.00",
        "prefix_cost": "25.07",
        "data_read_per_join": "960"
      },
      "used_columns": [
        "id",
        "weight"
      ],
      "attached_condition": "(`test`.`entries`.`id` <= (@`entryid`))"
    }
  }

优化器执行索引扫描，读取60行。

当ID = 60时，fenwick的好处是4次获取而不是60次。

现在，考虑一下如何缩放，例如，值高达64K。

使用fenwick，16位值最多可设置16位，因此要查找的元素数最多为16位。

如果没有fenwick，扫描最多可以读取64K个条目（平均读数为32K）。

问题2，找到给定重量的条目

OP的问题是为给定的体重找到一个条目。

例如

SET @search_weight := 35.123;

为了说明算法，这篇文章详细说明了查找的完成情况（对不起，如果这太详细了）

SET @found_id := 0;

首先，找出有多少条目。

SET @max_id := (select id from entries order by id desc limit 1);

在测试数据中，max_id为156。

因为128＆lt; = max_id＆lt; 256，开始搜索的最高位是128。

mysql> set @search_id := @found_id + 128;
mysql> select id, fenwick, @search_weight,
    ->    if (fenwick <= @search_weight, "keep", "discard") as action
    ->    from entries where id = @search_id;
+-----+---------+----------------+---------+
| id  | fenwick | @search_weight | action  |
+-----+---------+----------------+---------+
| 128 |  66.540 |         35.123 | discard |
+-----+---------+----------------+---------+

重量66.540大于我们的搜索，所以128被丢弃，继续下一位。

mysql> set @search_id := @found_id + 64;
mysql> select id, fenwick, @search_weight,
    ->    if (fenwick <= @search_weight, "keep", "discard") as action
    ->    from entries where id = @search_id;
+----+---------+----------------+--------+
| id | fenwick | @search_weight | action |
+----+---------+----------------+--------+
| 64 |  33.950 |         35.123 | keep   |
+----+---------+----------------+--------+

这里我们需要保持这个位（64），并计算找到的重量：

set @found_id := @search_id, @search_weight := @search_weight - 33.950;

然后继续下一位：

mysql> set @search_id := @found_id + 32;
mysql> select id, fenwick, @search_weight,
    ->    if (fenwick <= @search_weight, "keep", "discard") as action
    ->    from entries where id = @search_id;
+----+---------+----------------+---------+
| id | fenwick | @search_weight | action  |
+----+---------+----------------+---------+
| 96 |  16.260 |          1.173 | discard |
+----+---------+----------------+---------+

mysql> set @search_id := @found_id + 16;
mysql> select id, fenwick, @search_weight,
    ->    if (fenwick <= @search_weight, "keep", "discard") as action
    ->    from entries where id = @search_id;
+----+---------+----------------+---------+
| id | fenwick | @search_weight | action  |
+----+---------+----------------+---------+
| 80 |   7.394 |          1.173 | discard |
+----+---------+----------------+---------+

mysql> set @search_id := @found_id + 8;
mysql> select id, fenwick, @search_weight,
    ->    if (fenwick <= @search_weight, "keep", "discard") as action
    ->    from entries where id = @search_id;
+----+---------+----------------+---------+
| id | fenwick | @search_weight | action  |
+----+---------+----------------+---------+
| 72 |   3.995 |          1.173 | discard |
+----+---------+----------------+---------+

mysql> set @search_id := @found_id + 4;
mysql> select id, fenwick, @search_weight,
    ->    if (fenwick <= @search_weight, "keep", "discard") as action
    ->    from entries where id = @search_id;
+----+---------+----------------+---------+
| id | fenwick | @search_weight | action  |
+----+---------+----------------+---------+
| 68 |   1.915 |          1.173 | discard |
+----+---------+----------------+---------+

mysql> set @search_id := @found_id + 2;
mysql> select id, fenwick, @search_weight,
    ->    if (fenwick <= @search_weight, "keep", "discard") as action
    ->    from entries where id = @search_id;
+----+---------+----------------+--------+
| id | fenwick | @search_weight | action |
+----+---------+----------------+--------+
| 66 |   1.146 |          1.173 | keep   |
+----+---------+----------------+--------+

我们在这里发现了另一点

set @found_id := @search_id, @search_weight := @search_weight - 1.146;

mysql> set @search_id := @found_id + 1;
mysql> select id, fenwick, @search_weight,
    ->    if (fenwick <= @search_weight, "keep", "discard") as action
    ->    from entries where id = @search_id;
+----+---------+----------------+--------+
| id | fenwick | @search_weight | action |
+----+---------+----------------+--------+
| 67 |   0.010 |          0.027 | keep   |
+----+---------+----------------+--------+

还有一个

set @found_id := @search_id, @search_weight := @search_weight - 0.010;

最终搜索结果是：

mysql> select @found_id, @search_weight;
+-----------+----------------+
| @found_id | @search_weight |
+-----------+----------------+
|        67 |          0.017 |
+-----------+----------------+

验证

mysql> select sum(weight) from entries where id <= 67;        
+-------------+                                               
| sum(weight) |                                               
+-------------+                                               
|      35.106 |                                               
+-------------+                                               

mysql> select sum(weight) from entries where id <= 68;
+-------------+
| sum(weight) |
+-------------+
|      35.865 |
+-------------+

事实上，

35.106 (fenwick[67]) <= 35.123 (search) <= 35.865 (fenwick[68])

搜索查找值一次解析1位，每个查找结果决定要搜索的下一个ID的值。

此处给出的查询仅供参考。在实际应用程序中，代码应该只是一个包含以下内容的循环：

SELECT fenwick from entries where id = ?;

使用应用程序代码（或存储过程）实现与@found_id，@ search_id和@search_weight相关的逻辑。

一般性评论

Fenwick树用于前缀计算。如果要解决的问题首先涉及前缀，那么使用这些树才有意义。维基百科有一些应用程序的指针。不幸的是，OP没有描述fenwick树的用途。
Fenwick树是查找复杂性和更新复杂性之间的权衡，因此它们只对非静态数据有意义。对于静态数据，计算一次完整前缀将更有效。
执行的测试使用了一个INNODB表，主键索引对其进行了排序，因此计算max_id是一个简单的O（1）操作。如果max_id已在其他地方可用，我认为即使使用带有HASH索引ID的MEMORY表也可以，因为只使用主键查找。

P.S。

sqlfiddle今天已经关闭了，所以发布使用的原始数据（最初由concat提供），以便有兴趣的人可以重新运行测试。

INSERT INTO `entries` VALUES (1,0.480,0.480),(2,0.542,1.022),(3,0.269,0.269),(4,0.721,2.012),(5,0.798,0.798),(6,0.825,1.623),(7,0.731,0.731),(8,0.181,4.547),(9,0.711,0.711),(10,0.013,0.724),(11,0.930,0.930),(12,0.613,2.267),(13,0.276,0.276),(14,0.539,0.815),(15,0.867,0.867),(16,0.718,9.214),(17,0.991,0.991),(18,0.801,1.792),(19,0.033,0.033),(20,0.759,2.584),(21,0.698,0.698),(22,0.212,0.910),(23,0.965,0.965),(24,0.189,4.648),(25,0.049,0.049),(26,0.678,0.727),(27,0.245,0.245),(28,0.190,1.162),(29,0.214,0.214),(30,0.502,0.716),(31,0.868,0.868),(32,0.834,17.442),(33,0.566,0.566),(34,0.327,0.893),(35,0.939,0.939),(36,0.713,2.545),(37,0.747,0.747),(38,0.595,1.342),(39,0.733,0.733),(40,0.884,5.504),(41,0.218,0.218),(42,0.437,0.655),(43,0.532,0.532),(44,0.350,1.537),(45,0.154,0.154),(46,0.721,0.875),(47,0.140,0.140),(48,0.538,8.594),(49,0.271,0.271),(50,0.739,1.010),(51,0.884,0.884),(52,0.203,2.097),(53,0.361,0.361),(54,0.197,0.558),(55,0.903,0.903),(56,0.923,4.481),(57,0.906,0.906),(58,0.761,1.667),(59,0.089,0.089),(60,0.161,1.917),(61,0.537,0.537),(62,0.201,0.738),(63,0.397,0.397),(64,0.381,33.950),(65,0.715,0.715),(66,0.431,1.146),(67,0.010,0.010),(68,0.759,1.915),(69,0.763,0.763),(70,0.537,1.300),(71,0.399,0.399),(72,0.381,3.995),(73,0.709,0.709),(74,0.401,1.110),(75,0.880,0.880),(76,0.198,2.188),(77,0.348,0.348),(78,0.148,0.496),(79,0.693,0.693),(80,0.022,7.394),(81,0.031,0.031),(82,0.089,0.120),(83,0.353,0.353),(84,0.498,0.971),(85,0.428,0.428),(86,0.650,1.078),(87,0.963,0.963),(88,0.866,3.878),(89,0.442,0.442),(90,0.610,1.052),(91,0.725,0.725),(92,0.797,2.574),(93,0.808,0.808),(94,0.648,1.456),(95,0.817,0.817),(96,0.141,16.260),(97,0.256,0.256),(98,0.855,1.111),(99,0.508,0.508),(100,0.976,2.595),(101,0.353,0.353),(102,0.840,1.193),(103,0.139,0.139),(104,0.178,4.105),(105,0.469,0.469),(106,0.814,1.283),(107,0.664,0.664),(108,0.876,2.823),(109,0.390,0.390),(110,0.323,0.713),(111,0.442,0.442),(112,0.241,8.324),(113,0.881,0.881),(114,0.681,1.562),(115,0.760,0.760),(116,0.760,3.082),(117,0.518,0.518),(118,0.313,0.831),(119,0.008,0.008),(120,0.103,4.024),(121,0.488,0.488),(122,0.135,0.623),(123,0.207,0.207),(124,0.633,1.463),(125,0.542,0.542),(126,0.812,1.354),(127,0.433,0.433),(128,0.732,66.540),(129,0.358,0.358),(130,0.594,0.952),(131,0.897,0.897),(132,0.701,2.550),(133,0.815,0.815),(134,0.973,1.788),(135,0.419,0.419),(136,0.175,4.932),(137,0.620,0.620),(138,0.573,1.193),(139,0.004,0.004),(140,0.304,1.501),(141,0.508,0.508),(142,0.629,1.137),(143,0.618,0.618),(144,0.206,8.394),(145,0.175,0.175),(146,0.255,0.430),(147,0.750,0.750),(148,0.987,2.167),(149,0.683,0.683),(150,0.453,1.136),(151,0.219,0.219),(152,0.734,4.256),(153,0.016,0.016),(154,0.874,0.891),(155,0.325,0.325),(156,0.002,1.217);

P.S。 2

现在有一个完整的sqlfiddle：

http://sqlfiddle.com/#!9/d2c82/1

Answer 2

（使用的答案框，因为它可以选择格式化）。

如Rick所示，发动机是这种情况下的主要问题。您可以通过在索引创建中使用“USING BTREE”来影响创建的索引类型（在这种情况下，BTREE或HASH似乎并不重要：迭代一个范围：然后BTREE是最佳的。但是你可以检索它值，然后HASH是最佳的：您的查询有两种行为。）

当您切换到INNODB时，缓存将使查询可能与内存表中的查询一样快。然后，您可以获得索引的好处。为了保证BTREE索引，我将按如下方式创建模式：

CREATE TABLE `entries` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `weight` decimal(9,3) DEFAULT NULL,
  `fenwick` decimal(9,3) NOT NULL DEFAULT '0.000',
  PRIMARY KEY (`id`)
) ENGINE=INNODB DEFAULT CHARSET=latin1;

CREATE UNIQUE INDEX idx_nn_1 ON entries (id,fenwick) USING BTREE;

这在主计算中使用idx_nn_1索引（并且仅使用索引：由于所有数据都在索引中，因此根本不使用整个表）。然而，100条记录的样本量太小，无法给出关于性能的任何明确答案。与仅通过使用表访问的数据相比，索引构建的时间可能会使您根本没有任何性能提升。所以最后的答案将在你的测试中。

其他数据库引擎（SQL Server，Oracle，Postgres）：它们将显示类似的行为。因此切换到这些引擎中的任何一个都不会产生很大的不同，除了可能通常更好的处理（无法预测这一点）。

SQL Server在构建索引时可能会更好（=更快），因为它只会在id上使用唯一索引并包含fenwick值，因此无需真正索引该值。

Oracle可以真正强制索引，但是不建议这样做：在Oracle中，假设表中有序数据，读取表比读取索引更快，然后查找表。同样在这种情况下，您只需添加id，fenwick索引并永远不会访问该表。考虑到索引创建时间，Oracle无论如何都必须读取整个表，并且在那段时间内（或者更少取决于达到退出条件所需的记录数），它也会执行您的计算。

Answer 3

Fenwick树是否足够静态以预先计算某些东西？如果是这样，我可以给你一个几乎O（1）的解决方案：

构建一个2列表（n，cumulative_sum）
预填表：（1,0.851），（2,2.574），（3,2.916），（4,3.817），......
在FLOAT上创建索引

然后查找：

SELECT n FROM tbl WHERE cumulative_sum > 3.325
         ORDER BY cumulative_sum LIMIT 1;

如果@variables出现问题，那么让存储过程通过CONCAT，PREPARE和EXECUTE构建SQL。

<强>附加物

鉴于它是定期完全替换，请在重建表时计算累积总和。我的SELECT只查看一行，因此它是O（1）（忽略BTree查找）。

对于“完全替换”，我建议：

CREATE TABLE new LIKE real;
load the data into `new`                -- this is the slowest step
RENAME TABLE real TO old, new TO real;  -- atomic and fast, so no "downtime"
DROP TABLE old;

Answer 4

为了补充Marc的答案，对于存储过程或函数，无法直接通过函数参数传递要求的索引列表，我们可以在查询中生成索引，JOIN它到求和查询：

SELECT SUM(fenwick) FROM entries 
    CROSS JOIN (SELECT @n:=60) a 
    INNER JOIN (
          SELECT @n AS id, 0 
          UNION 
          SELECT 
              IF(@n&@bit<>0, @n:=@n-(@n&@bit), NULL) AS id,
              (@bit:=@bit<<1) 
          FROM entries 
          CROSS JOIN (SELECT @bit:=1) a 
          LIMIT 32
    ) dt ON dt.id=entries.id;

我希望性能类似，客户端不再需要生成索引。

MySQL：强制查询在WHERE子句中使用带有局部变量的索引

上下文

研究

问题

4 个答案:

声明

参考资料

问题1，找到给定条目的权重

问题2，找到给定重量的条目

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