在sql中的朴素贝叶斯计算

Question

在sql中的朴素贝叶斯计算

ʞɔı*_*ɔıu 9 mysql sql nlp machine-learning bayesian

我想使用朴素的贝叶斯将文档分类为相对大量的类.我想确认一篇文章中实体名称的提及是否真的是该实体,这取决于该文章是否与该实体已被正确验证的文章类似.

比如说,我们在一篇文章中找到了"通用汽车"的文字.我们有一组数据,其中包含文章和中提到的正确实体.因此,如果我们在新文章中找到"通用汽车",那么它是否属于先前数据中包含已知正版的那类文章提到"通用汽车"与没有提到该实体的文章类别？

(我不是为每个实体创建一个类,并试图将每个新文章分类到每个可能的类中.我已经有一个启发式方法来查找实体名称的合理提及,我只是想验证有限数量的实体的可信度.实体名称提到该方法已经检测到的每篇文章.)

鉴于潜在的类和文章的数量相当大,天真的贝叶斯相对简单,我想在sql中完成整个过程,但我在评分查询时遇到问题...

这是我到目前为止所拥有的:

CREATE TABLE `each_entity_word` (
  `word` varchar(20) NOT NULL,
  `entity_id` int(10) unsigned NOT NULL,
  `word_count` mediumint(8) unsigned NOT NULL,
  PRIMARY KEY (`word`, `entity_id`)
);

CREATE TABLE `each_entity_sum` (
  `entity_id` int(10) unsigned NOT NULL DEFAULT '0',
  `word_count_sum` int(10) unsigned DEFAULT NULL,
  `doc_count` mediumint(8) unsigned NOT NULL,
  PRIMARY KEY (`entity_id`)
);

CREATE TABLE `total_entity_word` (
  `word` varchar(20) NOT NULL,
  `word_count` int(10) unsigned NOT NULL,
  PRIMARY KEY (`word`)
);

CREATE TABLE `total_entity_sum` (
  `word_count_sum` bigint(20) unsigned NOT NULL,
  `doc_count` int(10) unsigned NOT NULL,
  `pkey` enum('singleton') NOT NULL DEFAULT 'singleton',
  PRIMARY KEY (`pkey`)
);

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标记数据中的每篇文章被分成不同的单词,并且对于每个实体的每篇文章,每个单词被添加到each_entity_word和/或其word_count递增并且相对于a doc_count递增.对于该文章中已知的每个实体重复这一过程.entity_word_sumentity_id

对于每篇文章而言,无论每个单词total_entity_word total_entity_word_sum中包含的实体是否同样递增.

P(词|任何文件)应该等于 word_count在total_entity_word过了这个词 doc_count在total_entity_sum
P(词|文件提到实体X)应等于word_count在 each_entity_word对这个词的entity_id X了doc_count在 each_entity_sum对entity_id X
P(词|文件并没有提及实体X)应该等于(在word_count中total_entity_word减去其word_count在each_entity_word该单词为单位)以上(在doc_count中total_entity_sum减去doc_count在该实体each_entity_sum)
P(文件提到实体X)应等于doc_count在each_entity_sum该实体ID通过doc_count在total_entity_word
P(文件未提及实体X)应该等于1负(doc_count在each_entity_sum对X的实体ID过doc_count的total_entity_word).

对于一篇新文章,将其拆分为单词,然后选择('我','想要','到','使用'...)中的任何一个each_entity_word或哪个单词total_entity_word.在我正在使用的数据库平台(mysql)IN子句中进行了相对较好的优化.

在sql中也没有product()聚合函数,所以你当然可以只用sum(log(x))或exp(sum(log(x))来得到product(x)的等价物.

所以,如果我收到一篇新文章,将其分成不同的单词,并将这些单词放入一个大的IN()子句和一个潜在的实体id进行测试,我怎样才能得到该文章落入该实体的幼稚贝叶斯概率sql中的id类？

编辑:

尝试#1:

set @entity_id = 1;

select @entity_doc_count = doc_count from each_entity_sum where entity_id=@entity_id;

select @total_doc_count = doc_count from total_entity_sum;

select 
            exp(

                log(@entity_doc_count / @total_doc_count) + 

                (
                    sum(log((ifnull(ew.word_count,0) + 1) / @entity_doc_count)) / 
                    sum(log(((aew.word_count + 1) - ifnull(ew.word_count, 0)) / (@total_doc_count - @entity_doc_count)))
                )

            ) as likelihood,
        from total_entity_word aew 
        left outer join each_entity_word ew on ew.word=aew.word and ew.entity_id=@entity_id

        where aew.word in ('I', 'want', 'to', 'use'...);

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Answer 1

ram*_*jan 6

使用R到Postgres(或MySQL等)界面

或者,我建议使用已建立的stats包以及db的连接器.如果您想从Naive Bayes切换到更复杂的东西,这将使您的应用程序更加灵活:

http://rpgsql.sourceforge.net/

bnd.pr> data(airquality)

bnd.pr> db.write.table(airquality, no.clobber = F)

bnd.pr> bind.proxy("airquality")

bnd.pr> summary(airquality)
Table name: airquality 
Database: test 
Host: localhost
Dimensions: 6 (columns) 153 (rows)


bnd.pr> print(airquality)
   Day Month Ozone Solar.R Temp
1    1     5    41     190   67
2    2     5    36     118   72
3    3     5    12     149   74
4    4     5    18     313   62
5    5     5    NA      NA   56
6    6     5    28      NA   66
7    7     5    23     299   65
8    8     5    19      99   59
9    9     5     8      19   61
10  10     5    NA     194   69
Continues for 143 more rows and 1 more cols...

bnd.pr> airquality[50:55, ]
   Ozone Solar.R Wind Temp Month Day
50    12     120 11.5   73     6  19
51    13     137 10.3   76     6  20
52    NA     150  6.3   77     6  21
53    NA      59  1.7   76     6  22
54    NA      91  4.6   76     6  23
55    NA     250  6.3   76     6  24

bnd.pr> airquality[["Ozone"]]
  [1]  41  36  12  18  NA  28  23  19   8  NA   7  16  11  14  18  14  34   6
 [19]  30  11   1  11   4  32  NA  NA  NA  23  45 115  37  NA  NA  NA  NA  NA
 [37]  NA  29  NA  71  39  NA  NA  23  NA  NA  21  37  20  12  13  NA  NA  NA
 [55]  NA  NA  NA  NA  NA  NA  NA 135  49  32  NA  64  40  77  97  97  85  NA
 [73]  10  27  NA   7  48  35  61  79  63  16  NA  NA  80 108  20  52  82  50
 [91]  64  59  39   9  16  78  35  66 122  89 110  NA  NA  44  28  65  NA  22
[109]  59  23  31  44  21   9  NA  45 168  73  NA  76 118  84  85  96  78  73
[127]  91  47  32  20  23  21  24  44  21  28   9  13  46  18  13  24  16  13
[145]  23  36   7  14  30  NA  14  18  20

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然后,您将要安装e1071软件包以执行Naive Bayes.在R提示符处:

[ramanujan:~/base]$R

R version 2.7.2 (2008-08-25)
Copyright (C) 2008 The R Foundation for Statistical Computing
ISBN 3-900051-07-0

R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.

R is a collaborative project with many contributors.
Type 'contributors()' for more information and
'citation()' on how to cite R or R packages in publications.

Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.


 ~/.Rprofile loaded.
Welcome at  Sun Apr 19 00:45:30 2009
> install.packages("e1071")  
> install.packages("mlbench")
> library(e1071)
> ?naiveBayes
> example(naiveBayes)

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更多信息:

http://cran.r-project.org/web/packages/e1071/index.html

Answer 2

Qua*_*noi 1

我没有时间计算公式的所有表达式NB，但主要思想如下：

SET @entity = 123;

SELECT  EXP(SUM(LOG(probability))) / (EXP(SUM(LOG(probability))) + EXP(SUM(LOG(1 - probability))))
FROM    (
        SELECT  @entity AS _entity,
                /* Above is required for efficiency, subqueries using _entity will be DEPENDENT and use the indexes */
                (
                SELECT  SUM(word_count)
                FROM    total_entity_word
                WHERE   word = d.word
                )
                /
                (
                SELECT  doc_count
                FROM    each_entity_sum
                WHERE   entity_id = _entity
                ) AS pwordentity,
                /* I've just referenced a previously selected field */
                (
                SELECT  1 - pwordentity
                ) AS pwordnotentity,
                /* Again referenced a previously selected field */
                ... etc AS probability
        FROM    total_entity_word
        ) q

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请注意，您可以通过在相关子查询中使用它们来轻松引用前一个字段SELECT（如示例所示）。

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