计算不同单词在列中出现的次数 Oracle 12c SQL

Posted 2023-05-09

【中文标题】计算不同单词在列中出现的次数 Oracle 12c SQL

【英文标题】：Count how many times distinct words appear in a column Oracle 12c SQL

【发布时间】：2020-11-23 13:14:45

【问题描述】：

如何计算列中所有不同单词出现的次数

以下是示例和预期输出

+--------+------------------------------+
| PERIOD |            STRING            |
+--------+------------------------------+
|        |                              |
| 1      | this is some text            |
|        |                              |
| 2      | more text                    |
|        |                              |
| 3      | this could be some more text |
+--------+------------------------------+

+-------+-------+
| WORD  | COUNT |
+-------+-------+
|       |       |
| this  | 2     |
|       |       |
| is    | 1     |
|       |       |
| some  | 2     |
|       |       |
| text  | 3     |
|       |       |
| more  | 2     |
|       |       |
| could | 1     |
|       |       |
| be    | 1     |
+-------+-------+

谢谢，

【问题讨论】：

你想用纯 SQL 来做，还是可以使用 PL/SQL 之类的语言？

或者，可以使用 PL/SQL 作为解决方案以及 SQL

顺便说一句，有大写字母和小写字母，例如This 和this 有关系吗？例如。两者是否相等？

@BarbarosÖzhan 案例无关紧要，所以两者都是平等的

几乎是***.com/q/38371989/1509264 的副本，只是在末尾添加了COUNT 步骤。

【参考方案1】：

您可以使用分层查询，例如

WITH t2 AS
(
 SELECT REGEXP_SUBSTR(LOWER(string),'[^[:space:]]+',1,level) AS word
   FROM t  
CONNECT BY level <= REGEXP_COUNT(LOWER(string),'[:space:]') + 1
    AND PRIOR SYS_GUID() IS NOT NULL
    AND PRIOR period = period
)    
SELECT word, COUNT(*) AS count
  FROM t2
 WHERE word IS NOT NULL
 GROUP BY word

Demo

附注应用LOWER() 函数是为了解决区分大小写的问题。

【参考方案2】：

诀窍是将字符串拆分为单词。一种方法使用递归 CTE：

with words(word, string, n) as (
      select regexp_substr(string, '[^ ]+', 1, 1) as word, string, 1 as n
      from t
      union all
      select regexp_substr(string, '[^ ]+', 1, n + 1), string, n + 1
      from words
      where regexp_substr(string, '[^ ]+', 1, n + 1) is not null
     )
select word, count(*)
from words
group by word;

Here 是一个 dbfiddle。

【参考方案3】：

您可以使用简单的字符串函数在没有（慢）正则表达式的情况下做到这一点：

WITH word_bounds ( string, start_pos, end_pos ) AS (
  SELECT string,
         1,
         INSTR( string, ' ', 1 )
  FROM   table_name
UNION ALL
  SELECT string,
         end_pos + 1,
         INSTR( string, ' ', end_pos + 1 )
  FROM   word_bounds
  WHERE  end_pos > 0
),
words ( word ) AS (
SELECT CASE end_pos
       WHEN 0
       THEN SUBSTR( string, start_pos )
       ELSE SUBSTR( string, start_pos, end_pos - start_pos )
       END
FROM   word_bounds
)
SELECT word,
       COUNT(*) AS frequency
FROM   words
GROUP BY
       word
ORDER BY
       frequency desc, word;

其中，对于样本数据：

CREATE TABLE table_name ( PERIOD, STRING ) AS
SELECT 1, 'this is some text' FROM DUAL UNION ALL
SELECT 2, 'more text' FROM DUAL UNION ALL
SELECT 3, 'this could be some more text' FROM DUAL

输出：

字 |频率 :---- | --------: 正文 | 3 更多 | 2 一些 | 2 这个 | 2 是| 1 可以| 1 是| 1

有一个关于分割分隔字符串here的不同方式的性能的讨论。

db小提琴here