Question

我正在寻找一种方法来找出字符串中最大字符重复的值。

例如：

 String         NMCR
-----------------------
akhsdjjjaajjj    6   
AABBDDDDDDD      7

Answer 1

我的尝试，以及CTE突出显示的步骤：

 with data as (select 'akhsdjjjaajjj' txt from dual
               union all
               select 'AABBDDDDDDD' txt from dual
               ),  
      chars as(select txt,substr(txt,lvl, 1) c, lvl
            from data join (select level lvl from dual connect by level < 1000) 
              on length(data.txt) >= lvl
              ),  
      counts as (select txt, c, count(*) cnt
            from chars
            group by txt, c
            )  
select txt, max(cnt)  
from counts  
group by txt;

结果：

TXT MAX（CNT）

akhsdjjjaajjj 6

AABBDDDDDDD 7

Answer 2

奇怪的要求，但这是一种方式：

create or replace
function max_repetetive_letter_count (string varchar2) return integer 
is
   letter_col SYS.KU$_VCNT := SYS.KU$_VCNT(); -- A handy collection type
   l_max_count integer;
begin
   letter_col.extend(length(string));

   for i in 1..length(string) loop
      letter_col(i) := substr(string,i,1);
   end loop;

   select max(letter_count) 
   into   l_max_count
   from
   ( select column_value, count(*) letter_count
     from   table(letter_col)
     group by column_value
   );

   return l_max_count;
end;
/

使用示例：

SQL> select string, max_repetetive_letter_count(string)
  2  from 
  3  ( select 'ajkhsdjjjaajjj' as string from dual
  4    union all
  5    select 'AABBDDDDDDD' as string from dual
  6  );

STRING         MAX_REPETETIVE_LETTER_COUNT(STRING)
-------------- -----------------------------------
ajkhsdjjjaajjj                                   7
AABBDDDDDDD                                      7

（注意你的例子中的6是不正确的！）

Answer 3

您也可以使用connect by子句：

SELECT cad,
       (    SELECT MAX (REGEXP_COUNT (cad, SUBSTR (cad, LEVEL, 1)))
              FROM DUAL
        CONNECT BY LEVEL < LENGTH (cad))
  FROM (SELECT 'akhsdjjjaajjj' cad FROM DUAL
        UNION ALL
        SELECT 'AABBDDDDDDD' FROM DUAL)

我希望这也有帮助。

Answer 4

基于问题的原始版本（仅计算重复的字符）：

Oracle安装程序：

CREATE TABLE test ( id, string ) AS
SELECT 1, 'ajkhsdjjjaajjj' FROM DUAL UNION ALL
SELECT 2, 'AABBDDDDDDD' FROM DUAL;

<强>查询：

SELECT id,
       matched_character,
       frequency
FROM   (
  SELECT id,
         matched_character,
         SUM( number_of_repeats ) AS frequency,
         RANK() OVER ( PARTITION BY ID ORDER BY SUM( number_of_repeats ) DESC ) AS rnk
  FROM   (       
    SELECT id,
           REGEXP_SUBSTR( string, '(.)\1+', 1, COLUMN_VALUE, NULL, 1 ) AS matched_character,
           LENGTH( REGEXP_SUBSTR( string, '(.)\1+', 1, COLUMN_VALUE ) ) AS number_of_repeats
    FROM   test t,
           TABLE(
             CAST(
               MULTISET(
                 SELECT LEVEL
                 FROM   DUAL
                 CONNECT BY LEVEL <= REGEXP_COUNT( t.string, '(.)\1+' )
                )
                AS SYS.ODCINUMBERLIST
              )
            ) v
  )
  GROUP BY id, matched_character
)
WHERE  rnk = 1;

<强>结果：

 ID MATCHED_CHARACTER   FREQUENCY
--- ------------------ ----------
  1 j                           6 
  2 D                           7

已更新 - 针对已修改的问题（统计所有字符）：

<强>查询：

SELECT id,
       matched_character,
       frequency
FROM   (
  SELECT id,
         SUBSTR( string, COLUMN_VALUE, 1 ) AS matched_character,
         COUNT(1) AS frequency,
         RANK() OVER ( PARTITION BY ID ORDER BY COUNT(1) DESC ) AS rnk
  FROM   test t,
         TABLE(
           CAST(
             MULTISET(
               SELECT LEVEL
               FROM   DUAL
               CONNECT BY LEVEL <= LENGTH( t.string )
              )
              AS SYS.ODCINUMBERLIST
            )
          ) v
  GROUP BY id, SUBSTR( string, COLUMN_VALUE, 1 )
)
WHERE  rnk = 1;

<强>结果：

        ID MATCHED_CHARACTER  FREQUENCY
---------- ----------------- ----------
         1 j                          7 
         2 D                          7

计算Oracle中最大重复字符的数量

4 个答案: