如何在SQL Server中計算字符串的所有trigrams

Question

我想計算和計算SQL Server中字符串的所有trigrams。

例如，如果字符串是hello我想下面的輸出：

 
Trigram Count 
------- ----- 
hel  1 
ell  1 
llo  1 
lo-  1 

Wikipedia page on n-gram

Answer 1

我仍然不知道是什麼的是N-gram但基於Ed的回答是這樣的你需要？

declare @string varchar(max) = 'hello' 
declare @n int = 3 

set @string = @string + REPLICATE('-',@n - (len(@string) % @n)) 

;with n as 
(
SELECT 1 AS i 
UNION ALL 
SELECT i+1 
FROM n 
WHERE i <= (LEN(@string)-@n) 
) 
select SUBSTRING(@string, i, @n), COUNT(*) 
from n 
group by SUBSTRING(@string, i, @n) 
option (maxrecursion 0) 

Answer 2

根據馬丁·史密斯的答案 - 由埃德和馬丁3

declare @string varchar(max) = 'hello' 

SET @string = (SELECT CASE LEN(@string) % 3 
          WHEN 1 THEN @string + '--' 
          WHEN 2 THEN @string + '-' 
          ELSE @string 
         END) 
;with n as 
(
SELECT 1 AS i 
UNION ALL 
SELECT i+1 
FROM n 
WHERE i < (LEN(@string)-2) 
) 
select SUBSTRING(@string, i, 3) AS Trigram, COUNT(*) AS Count 
from n 
group by SUBSTRING(@string, i, 3) 
option (maxrecursion 0) 

Answer 3

借款增加邏輯墊串出與-到數整除的字符，我認爲這是一個正確的實現：

declare @string varchar(max) = 'here kitty kitty' 

SET @string = replace(@string, ' ', '-') --Wikipedia says this should be underscore, not dash 
;with n as 
( 
    SELECT 1 AS i 
    UNION ALL 
    SELECT i + 1 
    FROM n 
    WHERE i < (LEN(@string)-2) 
) 
select SUBSTRING(@string, i, 3) AS Trigram, COUNT(*) AS Count 
from n 
group by SUBSTRING(@string, i, 3) 
option (maxrecursion 0)