5

我想计算每个用户在“5”内的行数。

例如，Don - 501 和 Don - 504 应该计算在内，而 Don - 501 和 Don - 1600 则不应计算在内。

开始：

Name        value
_________   ______________
Don         1235
Don         6012
Don         6014
Don         6300
James       9000
James       9502
James       9600
Sarah       1110
Sarah       1111
Sarah       1112
Sarah       1500
Becca       0500
Becca       0508
Becca       0709

结束：

Name            difference_5
__________      _____________
Don             1
James           0
Sarah           2
Becca           0
4

4 回答 4

2

结合子查询中的自联接使用ABS()函数：

所以，像：

SELECT name, COUNT(*) / 2 AS difference_5
FROM (
  SELECT a.name name, ABS(a.value - b.value) 
  FROM  tbl a JOIN tbl b USING(name)
  WHERE ABS(a.value - b.value) BETWEEN 1 AND 5
) AS t GROUP BY name

根据安德烈亚斯的评论进行编辑。

于 2012-08-01T16:21:02.120 回答
1

假设每个name->对都是唯一的，这将为您提供每个 namevalue值在 5 以内的次数：

SELECT    a.name, 
          COUNT(b.name) / 2 AS difference_5
FROM      tbl a
LEFT JOIN tbl b ON a.name = b.name AND 
                   a.value <> b.value AND
                   ABS(a.value - b.value) <= 5
GROUP BY  a.name

正如您会注意到的，我们还必须排除与自身相等的对。

但是，如果您想计算每个名称的值在 table 中任何值的5 以内的次数，您可以使用：

SELECT    a.name,
          COUNT(b.name) / 2 AS difference_5
FROM      tbl a
LEFT JOIN tbl b ON NOT (a.name = b.name AND a.value = b.value) AND
                   ABS(a.value - b.value) <= 5
GROUP BY  a.name

有关这两种解决方案，请参阅SQLFiddle 演示。

于 2012-08-01T19:32:15.357 回答
0

因为 OP 还想要零计数，所以我们需要一个自左连接。如果一个人有两个完全相同的值，则需要额外的逻辑，这些值也应该只计算一次。

WITH cnts AS (
        WITH pair AS (
                SELECT t1.zname,t1.zvalue
                FROM ztable t1
                JOIN ztable t2
                ON t1.zname = t2.zname
                WHERE ( t1.zvalue < t2.zvalue
                        AND t1.zvalue >= t2.zvalue - 5 )
                OR (t1.zvalue = t2.zvalue AND t1.ctid < t2.ctid)
                )
        SELECT DISTINCT zname
        , COUNT(*) AS znumber
        FROM pair
        GROUP BY zname
        )
, names AS (
        SELECT distinct zname  AS zname
        FROM ztable
        GROUP BY zname
        )
SELECT n.zname
        , COALESCE(c.znumber,0) AS znumber
FROM names n
LEFT JOIN cnts c ON n.zname = c.zname
        ;

结果：

DROP SCHEMA
CREATE SCHEMA
SET
CREATE TABLE
INSERT 0 14
 zname | znumber 
-------+---------
 Sarah |       3
 Don   |       1
 Becca |       0
 James |       0
(4 rows)

注意：对于 CTE 抱歉，我没有看到 mysql 标签，我只是喜欢这个问题 ;-)

于 2012-08-01T16:48:15.737 回答
0 
SELECT
    A.Name,
    SUM(CASE WHEN (A.Value < B.Value) AND (A.Value >= B.Value - 5) THEN 1 ELSE 0 END) Difference_5
FROM
    tbl A INNER JOIN
    tbl B USING(Name)
GROUP BY
    A.Name
于 2012-08-01T16:56:24.020 回答