Question

我想计算每个用户在彼此的'5'内有多少行。

例如，Don-501和Don-504应计算在内，而Don-501和Don-1600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

完成：

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

Answer 1

将ABS()函数与子查询中的自联接结合使用：

所以，比如：

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Andreas的评论编辑。

Answer 2

假设每个name - > value对是唯一的，这将为您提供值在5 每个名称 的范围内的次数：

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

正如您将注意到的那样，我们还必须排除与自己相等的对。

但是，如果您想计算每个名称的值在表中 任意值的5个范围内的次数，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 Demo。

Answer 3

因为OP也想要零计数，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：对不起CTE，我没见过mysql标签，我只是喜欢这个问题; - ）

Answer 4

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

SQL：查找行之间的差异

4 个答案: