Question

我正在寻找一种方法来获取一系列开始和结束日期的MEDIAN（很多和很多日期）。但是，它将特定于各种发票号码。＆＃34;请参阅下面的示例数据。

 invoice_no   invoice start date      invoice end date
 4006            11/14/2001               12/15/2004
 20071  11/29/2001               02/01/2003
 19893  11/30/2001               12/02/2001
 19894  11/30/2001               12/04/2001
 004             10/22/2002               10/31/2002

查找开始日期和结束日期之间的中位数。

中位数只是特定invoice_no的开始日期和结束日期之间的任何值。但是，尽可能地过滤数据。我意识到我也可以做WHERE STATUS＆lt;＆gt; ＆＃39; REJECTED＆＃39;它还应该有助于保持很多不确定的日期。另外，我只想在几个月之间过滤，所以我也加入了BETWEEN DATETIME。

到目前为止

代码：

  WITH
    tmp AS
    (
        SELECT invoice_no,
                invoice_start_date, invoice_end_date, check_date, status_code,
                cast(count(*) OVER (PARTITION BY invoice_no) as float) AS total,
                row_number() OVER (PARTITION BY invoice_no ORDER BY invoice_start_date, invoice_end_date, check_date) AS rn
        FROM    INVOICE_HEADER INNER JOIN
                      INVOICE_HEADER_CUSTOM ON INVOICE_HEADER.invoice_id = INVOICE_HEADER_CUSTOM.invoice_id

                  WHERE status_code <> 'REJECTED' AND 


Check_Date BETWEEN CONVERT(DATETIME, '2014-12-01 00:00:00', 102) AND CONVERT(DATETIME, '2014-12-31 00:00:00', 102)
    )

    SELECT *
    FROM tmp
    WHERE (total / 2.0 - 1) < rn and rn < (total / 2.0 + 1)

Answer 1

你很接近，只错过了PARTITION BY声明中的count条款。

WITH
    tmp AS
    (
        SELECT invoice_no,
                dates,
                cast(count(*) OVER (PARTITION BY invoice_no) as float) AS total,
                row_number() OVER (PARTITION BY invoice_no ORDER BY dates) AS rn
        FROM    INVOICE_HEADER
    )


SELECT *
FROM tmp
WHERE (total / 2.0 - 1) < rn
    and rn < (total / 2.0 + 1)

Answer 2

我会将中位数表示为：

SELECT invoice_no,
       (MIN(date) +
        (DATEDIFF(hour, MIN(DATE), MAX(DATE)) / 2.0)
       )
FROM (SELECT ih.invoice_no
             COUNT(*) OVER (PARTITION BY invoice_no) as cnt,
             ROW_NUMBER() OVER (PARTITION BY invoice_no ORDER BY dates) as seqnum
      FROM INVOICE_HEADER ih
     ) ih
WHERE 2*seqnum in (cnt, cnt + 1, cnt + 2)
GROUP BY invoice_no

大部分内容可能都是不言自明的。您需要partition by对每个发票号进行计算。您需要order by dates以正确的顺序放置值。 where子句是我处理中位数奇数/偶数问题的首选方法。

最后一步有点棘手。当存在偶数个值时，中位数是日期时间的平均值。这可能有点难以计算。而是采用最小日期并添加最大值和最小值之间的差值。当存在偶数个元素时，这会产生平均值。当存在奇数个元素时，这也会产生平均值，因为最小值和最大值是相同的值。

在SQL Server 2008中查找日期中位数

2 个答案: