Question

我有一个PostgreSQL表，如下所示，

portal   partner     location_quality 
38       10060       46
38       10060       46
38       5586        90
4        10060       90
23       11843       10
38       12594       90
38       10438       50
38       10060       90
23       10438       46
36       12660       90
23       12594       50

对于每对门户网站＆amp;合作伙伴location_quality的范围是0到100.我想要所有不同的门户网站＆amp;如果每对的location_quality的90％大于50，则为partner。

我想得到所有不同的门户网站＆amp;如果90％的location_quality大于50，那么合作伙伴对。有人可以帮忙吗？

Answer 1

对行进行分组
agregad with max

过滤

SELECT
 portal,
 partner,
 max(location_quality) as location_quality
FROM
 table01
GROUP BY
 portal,
 partner
HAVING
 max(location_quality) > 50

现在解决您的问题

统计所有
大于50时计算

取百分比并比较是否是90％

SELECT
 portal,
 partner,
 count(CASE WHEN location_quality > 50 THEN 1 END) * 100 /  count(location_quality ) as percent_above
FROM
 table01
GROUP BY
 portal,
 partner
HAVING
 count(CASE WHEN location_quality > 50 THEN 1 END) * 100 /  count(location_quality ) > 90;

Answer 2

我可能会误解，但这就是90％的位置质量的意思吗？

SELECT
 portal,
 partner,
 location_quality
FROM
 table01
GROUP BY
 portal,
 partner
WHERE
 (location_quality * 0.9) > 50

Answer 3

我提出了与Roger相似的解决方案，

SELECT DISTINCT partner, portal from (
    SELECT partner, portal, sum(case when location_quality > 50 then 1 else null end) as cnt_greater_50, CAST(count(*) AS FLOAT) as total_cnt
    FROM tb
    GROUP BY partner, portal
) WHERE cnt_greater_50/total_cnt > 0.9

还有更好的主意吗？感谢

具有百分比条件的不同SQL选择

3 个答案: