[英]How do I get the row number corresponding to each min and max in the table below
您可以使用:
SELECT t.id,
MIN(CASE WHEN t.h = sub.h1 THEN n1 END) AS n1,
MIN(CASE WHEN t.l = sub.l1 THEN n2 END) AS n2
FROM tab_name t
JOIN (SELECT id, MAX(h) AS h1, MIN(l) AS l1
FROM tab_name
GROUP BY id) sub
ON t.id = sub.id
AND (t.h = sub.h1 OR t.l = sub.l1)
GROUP BY t.id;
其他方式:
SELECT id,
min(h) as min_h,
max(h) as max_h,
max( case when low_rn = 1 then n1 end) as n1_for_min,
min( case when high_rn = 1 then n1 end) as n1_for_max
FROM (
SELECT *,
row_number() over (partition by id order by h) as low_rn,
row_number() over (partition by id order by h desc) as high_rn
FROM mytable
) x
WHERE 1 IN (low_rn, high_rn)
GROUP BY id
演示: https : //dbfiddle.uk/?rdbms=postgres_10&fiddle=35fcb41891a8a446bcb55f0a6fd0a774
Postgres具有非常方便的first_value()
函数,该函数可以执行您想要的操作。 。 。 几乎。 唯一的不便是它是窗口函数,而不是分析函数。 可以使用select distinct
来解决:
select distinct id,
first_value(n1) over (partition by id order by h desc) as n1_at_max_h,
first_value(n2) over (partition by id order by l desc) as n2_at_max_l
from t;
这可能是解决此问题的最简单方法。
如果您喜欢聚合,则可以使用:
select id,
( array_agg(n1 order by h desc) )[1] as n1_at_max_h,
( array_agg(n2 order by l desc) )[1] as n2_at_max_l
from t
group by id;
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