在postgres中用中位数填充缺失值

Question

在此如何用中位数计算替换 avg？

select *
, coalesce(val, avg(val) over (order by t rows between 3 preceding and 1 preceding)) as fixed
from (
    values
    (1, 10),
    (2, NULL),
    (3, 10),
    (4, 15),
    (5, 11),
    (6, NULL),
    (7, NULL),
    (8, NULL),
    (9, NULL)
) as test(t, val)
;

这有合法的版本吗？

percentile_cont(0.5) within group(order by val) over (order by t rows between 3 preceding and 1 preceding)

Answer 1

很遗憾，percentile_cont() 是一个聚合函数，没有等效的窗口函数。

一种解决方法是使用内联子查询进行聚合计算。

如果ids 一直在增加，那么你可以这样做：

select 
    t.*,
    coalesce(
        t.val, 
        (
            select percentile_cont(0.5) within group(order by t1.val)
            from test t1
            where t1.id between t.id - 3 and t.id - 1
        )
    ) fixed
from test t

否则，您需要额外的嵌套级别：

select 
    t.*,
    coalesce(
        t.val, 
        (
            select percentile_cont(0.5) within group(order by t1.val)
            from (select val from test t1 where t1.id < t.id order by t1.id desc limit 3) t1
        )
    ) fixed
from test t

Demo on DB Fiddle - 两个查询都产生：

编号 |值 |固定的 -: | ---: | :---- 1 | 10 | 10 2 | 空 | 10 3 | 10 | 10 4 | 15 | 15 5 | 11 | 11 6 | 空 | 11 7 | 空 | 13 8 | 空 | 11 9 | 空 | 空