基于相邻记录插值时间戳的SQL查询
基于相邻记录插值时间戳的SQL查询
提问于 2017-08-13 20:14:34
我使用Oracle并有下表：
create table test as
select to_date('02.05.2017 00:00', 'DD.MM.YYYY HH24:MI') as DT, 203.4 as VAL from dual union all
select to_date('02.05.2017 01:00', 'DD.MM.YYYY HH24:MI') as DT, 206.7 as VAL from dual union all
select to_date('02.05.2017 02:00', 'DD.MM.YYYY HH24:MI') as DT, 208.9 as VAL from dual union all
select to_date('02.05.2017 03:00', 'DD.MM.YYYY HH24:MI') as DT, 211.8 as VAL from dual union all
select to_date('02.05.2017 04:45', 'DD.MM.YYYY HH24:MI') as DT, 212.3 as VAL from dual union all
select to_date('02.05.2017 06:15', 'DD.MM.YYYY HH24:MI') as DT, 214.5 as VAL from dual union all
select to_date('02.05.2017 08:12', 'DD.MM.YYYY HH24:MI') as DT, 215 as VAL from dual
;
DT VAL
----------------------------
02.05.2017 00:00 203.4
02.05.2017 01:00 206.7
02.05.2017 02:00 208.9
02.05.2017 03:00 211.8
02.05.2017 04:45 212.3
02.05.2017 06:15 214.5
02.05.2017 08:12 215 我需要编写SQL查询(或PL/SQL过程)，以便为任何时间戳(DT)插入值(VAL)，假设该值在表中的两个相邻记录之间不断增加(即。线性插值)。
示例：
- 当我为时间戳“02.05.2017 00:00”选择值时，查询应该给我203.4 (表中有这样的时间戳记录)
- 当我为时间戳“02.05.2017 00:30”选择值时，查询应该给我205.05 (表中不存在这样的时间戳记录，所以我们在203.4到206.7之间取“中间”，因为想要的时间戳在它们的时间戳之间处于中间)
- 当我为时间戳“02.05.2017 00:15”选择值时，查询应该给我204.225 ( 203.4到206.7之间的“第四部分”)
解决这类任务的最简单方法是什么？
回答 4
Stack Overflow用户
回答已采纳
发布于 2017-08-14 06:06:30
我认为这个更紧凑，它避免了自我连接：
WITH t AS
(SELECT DT, VAL,
LEAD(DT, 1, DT) OVER (ORDER BY DT) AS FOLLOWING_DT,
LEAD(VAL, 1, VAL) OVER (ORDER BY VAL) AS FOLLOWING_VAL
FROM TEST)
SELECT VAL + (FOLLOWING_VAL - VAL) * ( (:timestamp - DT) / (FOLLOWING_DT - DT) )
FROM t
WHERE :timestamp BETWEEN DT AND FOLLOWING_DT;Stack Overflow用户
发布于 2017-08-13 20:59:43
我认为实现这一目标的最简单方法是使用PL/SQL函数，如下所示：
create or replace function get_val(dt in date) return number
is
cursor exact_cursor(dt in date) is
select t.val from t where t.dt = exact_cursor.dt;
cursor earlier_cursor(dt in date) is
select t.dt, t.val from t where t.dt < earlier_cursor.dt
order by t.dt desc;
cursor later_cursor(dt in date) is
select t.dt, t.val from t where t.dt > later_cursor.dt
order by t.dt asc;
result number;
factor number;
earlier_rec earlier_cursor%rowtype;
later_rec later_cursor%rowtype;
begin
open exact_cursor(dt);
fetch exact_cursor into result;
close exact_cursor;
if result is not null then
return result;
end if;
-- No exact match. Perform linear interpolation between values
-- from earlier and later records.
open earlier_cursor(dt);
fetch earlier_cursor into earlier_rec;
close earlier_cursor;
open later_cursor(dt);
fetch later_cursor into later_rec;
close later_cursor;
-- Return NULL unless earlier and later records found
if earlier_rec.dt is null or later_rec.dt is null then
return null;
end if;
factor := (dt - earlier_rec.dt) / (later_rec.dt - earlier_rec.dt);
result := earlier_rec.val + factor * (later_rec.val - earlier_rec.val);
return result;
end;
/Stack Overflow用户
发布于 2017-08-13 21:43:52
你不需要游标。你需要找到两个最近的记录，一个在上面，一个在下面，然后取它们的平均值。就像这样：
select :timestamp,
(case when lower.val = upper.val then val
else lower.val + (upper.val - lower.val) * ( (:timestamp - lower.dt) / (upper.dt - lower.dt) )
end) as imputed_val
from (select *
from (select dt, val
from t
where dt <= :timestamp
order by dt desc
)
where rownum = 1
) lower cross join
(select *
from (select dt, val
from t
where dt >= :timestamp
order by dt asc
)
where rownum = 1
) upper;页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/45664532
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