我有一个看起来像这样的大数据表
ID Marker Value1 Value2
================================
1 A 10 11
1 B 12 13
1 C 14 15
2 A 10 11
2 B 13 12
2 C
3 A 10 11
3 C 12 13
我想通过以下数据搜索此数据,这些数据是用户输入的,而不是存储在表中:
Marker Value1 Value2
==========================
A 10 11
B 12 13
C 14 14
结果应该是这样的:
ID Marker Value1 Value2 Match?
==========================================
1 A 10 11 true
1 B 12 13 true
1 C 14 15 false
2 A 10 11 true
2 B 13 12 true
2 C false
3 A 10 11 true
3 C 12 13 false
最终,这是必要的(上表不是必需的,它应该说明这些值是如何产生的):
ID Matches Percent
========================
1 2 66%
2 2 66%
3 1 33%
我正在寻找最有前途的方法来使它在SQL中工作(确切地说是PostgreSQL)。
我的想法:
我对这两种方法都不满意,因此是一个问题。如何有效地比较两个这样的表?
答案 0 :(得分:1)
您可以使用CTE预先计算匹配项。然后,简单的聚合即可解决问题。假设您的参数是:
Marker Value1 Value2
==========================
m1 x1 y1
m2 x2 y2
m3 x3 y3
您可以这样做:
with x as (
select
id,
case when
marker = :m1 and (value1 = :x1 and value2 = :y1 or value1 = :y1 and value2 = :x1)
or marker = :m2 and (value1 = :x2 and value2 = :y2 or value1 = :y2 and value2 = :x2)
or marker = :m3 and (value1 = :x3 and value2 = :y3 or value1 = :y3 and value2 = :x3)
then 1 else 0 end as matches
from t
)
select
id,
sum(matches) as matches,
100.0 * sum(matches) / count(*) as percent
from x
group by id
答案 1 :(得分:1)
尝试一下:
CREATE TABLE #Temp
(
Marker nvarchar(50),
Value1 nvarchar(50),
Value2 nvarchar(50)
)
INSERT INTO #Temp Values ('A', '10', '11')
INSERT INTO #Temp Values ('B', '12', '13')
INSERT INTO #Temp Values ('C', '14', '14')
SELECT m.Id, m.Marker, m.Value1, m.Value2,
(Select
CASE
WHEN COUNT(*) = 0 THEN 'False'
WHEN COUNT(*) <> 0 THEN 'True'
END
FROM #Temp t
WHERE t.Marker = m.Marker and t.Value1 = m.Value1 and t.Value2 = m.Value2) as Matches
FROM [Test].[dbo].[Markers] m
ORDER BY Matches DESC
Drop TABLE #Temp
如果这正是您想要的,我会尝试解决它的第二部分。
答案 2 :(得分:1)
可以使用公用表表达式中的VALUES子句提供用户输入,然后可以在与实际表的左联接中使用该用户输入。
with user_input (marker, value1, value2) as (
values
('A', 10, 11),
('B', 12, 13),
('C', 14, 14)
)
select d.id,
count(*) filter (where (d.marker, d.value1, d.value2) is not distinct from (u.marker, u.value1, u.value2)),
100 * count(*) filter (where (d.marker, d.value1, d.value2) is not distinct from (u.marker, u.value1, u.value2)) / cast(count(*) as numeric) as pct
from data d
left join user_input u on (d.marker, d.value1, d.value2) = (u.marker, u.value1, u.value2)
group by d.id
order by d.id;
返回:
id | count | pct
---+-------+------
1 | 2 | 66.67
2 | 2 | 66.67
3 | 1 | 50.00
在线示例:https://rextester.com/OBOOD9042
修改
如果值的顺序无关紧要(因此(12,13)被认为与(13,12)相同,则比较会变得更加复杂。
with user_input (marker, value1, value2) as (
values
('A', 10, 11),
('B', 12, 13),
('C', 14, 14)
)
select d.id,
count(*) filter (where (d.marker, least(d.value1, d.value2), greatest(d.value1, d.value2)) is not distinct from (u.marker, least(u.value1, u.value2), greatest(u.value1, u.value2)))
from data d
left join user_input u on (d.marker, least(d.value1, d.value2), greatest(d.value1, d.value2)) = (u.marker, least(u.value1, u.value2), greatest(u.value1, u.value2))
group by d.id
order by d.id;