通过选择最小差异连接多列和其中一个整数列

Question

我得到了 table t1，我想在下面的 a、b 和 c 列中加入 table t2

+---------+---------+---------+
|a        |b        |c        |
+---------+---------+---------+
|473200   |1        |1.-1-1   |
|472400   |10       |1.-1-1   |
|472800   |10       |1.-1-1   |
|473200   |93       |1.-1-1   |
|472800   |26240    |1.-1-1   |
+---------+---------+---------+

t2

+---------+---------+---------+
|a        |b        |c        |
+---------+---------+---------+
|473200   |1        |1.-1-1   |
|472400   |10       |1.-1-1   |
|472800   |10       |1.-1-1   |
|473200   |93       |1.-1-1   |
|472800   |26250    |1.-1-1   |
+---------+---------+---------+

当我只在 a 和 c 上加入时，结果是

+---------+---------+---------+---------+
|t1.b     |t2.b     |a        |c        |
+---------+---------+---------+---------+
|93       |1        |473200   |1.-1-1   |
|1        |1        |473200   |1.-1-1   |
|10       |10       |472400   |1.-1-1   |
|10       |10       |472800   |1.-1-1   |
|26240    |10       |472800   |1.-1-1   |
|93       |93       |473200   |1.-1-1   |
|1        |93       |473200   |1.-1-1   |
|10       |26250    |472800   |1.-1-1   |
|26240    |26250    |472800   |1.-1-1   |
+---------+---------+---------+---------+

我试图实现的是将列 b 添加到 'on' 子句，以便在 b 列的最小差异时发生连接。

想要的结果

+---------+---------+---------+---------+
|t1.b     |t2.b     |a        |c        |
+---------+---------+---------+---------+
|1        |1        |473200   |1.-1-1   |
|10       |10       |472400   |1.-1-1   |
|10       |10       |472800   |1.-1-1   |
|93       |93       |473200   |1.-1-1   |
|26240    |26250    |472800   |1.-1-1   |
+---------+---------+---------+---------+

我在这里看到了类似的东西

https://dba.stackexchange.com/questions/73804/how-to-retrieve-closest-value-based-on-look-up-table

但不确定如何适用于我的情况。

Answer 1

一个选项是横向连接：

select t1.*, t2.b b2
from t1
cross join lateral (
    select t2.*
    from t2
    where t2.a = t1.a and t2.c = t1.c
    order by abs(t2.b - t1.b)
    limit 1
)

另一种可能性是 distinct on - 但您需要 t1 的主键。假设 (a, c) 元组唯一标识 t1 中的每一行，你会去：

select distinct on (t1.a, t1.c) t1.*, t2.b b2
from t1
inner join t2 on t2.a = t1.a and t2.c = t1.c
order by t1.a, t1.c, abs(t2.b - t1.b)

Answer 2

加入表并计算列 c 的差异，然后使用 distinct on 到 return 每个 (a, c) 仅一行按差异排序。

with joined as (
  select t1.a, t1.c, t1.b as b1, t2.b as b2, t2.b - t1.b as b_diff
    from t1
         join t2 
           on t2.a = t1.a
          and t2.b = t1.b
          and t1.b <= t2.b
)
select distinct on (a, c) b1, b2, a, c
  from joined
 order by a, c, b_diff
;

通过选择最小差异连接多列和其中一个整数列

Join on multiple columns and in one of the integer columns join by choosing minimum difference

sql

postgresql

join

sql-order-by

greatest-n-per-group