Question

在oracle空间中，我有两个表（ AVALREGULACAO 和 ATROCOADUTOR ），分别代表点和线。

两个表的结构如下：

AVALREGULACAO（ 295点记录）
IPID [number（10）]
几何[MDSYS.SDO_GEOMETRY]

ATROCOADUTOR（ 12536行记录）
IPID [number（10）]
几何[MDSYS.SDO_GEOMETRY]

我需要从每个AVALREGULACAO中找到最近的ATROCOADUTOR邻居，并计算它们之间的距离

AVALREGULACAO_IPID | ATROCOADUTOR _IPID |距离

我使用了2个选项

1

SELECT /*+ ORDERED */ A.IPID, B.IPID, MIN(SDO_GEOM.SDO_DISTANCE(sdo_cs.make_2d(A.GEOMETRY), sdo_cs.make_2d(B.GEOMETRY), 0.005)) as DISTANCE
FROM AVALREGULACAO A, ATROCOADUTOR B 
GROUP BY c_b.IPID,c_d.IPID;

计算需要很长时间-它产生295 x 12536 = 3698120可能的组合（笛卡尔积）的巨大输出。此外，csv文件输出无法容纳所有这些记录（限制为1 048 576行）

我只需要295条对应于295条AVALREGULACAO的记录。

2
我也尝试使用/最接近的邻居（nn）运算符

PROMPT IPID, nearest_IPID, distance  
  select /*+ ORDERED USE_NL(s,s2)*/
         s.IPID,
         s2.IPID as nearest_IPID,
         TO_CHAR(REPLACE(mdsys.sdo_geom.sdo_distance(sdo_cs.make_2d(s.GEOMETRY),sdo_cs.make_2d(s2.GEOMETRY),0.05), ',','.')) as distance
    from AVALREGULACAO s,
         ATROCOADUTOR s2
   where s2.IPID in (select IPID
                from AVALREGULACAO s3
               where sdo_nn(s3.GEOMETRY,s.GEOMETRY,'sdo_batch_size=10',1) = 'TRUE'
                       and s3.IPID <> s.IPID
                       and rownum < 2)
 order by 1,2;

此查询将永远花费-我需要在进程结束之前关闭进程。

我想我缺少关于如何优化/过滤所需结果的要点。

任何有关如何有效解决此问题的技巧将不胜感激。

预先感谢，佩德罗

PS： @Boneist。非常感谢您的投入。

不幸的是，在应用您的查询后出现错误（仍然尝试使用新命令KEEP，deny_rank的语义/语法）

SELECT a.ipid a_ipid,
       MIN(b.ipid) KEEP (dense_rank FIRST order by sdo_nn(a.GEOMETRY,b.GEOMETRY,'sdo_batch_size=10',1)) b_ipid,
       MIN(sdo_geom.sdo_distance(sdo_cs.make_2d(a.geometry), sdo_cs.make_2d(b.geometry), 0.005)) AS distance
FROM   avalregulacao a
       INNER JOIN atrocoadutor b ON sdo_nn(a.GEOMETRY,b.GEOMETRY,'sdo_batch_size=10',1) = 'TRUE'
GROUP  BY a.ipid;

错误

Error starting at line : 1 in command -
SELECT a.ipid a_ipid,
MIN(b.ipid) KEEP (dense_rank FIRST order by sdo_nn(a.GEOMETRY,b.GEOMETRY,'sdo_batch_size=10',1)) b_ipid,
MIN(sdo_geom.sdo_distance(sdo_cs.make_2d(a.geometry), sdo_cs.make_2d(b.geometry), 0.005)) AS distance 
FROM avalregulacao a 
INNER JOIN atrocoadutor b ON sdo_nn(a.GEOMETRY,b.GEOMETRY,'sdo_batch_size=10',1) = 'TRUE'
GROUP  BY a.ipid
Error at Command Line : 2 Column : 45
Error report -
SQL Error: ORA-29907: foram encontradas etiquetas em duplicado em invocações primárias
29907. 00000 -  "found duplicate labels in primary invocations"
*Cause:    There are multiple primary invocations of operators with
           the same number as the label.
*Action:   Use distinct labels in primary invocations.

Answer 1

我认为您可能正在追求类似的东西：

SELECT a.ipid a_ipid,
       MIN(b.ipid) KEEP (dense_rank FIRST order by sdo_nn(a.GEOMETRY,b.GEOMETRY,'sdo_batch_size=10',1)) b_ipid,
       MIN(sdo_geom.sdo_distance(sdo_cs.make_2d(a.geometry), sdo_cs.make_2d(b.geometry), 0.005)) AS distance
FROM   avalregulacao a
       INNER JOIN atrocoadutor b ON sdo_nn(a.GEOMETRY,b.GEOMETRY,'sdo_batch_size=10',1) = 'TRUE'
GROUP  BY a.ipid;

这将两个表连接到最近的neighbor函数，这将减少返回的行数。

MIN(b.ipid) KEEP (dense_rank first order by sdo_nn(a.GEOMETRY,b.GEOMETRY,'sdo_batch_size=10',1))只是返回最低的b.ipid值以获得最小的差异。

（我认为此查询将按原样运行，但我无法对其进行测试。您可能必须进行连接并将sdo_nn(a.GEOMETRY,b.GEOMETRY,'sdo_batch_size=10',1)作为子查询中的列，然后在外部进行分组查询。）

最近的邻居和点与线之间的距离

1 个答案: