这个查询可以简化吗？计算线段的累积长度并将坐标折叠成线串

Question

我有一张road_vertices桌子：

create table road_vertices
(
road_id number,
vertex_index number,
x number,
y number
);

insert into road_vertices values ('100',1,0,5);
insert into road_vertices values ('100',2,10,10);
insert into road_vertices values ('100',3,30,0);
insert into road_vertices values ('100',4,50,10);
insert into road_vertices values ('100',5,60,10);

select * from road_vertices;

   ROAD_ID VERTEX_INDEX             X          Y
---------- --------------- ---------- ----------
       100 1                        0          5
       100 2                       10         10
       100 3                       30          0
       100 4                       50         10
       100 5                       60         10

我需要：

1. 计算线段的累积长度（如上图中灰色文本所示）。
2. 将坐标和累积长度折叠成线串。

这是最终目标：

ROAD_ID     LINESTRING
----------------------------------------------------------------------------
100         LINESTRING M ( 0 5 0, 10 10 11.18, 30 0 33.54, 50 10 55.9, 60 10 65.9)

我想出了一种方法来做到这一点：

--Step #3: Collapse the coordinates and cumulative lengths into a linestring
SELECT
    ROAD_ID,
    'LINESTRING M ( ' || LISTAGG(CUMULATIVE_LENGTH, ', ') 
        WITHIN GROUP (ORDER BY VERTEX_INDEX) || ')'  AS LINESTRING
FROM
    (
    --Step #2: Calculate each line segment's length using the Pythagorean theorem, and add together to get cumulative length
    SELECT  
        ROAD_ID,
        VERTEX_INDEX,
        X || ' '  || Y || ' ' || ROUND(SUM(NVL(SQRT(POWER((X - PREV_X),2) + POWER((Y - PREV_Y),2)),0)) 
            OVER (PARTITION BY ROAD_ID ORDER BY ROAD_ID,VERTEX_INDEX),2) 
            AS CUMULATIVE_LENGTH
    FROM   
        (
        --Step #1: Get the previous X and previous Y for Step #2's Pythagorean theorem calculation
        SELECT
            ROAD_ID,
            VERTEX_INDEX,
            ROUND(X,2) AS X,
            ROUND(Y,2) AS Y,
            LAG (X,1) OVER (PARTITION BY ROAD_ID ORDER BY VERTEX_INDEX) AS PREV_X,
            LAG (Y,1) OVER (PARTITION BY ROAD_ID ORDER BY VERTEX_INDEX) AS PREV_Y
        FROM
            INFRASTR.ROAD_VERTICES
        )
    ) 
GROUP BY 
    ROAD_ID;

然而，这个解决方案相当复杂。它可以简化/改进吗？

Answer 1

函数和查询/视图的组合可能是另一种选择。该函数满足您的第一个要求：“计算每个线段的长度......顶点之间的线的部分。” （该函数需要异常处理和测试！）

-- -----------------------------------------------------------------------------
--   function: calculate the segment length
-- -----------------------------------------------------------------------------
create or replace function seglength(
  x_ number
, oldx_ number
, y_ number
, oldy_ number
)
return number as
begin
  if oldx_ = 0 or oldy_ = 0 then  -- vertex_index 1, no "previous"/old values
    return 0;
  else 
    return round( 
        sqrt( 
          power( ( x_ - oldx_ ), 2 ) 
        + power( ( y_ - oldy_) , 2 ) 
        )
      , 2 
    );
  end if;
end seglength;
/

然后，我们可以使用原始查询的修改版本，如下所示：

select
  d.roadid
, 'LINESTRING M ( ' 
  || listagg( ( round(x,2) || ' '  || round(y,2) || ' ' 
           || seglength(x, d.old_x, y, d.old_y) ) , ', ' ) 
     within group ( order by d.vertexindex )
  || ')' linestring
from (
  select
    roadid
  , vertexindex
  , x
  , y
  , case 
      when vertexindex = 1 then 0 -- zero instead of NULL
      else ( lag (x,1) over ( partition by roadid order by vertexindex ) )
    end old_x
  , case 
      when vertexindex = 1 then 0
      else ( lag (y,1) over ( partition by roadid order by vertexindex ) )
    end old_y  
  from rdvx
  ) d 
group by d.roadid;

输出：

500100  LINESTRING M ( 670113.32 4863724.94 0, 670122.42 4863728.94 9.94, 670259.91 4863776.23 145.39)   
507200  LINESTRING M ( 670147.94 4863628.42 0, 670158.74 4863632.98 11.72, 670298.55 4863680.65 147.72) 

注意：“LINESTRING”中的最后一个值小于您问题中的值。您的原始查询是否实际上计算了顶点 1 和 3 之间的距离？我的理解是“段长度”应该是：距离v1-v1即0，距离v1-v2，距离v2-v3。dbfiddle 在这里

更新

功能：

create or replace function rlength(
  x number
, prev_x number
, y number
, prev_y number
)
return number as
begin
  if prev_x is null or prev_y is null then
    return 0 ;
  else 
    return round( 
      sqrt( 
        power( ( x - prev_x ), 2 ) 
      + power( ( y - prev_y ), 2 ) 
      )
    , 2 
    );
  end if;
end rlength;
/

询问：

with roads_ as (
  select
    road_id
  , vertex_index
  , round( x, 2 ) x
  , round( y, 2 ) y
  , sum ( rlen ) over ( partition by road_id order by road_id, vertex_index )  clength
  from (
    select
      road_id
    , vertex_index
    , x
    , y
    , rlength( 
        x 
      , lag( x,1 ) over ( partition by road_id order by vertex_index )
      , y
      , lag( y,1 ) over ( partition by road_id order by vertex_index )
      ) rlen
    from road_vertices 
    )
)
select 
   road_id
, 'LINESTRING M ( ' 
  || listagg( x || ' ' || y || ' ' || clength , ', ' )
     within group ( order by vertex_index )
  || ' )' linestring
from roads_
group by road_id;

测试数据：

create table road_vertices
(
road_id number,
vertex_index number,
x number,
y number
);

begin
  insert into road_vertices values ('100',1,0,5);
  insert into road_vertices values ('100',2,10,10);
  insert into road_vertices values ('100',3,30,0);
  insert into road_vertices values ('100',4,50,10);
  insert into road_vertices values ('100',5,60,10);
end;
/

输出：

ROAD_ID  LINESTRING                                                               
100      LINESTRING M ( 0 5 0, 10 10 11.18, 30 0 33.54, 50 10 55.9, 60 10 65.9 )