Question

想象一下跟踪多个物体位置的GPS跟踪系统。这些点存储在数据库中（PostgreSQL + PostGIS）。

每条路径由不同数量的点组成。这就是为什么，为了比较一对路径（比较整个路径），我想在一组100个点中划分每个路径。这就是问题。您知道任何已实现此算法的PostGIS功能吗？我找不到它。

如果没有，我想用Java解决它。在这种情况下，我想知道一个高效且易于实现的算法，将路径划分为N个点。

最简单的例子可能是将这四点路径分为八个点：

position 1 : x=1, y=2
position 2 : x=2, y=4
position 3 : x=3, y=6
position 4 : x=4, y=8

结果应该是：

position 1 : x=1, y=2 (starting point)
position 2 : x=1.5, y=3
position 2 : x=2, y=4
position 2 : x=2.5, y=5
position 2 : x=3, y=6
position 2 : x=3.5, y=7
position 2 : x=4, y=8 (ending point)

编辑：通过“比较一对路径”，我的意思是计算两条完整路径之间的距离。我计划将每条路径划分为100个点，并将每个点之间的欧氏距离相加作为两条路径之间的距离。

Answer 1

我不确定我到底知道你想要什么。尽管如此，这个PostGIS查询将占用一行并沿该行等距离吐出100个点：

SELECT ST_AsText(
  ST_Line_Interpolate_Point(
    ST_GeomFromText('LINESTRING(0 0, 1 2, 10 2)'),
    generate_series(0, 100):: double precision / 100
  )
);

显然你会使用真正的几何而不是ST_GeomFromText（..），并且不会包含ST_AsText（...）调用。结果如文字所示：

                  st_astext                  
---------------------------------------------
 POINT(0 0)
 POINT(0.0502492235949962 0.100498447189992)
 POINT(0.100498447189992 0.200996894379985)
 POINT(0.150747670784989 0.301495341569977)
 POINT(0.200996894379985 0.40199378875997)
 POINT(0.251246117974981 0.502492235949962)
 POINT(0.301495341569977 0.602990683139955)
 POINT(0.351744565164974 0.703489130329947)
 POINT(0.40199378875997 0.803987577519939)
 POINT(0.452243012354966 0.904486024709932)
 POINT(0.502492235949962 1.00498447189992)
 POINT(0.552741459544958 1.10548291908992)
 POINT(0.602990683139955 1.20598136627991)
 POINT(0.653239906734951 1.3064798134699)
 POINT(0.703489130329947 1.40697826065989)
 POINT(0.753738353924943 1.50747670784989)
 POINT(0.803987577519939 1.60797515503988)
 POINT(0.854236801114936 1.70847360222987)
 POINT(0.904486024709932 1.80897204941986)
 POINT(0.954735248304928 1.90947049660986)
 POINT(1.01114561800017 2)
 POINT(1.12350629777517 2)
 POINT(1.23586697755016 2)
 POINT(1.34822765732516 2)
 POINT(1.46058833710016 2)
 POINT(1.57294901687516 2)
 POINT(1.68530969665016 2)
 POINT(1.79767037642515 2)
 POINT(1.91003105620015 2)
 POINT(2.02239173597515 2)
 POINT(2.13475241575015 2)
 POINT(2.24711309552515 2)
 POINT(2.35947377530014 2)
 POINT(2.47183445507514 2)
 POINT(2.58419513485014 2)
 POINT(2.69655581462514 2)
 POINT(2.80891649440013 2)
 POINT(2.92127717417513 2)
 POINT(3.03363785395013 2)
 POINT(3.14599853372513 2)
 POINT(3.25835921350013 2)
 POINT(3.37071989327512 2)
 POINT(3.48308057305012 2)
 POINT(3.59544125282512 2)
 POINT(3.70780193260012 2)
 POINT(3.82016261237512 2)
 POINT(3.93252329215011 2)
 POINT(4.04488397192511 2)
 POINT(4.15724465170011 2)
 POINT(4.26960533147511 2)
 POINT(4.38196601125011 2)
 POINT(4.4943266910251 2)
 POINT(4.6066873708001 2)
 POINT(4.7190480505751 2)
 POINT(4.8314087303501 2)
 POINT(4.9437694101251 2)
 POINT(5.05613008990009 2)
 POINT(5.16849076967509 2)
 POINT(5.28085144945009 2)
 POINT(5.39321212922509 2)
 POINT(5.50557280900008 2)
 POINT(5.61793348877508 2)
 POINT(5.73029416855008 2)
 POINT(5.84265484832508 2)
 POINT(5.95501552810008 2)
 POINT(6.06737620787507 2)
 POINT(6.17973688765007 2)
 POINT(6.29209756742507 2)
 POINT(6.40445824720007 2)
 POINT(6.51681892697506 2)
 POINT(6.62917960675006 2)
 POINT(6.74154028652506 2)
 POINT(6.85390096630006 2)
 POINT(6.96626164607506 2)
 POINT(7.07862232585005 2)
 POINT(7.19098300562505 2)
POINT(7.30334368540005 2)
 POINT(7.41570436517505 2)
 POINT(7.52806504495005 2)
 POINT(7.64042572472504 2)
 POINT(7.75278640450004 2)
 POINT(7.86514708427504 2)
 POINT(7.97750776405004 2)
 POINT(8.08986844382504 2)
 POINT(8.20222912360003 2)
 POINT(8.31458980337503 2)
 POINT(8.42695048315003 2)
 POINT(8.53931116292503 2)
 POINT(8.65167184270003 2)
 POINT(8.76403252247502 2)
 POINT(8.87639320225002 2)
 POINT(8.98875388202502 2)
 POINT(9.10111456180002 2)
 POINT(9.21347524157501 2)
 POINT(9.32583592135001 2)
 POINT(9.43819660112501 2)
 POINT(9.55055728090001 2)
 POINT(9.66291796067501 2)
 POINT(9.77527864045 2)
 POINT(9.887639320225 2)
 POINT(10 2)

Answer 2

我对将路径分成100个部分的目的感到困惑。获得到路径末端的距离的明显解决方案是距离公式：

distance = Math.sqrt(Math.pow(2, (x1-x2)) + Math.pow(2, (y1-y2));

如果你想找到两点之间某条路线的距离（比如跟随街道和道路建筑物），那么你可以通过这样一系列的“目的地”来完成，这是一个完全不同的问题，但是您的示例将点显示为点之间的直线。（我假设你的意思是输入'位置2'是x = 9，y = 3）。

编辑：如果你的意思是给你一个点集合并需要找到沿路径的距离，你将使用如下相同的公式：

double distance = 0;
for(int i=1; i<numberOfLocations; i++)
{
    Location oldLoc = collection.get(i-1);
    Location nextLoc = collection.get(i);
    int x = nextLoc.getX() - oldLoc.getX();
    int y = nextLoc.getY() - oldLoc.getY();
    distance = distance + Math.sqrt(Math.pow(2, x) + Math.pow(2, y);
}

...距离是每个点之间距离的总和。

如果问题在于将路径划分为“N”部分，则可以使用以下内容：

//numberOfPoints must be greater than 1.
public ArrayList<Location> divideIntoPoints(Location pointA, Location pointB, int numberOfPoints)
{
    ArrayList<Location> locationList = new ArrayList<Location>();

    xStart = pointA.getX();
    xInterval = (pointB.getX() - pointA.getX()) / (numberOfPoints - 1);

    yStart = pointA.getY();
    yInterval = (pointB.getY() - pointA.getY()) / (numberOfPoints - 1);

    for(int i=0; i<numberOfPoints; i++)
    {
        locationList.add( new Location( (xStart + (i*xInterval)), (yStart + (i*yInterval))) );
    }

    return locationList;
}

将返回沿着从pointA到pointB的路径等距离的给定数量的Locations的ArrayList。

使用Java或PostgreSQL / PostGIS将路径划分为N个部分

2 个答案: