将经纬度坐标排序为顺时针四边形

时间：2023-03-15

本文介绍了将经纬度坐标排序为顺时针四边形的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着跟版网的小编来一起学习吧！

问题描述

问题

用户最多可以按任意顺序提供四个经纬度坐标.他们使用谷歌地图这样做.使用 Google 的 Polygon API (v3)，他们选择的坐标应该突出显示四个坐标之间的选定区域.

Users can provide up to four latitude and longitude coordinates, in any order. They do so with Google Maps. Using Google's Polygon API (v3), the coordinates they select should highlight the selected area between the four coordinates.

问题

如何按(逆)顺时针顺序对一组经纬度坐标进行排序?

How do you sort an array of latitude and longitude coordinates in (counter-)clockwise order?

解决方案和搜索

StackOverflow 问题

绘制可调整大小(不相交)的多边形
如何对 Google 地图多边形中的点进行排序以使线不会交叉?
按顺时针顺序对四个点进行排序

相关网站

http://www.daftlogic.com/projects-google-maps-area-calculator-tool.htm
http://en.literateprograms.org/Quickhull_%28Javascript%29
http://www.geocodezip.com/map-markers_ConvexHull_Polygon.asp
http://softsurfer.com/Archive/algorithm_0103/algorithm_0103.htm

已知算法

Graham 的扫描(太复杂)
Jarvis March 算法(处理 N 个点)
递归凸包(移除一个点)

代码

这是我目前所拥有的:

// Ensures the markers are sorted: NW, NE, SE, SW
function sortMarkers() {
  var ns = markers.slice( 0 );
  var ew = markers.slice( 0 );

  ew.sort( function( a, b ) {
    if( a.position.lat() < b.position.lat() ) {
      return -1;
    }
    else if( a.position.lat() > b.position.lat() ) {
      return 1;
    }

    return 0;
  });

  ns.sort( function( a, b ) {
    if( a.position.lng() < b.position.lng() ) {
      return -1;
    }
    else if( a.position.lng() > b.position.lng() ) {
      return 1;
    }

    return 0;
  });

  var nw;
  var ne;
  var se;
  var sw;

  if( ew.indexOf( ns[0] ) > 1 ) {
    nw = ns[0];
  }
  else {
    ne = ns[0];
  }

  if( ew.indexOf( ns[1] ) > 1 ) {
    nw = ns[1];
  }
  else {
    ne = ns[1];
  }

  if( ew.indexOf( ns[2] ) > 1 ) {
    sw = ns[2];
  }
  else {
    se = ns[2];
  }

  if( ew.indexOf( ns[3] ) > 1 ) {
    sw = ns[3];
  }
  else {
    se = ns[3];
  }

  markers[0] = nw;
  markers[1] = ne;
  markers[2] = se;
  markers[3] = sw;
}

谢谢.

推荐答案

给出要点:

   4  +        [d]            [g]                 
      |                             
   3 [a]            [e]                 
      |                             
   2  +                  [f]       [h]    
      |                             
   1  +   [b]                             
      |                             
   0  +----+---[c]---+----+----+----+
      0    1    2    3    4    5    6

你想找到以下绑定步行:

you want to find the following bound walk:

   4  +     ___[d]------------[g]                 
      |  __/                         
   3 [a]/           [e]__                
      |              \_ ```---      
   2  +                `[f]   \___[h]    
      |              __/            
   1  +   [b]      __/                   
      |          /                
   0  +----+--`[c]---+----+----+----+
      0    1    2    3    4    5    6

如果这是正确的，这里有一个方法:

If this is correct, here's a way:

在点集中找到最高点，P_top.如果出现平局，则选择 x 坐标最小的点

通过比较每对点的直线的斜率 m_i 和 m_j 对所有点进行排序(不包括 P_top！) P_i 和 P_j 在通过 P_top 时使

如果 m_i 和 m_j 相等，让点 P_i 或 P_j 最接近P_top先来
如果 m_i 为正且 m_j 为负(或零)，则 P_j 在前
如果 m_i 和 m_j 都是正数或负数，让属于斜率最大的线的点在前

find the upper most point, P_top, in the set of points. In case of a tie, pick the point with the smallest x coordinate

sort all points by comparing the slopes m_i and m_j of the lines each pair of points (excluding P_top!) P_i and P_j make when passing through P_top

if m_i and m_j are equal, let the point P_i or P_j closest to P_top come first
if m_i is positive and m_j is negative (or zero), P_j comes first
if both m_i and m_j are either positive or negative, let the point belonging to the line with the largest slope come first

这是地图的快速演示:

(我对 JavaScript 知之甚少，所以我可能或可能已经违反了一些 JavaScript 代码约定......):

(I know little JavaScript, so I might, or probably have, violated some JavaScript code conventions...):

var points = [
    new Point("Stuttgard", 48.7771056, 9.1807688),
    new Point("Rotterdam", 51.9226899, 4.4707867),
    new Point("Paris", 48.8566667, 2.3509871),
    new Point("Hamburg", 53.5538148, 9.9915752),
    new Point("Praha", 50.0878114, 14.4204598),
    new Point("Amsterdam", 52.3738007, 4.8909347),
    new Point("Bremen", 53.074981, 8.807081),
    new Point("Calais", 50.9580293, 1.8524129),
];
var upper = upperLeft(points);

print("points :: " + points);
print("upper  :: " + upper);
points.sort(pointSort);
print("sorted :: " + points);

// A representation of a 2D Point.
function Point(label, lat, lon) {

    this.label = label;
    this.x = (lon + 180) * 360;
    this.y = (lat + 90) * 180;

    this.distance=function(that) {
        var dX = that.x - this.x;
        var dY = that.y - this.y;
        return Math.sqrt((dX*dX) + (dY*dY));
    }

    this.slope=function(that) {
        var dX = that.x - this.x;
        var dY = that.y - this.y;
        return dY / dX;
    }

    this.toString=function() {
        return this.label;
    }
}

// A custom sort function that sorts p1 and p2 based on their slope
// that is formed from the upper most point from the array of points.
function pointSort(p1, p2) {
    // Exclude the 'upper' point from the sort (which should come first).
    if(p1 == upper) return -1;
    if(p2 == upper) return 1;

    // Find the slopes of 'p1' and 'p2' when a line is 
    // drawn from those points through the 'upper' point.
    var m1 = upper.slope(p1);
    var m2 = upper.slope(p2);

    // 'p1' and 'p2' are on the same line towards 'upper'.
    if(m1 == m2) {
        // The point closest to 'upper' will come first.
        return p1.distance(upper) < p2.distance(upper) ? -1 : 1;
    }

    // If 'p1' is to the right of 'upper' and 'p2' is the the left.
    if(m1 <= 0 && m2 > 0) return -1;

    // If 'p1' is to the left of 'upper' and 'p2' is the the right.
    if(m1 > 0 && m2 <= 0) return 1;

    // It seems that both slopes are either positive, or negative.
    return m1 > m2 ? -1 : 1;
}

// Find the upper most point. In case of a tie, get the left most point.
function upperLeft(points) {
    var top = points[0];
    for(var i = 1; i < points.length; i++) {
        var temp = points[i];
        if(temp.y > top.y || (temp.y == top.y && temp.x < top.x)) {
            top = temp;
        }
    }
    return top;
}

注意:您应该双重或三重检查从 lat,lon 到 x,y 的转换，因为我是 GIS 新手！！！但也许你甚至不需要转换任何东西.如果不这样做，upperLeft 函数可能只返回最低点而不是最高点，具体取决于相关点的位置.再次:三重检查这些假设！

Note: your should double, or triple check the conversions from lat,lon to x,y as I am a novice if it comes to GIS!!! But perhaps you don't even need to convert anything. If you don't, the upperLeft function might just return the lowest point instead of the highest, depending on the locations of the points in question. Again: triple check these assumptions!

执行上面的代码片段时，会打印以下内容:

When executing the snippet above, the following gets printed:

points :: Stuttgard,Rotterdam,Paris,Hamburg,Praha,Amsterdam,Bremen,Calais
upper  :: Hamburg
sorted :: Hamburg,Praha,Stuttgard,Paris,Bremen,Calais,Rotterdam,Amsterdam

交替距离函数

function distance(lat1, lng1, lat2, lng2) {
  var R = 6371; // km
  var dLat = (lat2-lat1).toRad();
  var dLon = (lng2-lng1).toRad();
  var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
          Math.cos(lat1.toRad()) * Math.cos(lat2.toRad()) *
          Math.sin(dLon/2) * Math.sin(dLon/2);
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
  return R * c;
}

这篇关于将经纬度坐标排序为顺时针四边形的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！

上一篇：根据管端法线向圆柱体应用旋转 下一篇：圆形按钮 css

将经纬度坐标排序为顺时针四边形

问题描述

推荐答案

相关文章