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本文介紹了將經緯度坐標排序為順時針四邊形的處理方法，對大家解決問題具有一定的參考價值，需要的朋友們下面隨著小編來一起學習吧！

問題描述

問題

用戶最多可以按任意順序提供四個經緯度坐標.他們使用谷歌地圖這樣做.使用 Google 的 Polygon API (v3)，他們選擇的坐標應該突出顯示四個坐標之間的選定區域.

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.

問題

如何按(逆)順時針順序對一組經緯度坐標進行排序?

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

解決方案和搜索

StackOverflow 問題

繪制可調整大小(不相交)的多邊形
如何對 Google 地圖多邊形中的點進行排序以使線不會交叉?
按順時針順序對四個點進行排序

相關網站

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 的掃描(太復雜)
Jarvis March 算法(處理 N 個點)
遞歸凸包(移除一個點)

代碼

這是我目前所擁有的:

// 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;
}

謝謝.

推薦答案

給出要點:

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

你想找到以下綁定步行:

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

如果這是正確的，這里有一個方法:

If this is correct, here's a way:

在點集中找到最高點，P_top.如果出現平局，則選擇 x 坐標最小的點

通過比較每對點的直線的斜率 m_i 和 m_j 對所有點進行排序(不包括 P_top！) P_i 和 P_j 在通過 P_top 時使

如果 m_i 和 m_j 相等，讓點 P_i 或 P_j 最接近P_top先來
如果 m_i 為正且 m_j 為負(或零)，則 P_j 在前
如果 m_i 和 m_j 都是正數或負數，讓屬于斜率最大的線的點在前

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

這是地圖的快速演示:

(我對 JavaScript 知之甚少，所以我可能或可能已經違反了一些 JavaScript 代碼約定......):

(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:您應該雙重或三重檢查從 lat,lon 到 x,y 的轉換，因為我是 GIS 新手！！！但也許你甚至不需要轉換任何東西.如果不這樣做，upperLeft 函數可能只返回最低點而不是最高點，具體取決于相關點的位置.再次:三重檢查這些假設！

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!

執行上面的代碼片段時，會打印以下內容:

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

交替距離函數

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;
}

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將經緯度坐標排序為順時針四邊形

問題描述

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