2

我在计算地图上几个点之间的距离时遇到了困难:

我有一组坐标,其中第一个坐标 ["30.327547", "59.919676"] 是旅行的开始,其他的是进站:

var J = [
    ["30.327547", "59.919676"],
    ["29.84964", "58.737619"],
    ["28.250252", "57.785994"],
    ["30.098912", "55.175885"],
    ["30.37357", "54.503783"],
    ["27.572056", "53.898325"],
    ["26.000193", "53.11856"]

];

接下来,要从这个坐标制作地图上的地理点,我应该使用特殊的 Yandex Maps API 函数 YMaps.Geopoint:

var a=new YMaps.GeoPoint(30.327547,59.919676);
var b=new YMaps.GeoPoint(29.84964,58.737619);
var c=new YMaps.GeoPoint(28.250252,57.785994);
var d=new YMaps.GeoPoint(30.098912,55.175885);
var e=new YMaps.GeoPoint(30.37357,54.503783);
var f=new YMaps.GeoPoint(27.572056,53.898325);
var g=new YMaps.GeoPoint(26.000193,53.11856);

最后,为了计算点之间的距离,我使用了另一个 API 函数“point1.distance(point2)”:

var d1=a.distance(b);        //distance1
var d2=a.distance(b)+b.distance(c);    //distance2
var d3=a.distance(b)+b.distance(c)+c.distance(d);  //distance3
var d4=a.distance(b)+b.distance(c)+c.distance(d)+d.distance(e);   //distance4
var d5=a.distance(b)+b.distance(c)+c.distance(d)+d.distance(e)+e.distance(f);   //distance5
var d6=a.distance(b)+b.distance(c)+c.distance(d)+d.distance(e)+e.distance(f)+f.distance(g);   //distance6

这很好用(我还将每个结果转换为格式 (result km) ),结果是:

    console.log(YMaps.humanDistance(d1));console.log(YMaps.humanDistance(d2));console.log(YMaps.humanDistance(d3));
    console.log(YMaps.humanDistance(d4));console.log(YMaps.humanDistance(d5));console.log(YMaps.humanDistance(d6));

//{"Point1":"134 km.","Point2":"275 km.","Point3":"586 km.","Point4":"663 km.","Point5":"857 km.","Point6":"992 km."}

我实际上想在循环中进行此操作:

for(var i=0;i<J.length;i++){

    // Iteratting through the array of points of J and creating Geoobjects on the map, dynamically putting them into public variables "temp_*"
    window["temp_" + i]=new YMaps.GeoPoint(J[i][0],J[i][1]);

    //next point
    var next=i+1;
    //master point (the begin of trip)
    var master=window["temp_0"];


    //calculating the distance between the master point and actual [i] point in the loop
    var formula=master.distance(window["temp_"+i]);

    // calculating the distance between the actual [i] point and the next in the loop
    var formula2=window["temp_" + i].distance(window["temp_"+next]);


    //summing and converting into human format and dinamically putting them into variables "result_*"
    window["result_"+i]=YMaps.humanDistance(formula+formula2);

    //logging the results
    console.log(YMaps.humanDistance(window["result_"+i]));
    }

此循环有效,但返回错误结果。有人可以建议循环中有什么问题吗?我想可能我需要在这个循环中使用另一个循环,它将根据需要(需要)多次返回其他点的总和。谢谢。

4

1 回答 1

3

正如您声明的那样,该J变量是一个二维字符串数组,据我所知,该YMaps.GeoPoint函数需要一个数字。所以尝试解析:

window['temp_' + i] = new YMaps.GeoPoint(
    parseFloat(J[i][0]), 
    parseFloat(J[i][1])
);

更新:

这是您的代码有什么问题:

var formula=master.distance(window["temp_"+i]);
var formula2=window["temp_" + i].distance(window["temp_"+next]);
window["result_"+i]=YMaps.humanDistance(formula+formula2);

在每次迭代中,您重新声明formula2变量,并在计算当前点之间的人类距离时。您需要累积总和,因此在循环外部声明变量并在内部添加:

var sum = 0;
for(var i = 0; i < J.length - 1; i++) {
    var p1 = new YMaps.GeoPoint(J[i][0], J[i][1]);
    var p2 = new YMaps.GeoPoint(J[i + 1][0], J[i + 1][1]);
    sum += p1.distance(p2);
    console.log(YMaps.humanDistance(sum);
}

请注意,循环结束于J.length - 1:ex。对于 7 个点,您有 6 个距离。

于 2010-02-20T11:00:55.737 回答