Question

我想对一个数组进行排序，使每个元素与前一个位置的距离最短。

数组就像那样

locations=[{"loc1",lat,long},{"loc2",lat,long},{"loc3",lat,long},{"loc4",lat,long},{"loc5",lat,long}]

计算距离的函数是：

var distance = function(lat1, lon1, lat2, lon2)
{
  var radlat1 = Math.PI * lat1/180;
  var radlat2 = Math.PI * lat2/180;
  var theta = lon1-lon2;
  var radtheta = Math.PI * theta/180;
  var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
  dist = Math.acos(dist);
  dist = dist * 180/Math.PI;
  dist = dist * 60 * 1.1515;
  dist = dist * 1.609344 ;

  return dist;
}

此函数在传递值时提供两个位置之间的距离。

起点是位置数组的第一个元素现在我想要一个将采用数组并返回已排序数组的函数。

Answer 1

您需要在排序函数中执行距离计算：

由于距离计算非常密集，因此对象中的缓存距离会更快，因此您在排序方法中不会多次计算相同的距离：

var startingLoc = {lat:0.000,lng:0.000};//your starting location;
var distanceCache = {}; //a cache to hold your distance calculations

//sort the locations - assumes loc1 is {lat:some number, lng: some number}
locations.sort(function(loc1,loc2) {
    var loc1Key = loc1.lat+'-'+loc1.lng;
    var loc2Key = loc2.lat+'-'+loc2.lng;

    if(!distanceCache.hasOwnProperty(loc1Key)) {
      distanceCache[loc1Key] = distance(startingLoc.lat,startingLoc.lng,loc1.lat,loc1.lng);
    }

    if(!distanceCache.hasOwnProperty(loc2Key)) {
      distanceCache[loc2Key] = distance(startingLoc.lat,startingLoc.lng,loc2.lat,loc2.lng);
    }

    return distanceCache[loc1Key] - distanceCache[loc2Key];

 });

distanceCache = null; //done with distanceCache

Answer 2

你可能会在这个上面找个漂亮的大兔子洞。我先来看看你是否可以使用Maps API Google Map V3, how to get list of best driving route for multiple destinations?。 optimizeWaypoints设置旨在以最佳驾驶顺序返回位置列表。

显然这与纯距离比较不同，但它可能适合您的需要。

Answer 3

您可以为Array原型上的sort方法提供自定义函数，如下所示：

＆＃13;

locations = [
  ["loc1", 1, 1],
  ["loc2", 3, 3],
  ["loc3", 2, 2],
  ["loc4", 5, 4],
  ["loc5", 3, 5]
];

var distance = function(lat1, lon1, lat2, lon2) {
  var radlat1 = Math.PI * lat1 / 180;
  var radlat2 = Math.PI * lat2 / 180;
  var theta = lon1 - lon2;
  var radtheta = Math.PI * theta / 180;
  var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
  dist = Math.acos(dist);
  dist = dist * 180 / Math.PI;
  dist = dist * 60 * 1.1515;
  dist = dist * 1.609344;

  return dist;
};

locations.sort(function(a, b) {
  var origLat = 0,
    origLong = 0;

  return distance(origLat, origLong, a[1], a[2]) - distance(origLat, origLong, b[1], b[2]);
});

console.log(locations)

＆＃13;

Answer 4

如果没有真正的位置，就不会测试，但下面的内容应该可以完成这项工作：

（我不知道谷歌地图API，也许你可以找到更好的方法来做到这一点......）

var locations = [{
		name : "loc1",
		lat : 1001,
		long : 2001
	}, {
		name : "loc2",
		lat : 150,
		long : 630
	}, {
		name : "loc3",
		lat : 151,
		long : 631
	}, {
		name : "loc4",
		lat : 850,
		long : 56
	}, {
		name : "loc5",
		lat : 960,
		long : 698
	}
];

var distance = function (lat1, lon1, lat2, lon2) {
	var radlat1 = Math.PI * lat1 / 180;
	var radlat2 = Math.PI * lat2 / 180;
	var theta = lon1 - lon2;
	var radtheta = Math.PI * theta / 180;
	var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
	dist = Math.acos(dist);
	dist = dist * 180 / Math.PI;
	dist = dist * 60 * 1.1515;
	dist = dist * 1.609344;

	return dist;
}

var locationWithDistFromPrevious = locations.map(function (l, i) {
		if (i === 0) {
			l.dist = 0;
		} else {
			l.dist = distance(l.lat, l.long, locations[i - 1].lat, locations[i - 1].long)
		}
		return l;
	}).sort(function (a, b) {
		return a.dist - b.dist
	});

var locationWithDistFromFirst = locations.map(function (l, i) {
		if (i === 0) {
			l.dist = 0;
		} else {          
			l.dist = distance(l.lat, l.long, locations[0].lat, locations[0].long)
		}
		return l;
	}).sort(function (a, b) {
		return a.dist - b.dist
	});


document.getElementById("resultFromPrev").textContent = JSON.stringify(locationWithDistFromPrevious, null, 4);
document.getElementById("resultFromFirst").textContent = JSON.stringify(locationWithDistFromFirst, null, 4);

<body>
  Sort by previous item<br/>
  <pre id="resultFromPrev"></pre><br/>
  Sort by first item dist <br/>
  <pre id="resultFromFirst"></pre><br/>
</body>

如何根据短距离Javascript对数组进行排序

4 个答案: