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[英]Separate the repeated occurring elements and unique elements from a given array to two new arrays containing unique elements and recurring elements
[英]Create a new array from the elements that are unique for several arrays
这是我的一段代码:
var things = new Object();
things.location = new Array();
for (var i = 0; i < clickedMarkers.length; i++) {
var mark = clickedMarkers[i],
markLatitude = mark.position.lat(),
markLongitude = mark.position.lng();
things.location.push({
'lat': markLatitude,
'lng': markLongitude
});
}
因此,我们有以下几对:
{lat:53.5412,lng:9.994180000000028},{lat:53.0862838,lng:8.789075099999991}, {lat:53.0738902,lng:8.805211800000052},{lat:53.5412,lng:9.994180000000028},{lat:53.5412,lng:9.994180000000028}
我使用以下代码通过“ lat”比较每对元素:
var arr = {};
for ( var j=0, len=things.location.length; j < len; j++ )
arr[things.location[j]['lat']] = things.location[j];
things.location = new Array();
for ( var key in arr )
things.location.push(arr[key]);
但是正确的方法是比较lat和lng对中的两个元素是否不同,因为对于大对的数组-lat可以相同,但是lng-不同。
如何比较每对的两个元素?
选项:
google.maps.LatLng
.equals
方法 google.maps.geometry.spherical.computeDistanceBetween
方法,如果距离小于某个小数(例如1.0米),则将它们视为相同(纬度和经度是浮点数,比较它们是否相等是有问题的)。 基于geocodezip的答案。 使用以下代码:
//get all markers for the cluster;
var clickedMarkers = cluster.getMarkers();
//get the base point to compare all other markers
var base = new google.maps.LatLng(clickedMarkers[0].position.lat(), clickedMarkers[0].position.lng());
var total = 0;
for (var i = 0; i < clickedMarkers.length; i++) {
var mark = clickedMarkers[i],
markLatitude = mark.position.lat(),
markLongitude = mark.position.lng();
var _coordinates = new google.maps.LatLng(markLatitude, markLongitude);
var point = google.maps.geometry.spherical.computeDistanceBetween(_coordinates, base);
total= total + point;
}
console.log('total', total);
if (total == 0)
{some action}
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