经度\纬度点的排序名单，先从最近 [英] Sort list of lon\lat points, start with nearest

问题描述

我从GPS有位置（lon_base，lat_base）。 我有位置列表（lon1，LAT1 | lon2，LAT2 | lon3，lat3 ...） 这份名单很长，而且是世界各地。

I have location from GPS (lon_base, lat_base). I have a list of locations (lon1, lat1|lon2, lat2|lon3, lat3...) This list is very long and is around the world.

我的问题是： 1.我如何从该列表中只有经度\纬度是从我lon_base \ lat_base1英里？ 2.如何对它们进行排序，从最接近最远？

My questions are: 1. How do I get from that list only the lon\lat that are 1 mile from my lon_base\lat_base? 2. How do I sort them from closest to farthest?

在此先感谢！

推荐答案

您需要定义自己的比较，在一般情况下，看起来是这样的：

You want to define your own Comparator that, in general, looks something like this:

LonLat myHouse = /* whatever */ ;
Comparable comp = new Comparable () {
    LonLat a;
    int compareTo (Object b) {
        int aDist = calcDistance(a, myHouse) ;
        int bDist = calcDistance(b, myHouse) ;
        return aDist - bDist;
    }
};
myLonLatList.sort(lonLatList, comp);

其中， calcDistance（）简单地计算两个点之间的距离。如果你在Android上，我认为谷歌地图具有这样的功能的地方在其API，它会为你做这个。

where calcDistance() simply calculates the distance between the two points. If you're on Android, I think Google Maps has a function somewhere in their API that will do this for you.

修改：你希望你的 calcDistance（）函数看起来像ChrisJ的距离功能。

EDIT : You'll want your calcDistance() function to look like ChrisJ's distance function.

-tjw

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