如何使用leaflet-knn计算到给定点的最接近点? [英] how to calculate the closest points to a given point using leaflet-knn?

问题描述

我有两个点层(客户，供应商)，我想在用户输入的距离内找到特定客户的所有最近的供应商，我听说了传单knn，但找不到完整的示例来理解它好吧，有什么建议吗? 这是我的代码，是通过ajax获取客户数据的部分

I have two point Layers (Customers,Suppliers), i want to get all the nearest suppliers to a specific customer within a distance entered by user,i heared about leaflet knn but i couldn't find a full example to understand it well,any suggestions? This is my code in the part of getting the customers data through ajax

    $.ajax({
            type: "POST",
            url: 'Customers_geojson.php',
            dataType: 'json',
            success: function (response) {
                    geojsonLayer = L.geoJson(
                        response,
                        {
                        pointToLayer: function (feature, latlng) 
                            {
                             return L.circleMarker(latlng, geojsonMarkerOptions1);
                            }
                            },
                            {   
                            onEachFeature: function (feature, layer) 
                            {
                            layer.bindPopup('<label>Nick Name:</label>' + feature.properties.nick_name_);
                            }
                    }).addTo(mymap);
                    $("#info").fadeOut(500);
                    var gj = L.geoJson(response);
                    var nearest = leafletKnn(gj).nearest(L.latLng(8.71224, 125.692), 10000);
                    alert(nearest);

                    }
                });

推荐答案

如何使用leaflet-knn计算到给定点的最接近点?

How to calculate the closest points to a given point using leaflet-knn?

答案是你不能".

一种最近的邻居搜索算法，例如

A nearest neighbour search algorithm, such as the one that rbush-knn (and the much older leaflet-knn) implements, helps you answer queries like:

我住在(x，y);离我家最近的 5家商店是什么?

或

我在(x，y)； 最近地铁站在哪里?

甚至

我在(x，y)； 最近地铁站在哪里? 将限制在500米以内，因为我不会走那么远.

I'm at (x,y); where is the closest subway station? Limit the search to 500 meters because I won't bother walking that far.

...但是您要回答的查询 非常不同(强调我的意思):

...but the query you want to answer is very different (emphasis mine):

我想所有在用户输入的距离内最接近特定客户的供应商.

I want to get all the nearest suppliers to a specific customer within a distance entered by user.

使用最近的搜索算法是该工作的错误工具.您想要的是:

Using a nearest-neighbour search algorithm is the wrong tool for the job. What you want is either:

• 给出一个点，生成一个以该点为中心的圆，半径为N.该圆将包含与中心的距离等于或小于N的所有所有点，
• 给出一个点，生成一个以该点为中心的正方形，其边长为N*2.该正方形将包含所有个点，这些点与中心的距离等于或小于N，而加则不是.

• Given a point, generate a circle centred on that point, with radius N. That circle will contain all points that are at a distance equal or less than N from the centre, and nothing else.
• Given a point, generate a square centred on that point, with sides of length N*2. That square will contain all points that are at a distance equal or less than N from the centre, plus some which are not.

在GIS行话中，做一个圆被称为制作一个缓冲区，然后找到缓冲区的交点和您的点集.

In GIS jargon, making a circle is known as making a buffer, then finding the intersection of the buffer and your points set.

想法，制作一个正方形称为制作边界框，并在边界框内查询点.

Idem, making a square is known as making a bounding box, and querying points inside the bounding box.

制作缓冲区听起来最好，但是在计算上更昂贵.非常非常.创建一个边界框并查询一组点非常便宜(读为"fast")，并且可以通过对r-tree结构的简单查询来实现(请参见例如

Making a buffer sounds best, but is computationally more expensive. Very much so. Creating a bounding box and querying a set of points is very very cheap (read "fast") and can be achieved with a simple query to a r-tree structure (see e.g. the documentation and examples for rbush).

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