distanceTo: function(point) { var distance = 0.0; if ((this.x != null) && (this.y != null) && (point != null) && (point.x != null) && (point.y != null)) { var dx2 = Math.pow(this.x - point.x, 2); var dy2 = Math.pow(this.y -

在地图上显示点数据是最常用的地图展示功能之一,但是如果很多点在地图上显示,或造成密密麻麻的一片,无法正常看清楚,这个时候,一般有两种解决方案,一种是根据数据重要程度进行标注,重要的显示大一些,不重要的显示小点,比如百度地图就是这样的:另一种方法是使用聚合,让相邻的点聚合成一个点,也能解决这个问题. 使用openlayers 3 地图组件比较容易解决这个问题,关键是  ol.source.Cluster 对象,这个对象有两个参数,一个是聚合距离,一个是原始的点数据.代码片段如下: 1:初始化ol3

OpenLayers 是由MetaCarta公司开发的,用于WebGIS客户端的JavaScript包,目前的最高版本是2.7 V,通过BSD License 发行.它实现访问地理空间数据的方法都符合行业标准,比如OpenGIS的WMS和WFS规范, OpenLayers采用纯面向对象的JavaScript方式开发,同时借用了Prototype框架和Rico库的一些组件. 采用OpenLayers作为客户端不存在浏览器依赖性.由于OpenLayers采用JavaScript语言实现,而应用于We

路径分析-QGIS+PostgreSQL+PostGIS+pgRouting(一) 路径分析-PostgreSQL+GeoServer+Openlayers(二) 前言 上一篇文章中实现数据库层面的路径分析了,可以在数据库里面通过 SQL 查询到结果. 本篇文章实现了从前端页面直接可视化操作点选起点.终点,并返回最短路径进行展示. 一.数据库函数 在 PostgreSQL 数据库中创建函数,该函数实现的功能是:传入表名.起点.终点经纬度.距离等参数,返回对应的最短距离 geometry. 创建的

在前面<百度地图.ECharts整合HT for Web网络拓扑图应用>我们有介绍百度地图和 HT for Web 的整合,我们今天来谈谈 OpenLayers 和 HT for Web 的整合. HT for Web作为逻辑拓扑图形组件自身没有GIS功能,但可以与各种GIS引擎即其客户端组件进行融合,各取所长实现逻辑拓扑和物理拓扑的无缝融合,本章将具体介绍HT for Web与开发免费的OpenLayers地图结合应用的关键技术点,该文介绍的结合的原理,其实还可推广到与ArcGIS.百度地图

The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Now your job is to find the total Hamming distance between all pairs of the given numbers. Example: Input: 4, 14, 2 Output: 6 Explanat

The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Note:0 ≤ x, y < 231. Example: Input: x = 1, y = 4 Output: 2 Explanation: 1

Given a non-empty string str and an integer k, rearrange the string such that the same characters are at least distance k from each other. All input strings are given in lowercase letters. If it is not possible to rearrange the string, return an empt

You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where: Each 0 marks an empty land which you can pass b

This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same as word2. Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list. word1 and word2 may be

This is a follow up of Shortest Word Distance. The only difference is now you are given the list of words and your method will be called repeatedly many times with different parameters. How would you optimize it? Design a class which receives a list

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list. For example,Assume that words = ["practice", "makes", "perfect", "coding", "makes"]. G

Given two strings S and T, determine if they are both one edit distance apart. 这道题是之前那道Edit Distance的拓展,然而这道题并没有那道题难,这道题只让我们判断两个字符串的编辑距离是否为1,那么我们只需分下列三种情况来考虑就行了: 1. 两个字符串的长度之差大于1,那么直接返回False 2. 两个字符串的长度之差等于1,那么长的那个字符串去掉一个字符,剩下的应该和短的字符串相同 3. 两个字符串的长度之

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.) You have the following 3 operations permitted on a word: a) Insert a characterb) Delete a characterc) Replace

Levenshtein distance,中文名为最小编辑距离,其目的是找出两个字符串之间需要改动多少个字符后变成一致.该算法使用了动态规划的算法策略,该问题具备最优子结构,最小编辑距离包含子最小编辑距离,有下列的公式. 其中d[i-1,j]+1代表字符串s2插入一个字母才与s1相同,d[i,j-1]+1代表字符串s1删除一个字母才与s2相同,然后当xi=yj时,不需要代价,所以和上一步d[i-1,j-1]代价相同,否则+1,接着d[i,j]是以上三者中最小的一项. 算法实现(C#): 假设两个

1.首先将word文档解压缩为zip /** * 修改后缀名 */ public static String reName(String path){ File file=new File(path); String filename=file.getAbsolutePath(); if(filename.indexOf(".")>=0){ filename=filename.substring(0,filename.lastIndexOf(".")); }

The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Now your job is to find the total Hamming distance between all pairs of the given numbers. Example: Input: 4, 14, 2 Output: 6 Explanat

The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Note:0 ≤ x, y < 231. Example: Input: x = 1, y = 4 Output: 2 Explanation: 1

题目: The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Note:0 ≤ x, y < 231. Example: Input: x = 1, y = 4 Output: 2 Explanatio

从网上下载openlayers2,解压取得img theme 文件夹和openlayes.js文件.放在同一文件夹下用phpstorm打开. 创建index.html-------------每一行都有注释 <!DOCTYPE html> <html lang='en'> <head> <meta charset='utf-8'/> <title>My OpenLayers Map</title> <script type='t

马氏距离(Mahalanobis distance)是由印度统计学家马哈拉诺比斯(P. C. Mahalanobis)提出的,表示数据的协方差距离.它是一种有效的计算两个未知样本集的相似度的方法.与欧氏距离不同的是它考虑到各种特性之间的联系(例如:一条关于身高的信息会带来一条关于体重的信息,因为两者是有关联的)并且是尺度无关的(scale-invariant),即独立于测量尺度. 对于一个均值为,协方差矩阵为Σ的多变量矢量,其马氏距离为 马氏距离也可以定义为两个服从同一分布并且其协方差矩阵为Σ的