lng

python使用列表按照一定格式批量输出，配合地图热力图程序“食用”更佳哦~ import re import random posi = [ '113.936539169252,22.5385001957843', '126.541615090316,45.8088258279521', '126.675638499523,45.7658661637579', '125.296653763226,43.8524275671504', '125.295401255854,43.8406414749619', '125.330602075906,43.8219535010431', '123.465461035375,41.8022741388046', '123.471096644822,41.6838300691906', ] count = [ '549', '36', '243', '713', '697', '97', '883', '59' ] listlng = [] listlat = [] for i in posi: lng = i.split(',')[0] lat = i.split(',')[-1] listlng.append(lng) listlat.append(lat) #print(listlng) for j in range(0,len

function bMapToQQMap(lng, lat) { if (lng == null || lng == '' || lat == null || lat == '') return [lng, lat]; var x_pi = 3.14159265358979324; var x = parseFloat(lng) - 0.0065; var y = parseFloat(lat) - 0.006; var z = Math.sqrt(x * x + y * y) - 0.00002 * Math.sin(y * x_pi); var theta = Math.atan2(y, x) - 0.000003 * Math.cos(x * x_pi); var lng = (z * Math.cos(theta)).toFixed(7); var lat = (z * Math.sin(theta)).toFixed(7); return [lng, lat]; } /** * 坐标转换，腾讯地图转换成百度地图坐标 * lng 腾讯经度（pointy） * lat 腾讯纬度（pointx） * 经度>纬度 */ function qqMapToBMap(lng, lat) { if (lng == null || lng == '' || lat == null ||

jq进行json数据解析 学习了： https://www.jianshu.com/p/6de3cfdbdb0e https://blog.csdn.net/u011641885/article/details/45559031 https://www.cnblogs.com/tinywan/p/7684414.html 使用man可以查看详细帮助，太强大了； ➜ yiqing cat a.json|jq '.' { "another": 1, "all": { "a0": { "a1": { "a2": [ { "a3": "a3", "lng": 1, "lat": 2 }, { "a32": "a32", "lng": 1, "lat": 2 } ] } }, "b0": { "b1": { "b2": [ { "b3": "b3", "lng": 1, "lat": 2 }, { "b32": "b32", "lng": 1, "lat": 2 } ] } } } } a获取多个属性，逗号分隔，可以列出不包含的 ➜ yiqing cat a.json|jq '.all|.a0,.b0,.c0' { "a1": { "a2": [ { "a3": "a3", "lng": 1, "lat": 2 }, { "a32": "a32", "lng": 1, "lat": 2 }

JSON的具体形式 　　1、对象是一个无序的“‘名称/值’对”集合。一个对象以 “{” 开始，以 “}” 结束。每个“名称”后跟一个“ : ”，“‘名称/值’对”之间使用“ ， ”分隔。 举个例子： { 　　name:"张三" , 　　password:"123456" , 　　department:"技术部" , 　　sex:"男", 　　old:30 　　} 这种格式的就是json格式 　　2、数组是值（value）的有序集合。一个数组以“[”开始，“]”结束。值之间使用“，”分隔。 例如： hotellist: { point:{name:"复旦大学",lat:31.2965930000,lng:121.5035790000}, hotels: { P2472:[1977,"上海复旦皇冠假日酒店",2472,"|IHG|FD|HGJR|FDHG|SHFDHGJRJD|",5,0,"酒店位于杨浦区，正对着享有盛名的复旦大学。紧邻新近开业的万达商业广场，巴黎春天，沃尔玛大型超级市场及各种餐饮娱乐设施。便利的交通让您从容地到达上海的金融区、外滩、浦东和旅游胜地。",788.00,906,"UploadPhotos/200823111239573.bmp","上海杨浦区邯郸路199号","310000",24,48,"121.50185823440552","31

http://python-data.dr-chuck.net/geojson?sensor=false&address=Vilnius+University 1 { 2 "results" : [ 3 { 4 "address_components" : [ 5 { 6 "long_name" : "Vilnius", 7 "short_name" : "Vilnius", 8 "types" : [ "locality", "political" ] 9 }, 10 { 11 "long_name" : "Vilnius city municipality", 12 "short_name" : "Vilnius city municipality", 13 "types" : [ "administrative_area_level_2", "political" ] 14 }, 15 { 16 "long_name" : "Vilnius County", 17 "short_name" : "Vilnius County", 18 "types" : [ "administrative_area_level_1", "political" ] 19 }, 20 { 21 "long_name" : "Lithuania", 22 "short_name" : "LT",

package com.rxdatainfo.modules.data; import java.util.ArrayList; import java.util.List; import com.alibaba.fastjson.JSONArray; public class LatLonUtil { private static double PI = 3.14159265; private static double EARTH_RADIUS = 6378137; private static double RAD = Math.PI / 180.0; /** * @Description: 获取四至信息(最大最小经纬度) * @author 董泽琦 * @param lat * @param lng * @date 2019年4月25日 下午5:50:54 */ public static double[] getFourBounds(String lng, String lat, String[] pointArr) throws Exception { if (pointArr.length>1) { return getPolygonAround(pointArr); } return getCircleAround(Double.parseDouble(lng),

$lat = trim($_POST['lat']); $lng = trim($_POST['lng']); $distance = " ROUND( 6378.138 * 2 * ASIN( SQRT( POW( SIN( ( '$lat' * PI() / 180 - lat * PI() / 180 ) / 2 ), 2 ) + COS(40.0497810000 * PI() / 180) * COS(lat * PI() / 180) * POW( SIN( ( '$lng' * PI() / 180 - lng * PI() / 180 ) / 2 ), 2 ) ) ) * 1000 ) AS distance "; // 内容 $sql = <<<doc select id,title as shop_name,create_at,master_title,lng,lat,reg_address,province,city,area,address,start_at,end_at,phone,business_name,business_phone,pics,club_id,{$distance} from jiazhen_shop_info as jsi where 1=1 order by distance ASC,id desc doc; 来源： https:

/** * 百度地图---->腾讯地图 * @param double $lat 纬度 * @param double $lng 经度 * @return array(); */ function Convert_BD09_To_GCJ02($lat,$lng){ $x_pi = 3.14159265358979324 * 3000.0 / 180.0; $x = $lng - 0.0065; $y = $lat - 0.006; $z = sqrt($x * $x + $y * $y) - 0.00002 * sin($y * $x_pi); $theta = atan2($y, $x) - 0.000003 * cos($x * $x_pi); $lng = $z * cos($theta); $lat = $z * sin($theta); return array('lng'=>$lng,'lat'=>$lat); } /** * 腾讯地图---->百度地图 * @param double $lat 纬度 * @param double $lng 经度 */ function Convert_GCJ02_To_BD09($lat,$lng){ $x_pi = 3.14159265358979324 * 3000.0 / 180.0; $x = $lng; $y = $lat

/// <summary> /// 判断点是否在多边形内 /// </summary> /// <param name="checkPoint">需要判断的点</param> /// <param name="polygonPoints">组成多边形点的集合</param> /// <returns></returns> public static bool IsInPolygon2(PointLatLng checkPoint, List<PointLatLng> polygonPoints) { int counter = 0; int i; double xinters; PointLatLng p1, p2; int pointCount = polygonPoints.Count; p1 = polygonPoints[0]; for (i = 1; i <= pointCount; i++) { p2 = polygonPoints[i % pointCount]; if (checkPoint.Lng > Math.Min(p1.Lng, p2.Lng)//校验点的Y大于线段端点的最小Y && checkPoint.Lng <= Math.Max(p1.Lng, p2.Lng))//校验点的Y小于线段端点的最大Y { if (checkPoint.Lat <=

#!/usr/bin/env python # encoding: utf-8 """ @version: v1.0 @author: xag @license: Apache Licence @contact: xinganguo@gmail.com @site: http://www.xingag.top @software: PyCharm @file: meta_picture.py.py @time: 2019-08-23 16:23 @description：高德坐标拾取网站：https://lbs.amap.com/console/show/picker """ import os import exifread from decimal import Decimal from position_utils import * import requests import json import datetime # pip3 install exifread class Location(object): def __init__(self, image_path): self.img_path = image_path self.api_key = "你申请的key" self.url_get_position = 'https://restapi.amap.com