代码分析Python地图坐标转换
最近做项目正好需要坐标的转换
- 各地图API坐标系统比较与转换;
- WGS84坐标系:即地球坐标系,国际上通用的坐标系。设备一般包含GPS芯片或者北斗芯片获取的经纬度为WGS84地理坐标系,
- 谷歌地图采用的是WGS84地理坐标系(中国范围除外);
- GCJ02坐标系:即火星坐标系,是由中国国家测绘局制订的地理信息系统的坐标系统。由WGS84坐标系经加密后的坐标系。
- 谷歌中国地图和搜搜中国地图采用的是GCJ02地理坐标系; BD09坐标系:即百度坐标系,GCJ02坐标系经加密后的坐标系;
- 搜狗坐标系、图吧坐标系等,估计也是在GCJ02基础上加密而成的.
然后在csv中将其转化。
最后再在百度地图API上进行检验成功
#http://bbs.lbsyun.baidu.com/forum.php?mod=viewthread&tid=10923 #代码原地址 import csv import string import time import math #系数常量 a = 6378245.0 ee = 0.00669342162296594323 x_pi = 3.14159265358979324 * 3000.0 / 180.0; #转换经度 def transformLat(lat,lon): ret = -100.0 + 2.0 * lat + 3.0 * lon + 0.2 * lon * lon + 0.1 * lat * lon +0.2 * math.sqrt(abs(lat)) ret += (20.0 * math.sin(6.0 * lat * math.pi) + 20.0 * math.sin(2.0 * lat * math.pi)) * 2.0 / 3.0 ret += (20.0 * math.sin(lon * math.pi) + 40.0 * math.sin(lon / 3.0 * math.pi)) * 2.0 / 3.0 ret += (160.0 * math.sin(lon / 12.0 * math.pi) + 320 * math.sin(lon * math.pi / 30.0)) * 2.0 / 3.0 return ret #转换纬度 def transformLon(lat,lon): ret = 300.0 + lat + 2.0 * lon + 0.1 * lat * lat + 0.1 * lat * lon + 0.1 * math.sqrt(abs(lat)) ret += (20.0 * math.sin(6.0 * lat * math.pi) + 20.0 * math.sin(2.0 * lat * math.pi)) * 2.0 / 3.0 ret += (20.0 * math.sin(lat * math.pi) + 40.0 * math.sin(lat / 3.0 * math.pi)) * 2.0 / 3.0 ret += (150.0 * math.sin(lat / 12.0 * math.pi) + 300.0 * math.sin(lat / 30.0 * math.pi)) * 2.0 / 3.0 return ret #Wgs transform to gcj def wgs2gcj(lat,lon): dLat = transformLat(lon - 105.0, lat - 35.0) dLon = transformLon(lon - 105.0, lat - 35.0) radLat = lat / 180.0 * math.pi magic = math.sin(radLat) magic = 1 - ee * magic * magic sqrtMagic = math.sqrt(magic) dLat = (dLat * 180.0) / ((a * (1 - ee)) / (magic * sqrtMagic) * math.pi) dLon = (dLon * 180.0) / (a / sqrtMagic * math.cos(radLat) * math.pi) mgLat = lat + dLat mgLon = lon + dLon loc=[mgLat,mgLon] return loc #gcj transform to bd2 def gcj2bd(lat,lon): x=lon y=lat z = math.sqrt(x * x + y * y) + 0.00002 * math.sin(y * x_pi) theta = math.atan2(y, x) + 0.000003 * math.cos(x * x_pi) bd_lon = z * math.cos(theta) + 0.0065 bd_lat = z * math.sin(theta) + 0.006 bdpoint = [bd_lon,bd_lat] return bdpoint #wgs transform to bd def wgs2bd(lat,lon): wgs_to_gcj = wgs2gcj(lat,lon) gcj_to_bd = gcj2bd(wgs_to_gcj[0], wgs_to_gcj[1]) return gcj_to_bd; for i in range (3,4): n = str('2017.040'+ str(i)+'.csv') m = str('2017040' + str(i)+'.csv') csvfile = open(m,'w',encoding='UTF-8',newline='') nodes = csv.writer(csvfile) nodes.writerow(['md5','content','phone','conntime','recitime','lng','lat']) l=[] with open(n,newline='',encoding='UTF-8') as f: reader = csv.DictReader(f) for row in reader: if row['md5'] == 'md5': continue else: y=float(row['lng']) x=float(row['lat']) loc=wgs2bd(x,y) l.append([row['md5'],row['content'],row['phone'],row['conntime'],row['recitime'],loc[0],loc[1]]) nodes.writerows(l) csvfile.close() print("转换成功")
您可能感兴趣的文章:
- python通过百度地图API获取某地址的经纬度详解
- Python爬虫实例_利用百度地图API批量获取城市所有的POI点
- Python学习之用pygal画世界地图实例
- Python和Perl绘制中国北京跑步地图的方法
- 用Python制作在地图上模拟瘟疫扩散的Gif图
赞 (0)