gps 模块返回的 经纬度 怎么对应上 百度或者 高德 gps经纬度 ：报错

有做gps 模块的 数据便宜 纠正的 给点 意见啊######不能沉了啊, 有谁做过 gps模块数据解析的, 给个 方向啊######

这是几年前用的办法，不保证能100%用。

#!/usr/bin/env python
#coding=utf-8

import math

pi = 3.14159265358979324
a = 6378245.0
ee = 0.00669342162296594323

#纠偏函数
def transform(wgLat,wgLon):
    latlng = [0]*2
    if outOfChina(wgLat, wgLon):
        latlng[0] = wgLat;  
        latlng[1] = wgLon; 

    dLat = transformLat(wgLon - 105.0, wgLat - 35.0)
    dLon = transformLon(wgLon - 105.0, wgLat - 35.0)
    radLat = wgLat / 180.0 * pi
    magic = math.sin(radLat)
    magic = 1 - ee * magic * magic
    sqrtMagic = math.sqrt(magic)
    dLat = (dLat * 180.0) / ((a * (1 - ee)) / (magic * sqrtMagic) * pi)
    dLon = (dLon * 180.0) / (a / sqrtMagic * math.cos(radLat) * pi)
    latlng[0] = wgLat + dLat
    latlng[1] = wgLon + dLon
    return latlng

def outOfChina(lat,lon):
    if (lon < 72.004 or lon > 137.8347): 
        return True  
    if (lat < 0.8293 or lat > 55.8271):
        return True  
    return False

def transformLat(x,y):

    ret = -100.0 + 2.0 * x + 3.0 * y + 0.2 * y * y + 0.1 * x * y + 0.2 * math.sqrt(abs(x)) 
    ret += (20.0 * math.sin(6.0 * x * pi) + 20.0 * math.sin(2.0 * x * pi)) * 2.0 / 3.0
    ret += (20.0 * math.sin(y * pi) + 40.0 * math.sin(y / 3.0 * pi)) * 2.0 / 3.0
    ret += (160.0 * math.sin(y / 12.0 * pi) + 320 * math.sin(y * pi / 30.0)) * 2.0 / 3.0 
    return ret

def transformLon(x,y):

    ret = 300.0 + x + 2.0 * y + 0.1 * x * x + 0.1 * x * y + 0.1 * math.sqrt(abs(x)) 
    ret += (20.0 * math.sin(6.0 * x * pi) + 20.0 * math.sin(2.0 * x * pi)) * 2.0 / 3.0
    ret += (20.0 * math.sin(x * pi) + 40.0 * math.sin(x / 3.0 * pi)) * 2.0 / 3.0
    ret += (150.0 * math.sin(x / 12.0 * pi) + 300.0 * math.sin(x / 30.0 * pi)) * 2.0 / 3.0
    return ret