python提取等值线经纬度

‘壹’ 如何用python批量获取经纬度坐标

python根据地址获取经纬度方法一：
from geopy.geocoders import Nominatim
#使用geopy查询
def geocodeN(address):
gps=Nominatim()
location=gps.geocode(address)
return location.longitude,location.latitude

使用Geopy包 ： github.com/geopy/geopy （仅能精确到城镇，具体街道无结果返回）
另外还有一种使用高德地图或网络地图API的方法，有兴趣的朋友可以参考下。
参考链接：https://panxu.net/article/8382.html

‘贰’ python将高斯坐标转换经纬度 经纬度坐标与高斯坐标的转换代码

#网上搜来的

# 高斯坐标转经纬度算法 # B=大地坐标X # C=大地坐标Y # IsSix=6度带或3度带

import math
def GetLatLon2(B, C,IsSix):
#带号
D = math.trunc( C/ 1000000)

#中央经线(单位：弧度)
K = 0
if IsSix:
K = D * 6 - 3 #6度带计算
else:
K = D * 3 #3度带计算
L = B/(6378245*(1-0.006693421623)*1.0050517739)
M = L +(0.00506237764 * math.sin(2*L)/2-0.00001062451*math.sin(4*L)/4+0.0000002081*math.sin(6*L)/6)/1.0050517739
N = L +(0.00506237764 * math.sin(2*M)/2-0.00001062451*math.sin(4*M)/4+0.0000002081*math.sin(6*M)/6)/1.0050517739
O = L +(0.00506237764 * math.sin(2*N)/2-0.00001062451*math.sin(4*N)/4+0.0000002081*math.sin(6*N)/6)/1.0050517739
P = L +(0.00506237764 * math.sin(2*O)/2-0.00001062451*math.sin(4*O)/4+0.0000002081*math.sin(6*O)/6)/1.0050517739
Q = L +(0.00506237764 * math.sin(2*P)/2-0.00001062451*math.sin(4*P)/4+0.0000002081*math.sin(6*P)/6)/1.0050517739
R = L +(0.00506237764 * math.sin(2*Q)/2-0.00001062451*math.sin(4*Q)/4+0.0000002081*math.sin(6*Q)/6)/1.0050517739
S = math.tan(R)
T = 0.006738525415*(math.cos(R))**2
U = 6378245/math.sqrt(1-0.006693421623*(math.sin(R))**2)
V = 6378245*(1-0.006693421623)/(math.sqrt((1-0.006693421623*(math.sin(R))**2)))**3
W = 5+3*S**2+T-9*T*S**2
X = 61+90*S**2+45*S**4
Y = 1+2*S**2+T**2
Z = 5+28*S**2+24*S**4+6*T+8*T*S**2
Lat= (180/math.pi)*(R-(C-D*1000000-500000)**2*S/(2*V*U)+(C-D*1000000-500000)**4*W/(24*U**3*V)-(C-D*1000000-500000)**6*X/(7200*U**5*V))
Lon= (180/math.pi)*(C-D*1000000-500000)*(1-(C-D*1000000-500000)**2*Y/(6*U**2)+(C-D*1000000-500000)**4*Z/(120*U**4))/(U*math.cos(P))
Lat = Lat
Lon = K + Lon
return (Lon, Lat)