cs2000坐标系转GCJ02的python代码
时间: 2023-10-03 15:05:55 浏览: 43
以下是一个将 CS2000 坐标系转换为 GCJ02 坐标系的 Python 代码示例:
```python
import math
def transform(x, y):
pi = 3.1415926535897932384626
a = 6378245.0
ee = 0.00669342162296594323
dLat = transformLat(x - 105.0, y - 35.0)
dLon = transformLon(x - 105.0, y - 35.0)
radLat = y / 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)
mgLat = y + dLat
mgLon = x + dLon
return (mgLon, mgLat)
def transformLat(x, y):
pi = 3.1415926535897932384626
a = 6378245.0
ee = 0.00669342162296594323
dLat = -100.0 + 2.0 * x + 3.0 * y + 0.2 * y * y + 0.1 * x * y + 0.2 * math.sqrt(abs(x))
dLat += (20.0 * math.sin(6.0 * x * pi) + 20.0 * math.sin(2.0 * x * pi)) * 2.0 / 3.0
dLat += (20.0 * math.sin(y * pi) + 40.0 * math.sin(y / 3.0 * pi)) * 2.0 / 3.0
dLat += (160.0 * math.sin(y / 12.0 * pi) + 320 * math.sin(y * pi / 30.0)) * 2.0 / 3.0
return dLat
def transformLon(x, y):
pi = 3.1415926535897932384626
a = 6378245.0
ee = 0.00669342162296594323
dLon = 300.0 + x + 2.0 * y + 0.1 * x * x + 0.1 * x * y + 0.1 * math.sqrt(abs(x))
dLon += (20.0 * math.sin(6.0 * x * pi) + 20.0 * math.sin(2.0 * x * pi)) * 2.0 / 3.0
dLon += (20.0 * math.sin(x * pi) + 40.0 * math.sin(x / 3.0 * pi)) * 2.0 / 3.0
dLon += (150.0 * math.sin(x / 12.0 * pi) + 300.0 * math.sin(x / 30.0 * pi)) * 2.0 / 3.0
return dLon
# 示例用法
x = 1234567.89 # CS2000 坐标系中的 X 坐标
y = 9876543.21 # CS2000 坐标系中的 Y 坐标
gcj_lng, gcj_lat = transform(x, y)
print(gcj_lng, gcj_lat) # 输出 GCJ02 坐标系中的经纬度
```
请注意,这只是一个示例代码,并不能覆盖所有情况,也可能存在精度问题。如果您需要更加精确的坐标转换,建议使用专业的坐标转换库。