python实现两个经纬度点之间的距离和方位角的方法

最近做有关GPS轨迹上有关的东西，花费心思较多，对两个常用的函数总结一下，求距离和求方位角，比较精确，欢迎交流！

1. 求两个经纬点的方位角，P0(latA, lonA), P1(latB, lonB)（很多博客写的不是很好，这里总结一下）

def getDegree(latA, lonA, latB, lonB):
 """
 Args:
   point p1(latA, lonA)
   point p2(latB, lonB)
 Returns:
   bearing between the two GPS points,
   default: the basis of heading direction is north
 """
 radLatA = radians(latA)
 radLonA = radians(lonA)
 radLatB = radians(latB)
 radLonB = radians(lonB)
 dLon = radLonB - radLonA
 y = sin(dLon) * cos(radLatB)
 x = cos(radLatA) * sin(radLatB) - sin(radLatA) * cos(radLatB) * cos(dLon)
 brng = degrees(atan2(y, x))
 brng = (brng + 360) ％ 360
 return brng

2. 求两个经纬点的距离函数：P0(latA, lonA), P1(latB, lonB)

def getDistance(latA, lonA, latB, lonB):
 ra = 6378140 # radius of equator: meter
 rb = 6356755 # radius of polar: meter
 flatten = (ra - rb) / ra # Partial rate of the earth
 # change angle to radians
 radLatA = radians(latA)
 radLonA = radians(lonA)
 radLatB = radians(latB)
 radLonB = radians(lonB)

pA = atan(rb / ra * tan(radLatA))
 pB = atan(rb / ra * tan(radLatB))
 x = acos(sin(pA) * sin(pB) + cos(pA) * cos(pB) * cos(radLonA - radLonB))
 c1 = (sin(x) - x) * (sin(pA) + sin(pB))**2 / cos(x / 2)**2
 c2 = (sin(x) + x) * (sin(pA) - sin(pB))**2 / sin(x / 2)**2
 dr = flatten / 8 * (c1 - c2)
 distance = ra * (x + dr)
 return distance

来源：https://blog.csdn.net/zhuqiuhui/article/details/53180395