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
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