python实现两个经纬度点之间的距离和方位角

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

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