如何在pandas dataframe Python中查找GPS坐标之间的夹角

我有测量坐标和单元坐标的数据框

我需要找出连接这两点和北极的直线之间的每一行角度（方位角）

df:

我在网上找到了一些代码，但如果这真的能帮助我更接近解决方案的话，就没有了

我用过函数，但不确定是否正确，我想有更简单的解决方案

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欢迎提供任何帮助或提示，提前感谢。

此问题最棘手的部分是将大地坐标（纬度、经度）转换为笛卡尔坐标（x、y、z）。如果你看一下，你可以看到如何做到这一点，这涉及到选择一个参考系统。假设我们选择ECEF（），下面的代码将计算您要查找的角度：

def vector_calc(lat, long, ht):
    '''
    Calculates the vector from a specified point on the Earth's surface to the North Pole.
    '''
    a = 6378137.0  # Equatorial radius of the Earth
    b = 6356752.314245  # Polar radius of the Earth

    e_squared = 1 - ((b ** 2) / (a ** 2))  # e is the eccentricity of the Earth
    n_phi = a / (np.sqrt(1 - (e_squared * (np.sin(lat) ** 2))))

    x = (n_phi + ht) * np.cos(lat) * np.cos(long)
    y = (n_phi + ht) * np.cos(lat) * np.sin(long)
    z = ((((b ** 2) / (a ** 2)) * n_phi) + ht) * np.sin(lat)

    x_npole = 0.0
    y_npole = 6378137.0
    z_npole = 0.0

    v = ((x_npole - x), (y_npole - y), (z_npole - z))

    return v

def angle_calc(lat1, long1, lat2, long2, ht1=0, ht2=0):
    '''
    Calculates the angle between the vectors from 2 points to the North Pole.
    '''
    # Convert from degrees to radians
    lat1_rad = (lat1 / 180) * np.pi
    long1_rad = (long1 / 180) * np.pi
    lat2_rad = (lat2 / 180) * np.pi
    long2_rad = (long2 / 180) * np.pi

    v1 = vector_calc(lat1_rad, long1_rad, ht1)
    v2 = vector_calc(lat2_rad, long2_rad, ht2)

    # The angle between two vectors, vect1 and vect2 is given by:
    # arccos[vect1.vect2 / |vect1||vect2|]
    dot = np.dot(v1, v2)  # The dot product of the two vectors
    v1_mag = np.linalg.norm(v1)  # The magnitude of the vector v1
    v2_mag = np.linalg.norm(v2)  # The magnitude of the vector v2

    theta_rad = np.arccos(dot / (v1_mag * v2_mag))
    # Convert radians back to degrees
    theta = (theta_rad / np.pi) * 180

    return theta

angles = []
for row in range(df.shape[0]):
    cell_lat = df.iloc[row]['cell_lat']
    cell_long = df.iloc[row]['cell_long']
    meas_lat = df.iloc[row]['meas_lat']
    meas_long = df.iloc[row]['meas_long']

    angle = angle_calc(cell_lat, cell_long, meas_lat, meas_long)

    angles.append(angle)

这将读取数据帧中的每一行，计算角度并将其附加到列表角度。显然，在计算完这些角度后，你可以用它们做你喜欢的事情

希望有帮助

可能重复的See答案-包括2d和3d（考虑到距离更重要，因为您在计算中使用北极）。

def vector_calc(lat, long, ht):
    '''
    Calculates the vector from a specified point on the Earth's surface to the North Pole.
    '''
    a = 6378137.0  # Equatorial radius of the Earth
    b = 6356752.314245  # Polar radius of the Earth

    e_squared = 1 - ((b ** 2) / (a ** 2))  # e is the eccentricity of the Earth
    n_phi = a / (np.sqrt(1 - (e_squared * (np.sin(lat) ** 2))))

    x = (n_phi + ht) * np.cos(lat) * np.cos(long)
    y = (n_phi + ht) * np.cos(lat) * np.sin(long)
    z = ((((b ** 2) / (a ** 2)) * n_phi) + ht) * np.sin(lat)

    x_npole = 0.0
    y_npole = 6378137.0
    z_npole = 0.0

    v = ((x_npole - x), (y_npole - y), (z_npole - z))

    return v

def angle_calc(lat1, long1, lat2, long2, ht1=0, ht2=0):
    '''
    Calculates the angle between the vectors from 2 points to the North Pole.
    '''
    # Convert from degrees to radians
    lat1_rad = (lat1 / 180) * np.pi
    long1_rad = (long1 / 180) * np.pi
    lat2_rad = (lat2 / 180) * np.pi
    long2_rad = (long2 / 180) * np.pi

    v1 = vector_calc(lat1_rad, long1_rad, ht1)
    v2 = vector_calc(lat2_rad, long2_rad, ht2)

    # The angle between two vectors, vect1 and vect2 is given by:
    # arccos[vect1.vect2 / |vect1||vect2|]
    dot = np.dot(v1, v2)  # The dot product of the two vectors
    v1_mag = np.linalg.norm(v1)  # The magnitude of the vector v1
    v2_mag = np.linalg.norm(v2)  # The magnitude of the vector v2

    theta_rad = np.arccos(dot / (v1_mag * v2_mag))
    # Convert radians back to degrees
    theta = (theta_rad / np.pi) * 180

    return theta

angles = []
for row in range(df.shape[0]):
    cell_lat = df.iloc[row]['cell_lat']
    cell_long = df.iloc[row]['cell_long']
    meas_lat = df.iloc[row]['meas_lat']
    meas_long = df.iloc[row]['meas_long']

    angle = angle_calc(cell_lat, cell_long, meas_lat, meas_long)

    angles.append(angle)