如何在 pandas 数据帧 Python 中找到 GPS 坐标之间的角度
How to find angle between GPS coordinates in pandas dataframe Python
我有包含测量坐标和单元格坐标的数据框。
我需要找到连接这两个点的线与北极之间的每一行角度(方位角)。
df:
id cell_lat cell_long meas_lat meas_long
1 53.543643 11.636235 53.44758 11.03720
2 52.988823 10.0421645 53.03501 9.04165
3 54.013442 9.100981 53.90384 10.62370
我在网上找到了一些代码,但是 none 如果这真的能帮助我更接近解决方案的话。
我使用了 this 函数,但不确定是否正确,我想有更简单的解决方案。
欢迎任何帮助或提示,在此先感谢。
这个问题最棘手的部分是将大地(纬度、经度)坐标转换为笛卡尔(x、y、z)坐标。如果你看https://en.wikipedia.org/wiki/Geographic_coordinate_conversion you can see how to do this, which involves choosing a reference system. Assuming we choose ECEF (https://en.wikipedia.org/wiki/ECEF),下面的代码计算你要找的角度:
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)
这将从数据框中读取每一行,计算角度并将其附加到角度列表中。显然,在计算出这些角度后,您可以随心所欲地使用这些角度。
希望对您有所帮助!
我有包含测量坐标和单元格坐标的数据框。
我需要找到连接这两个点的线与北极之间的每一行角度(方位角)。
df:
id cell_lat cell_long meas_lat meas_long
1 53.543643 11.636235 53.44758 11.03720
2 52.988823 10.0421645 53.03501 9.04165
3 54.013442 9.100981 53.90384 10.62370
我在网上找到了一些代码,但是 none 如果这真的能帮助我更接近解决方案的话。
我使用了 this 函数,但不确定是否正确,我想有更简单的解决方案。
欢迎任何帮助或提示,在此先感谢。
这个问题最棘手的部分是将大地(纬度、经度)坐标转换为笛卡尔(x、y、z)坐标。如果你看https://en.wikipedia.org/wiki/Geographic_coordinate_conversion you can see how to do this, which involves choosing a reference system. Assuming we choose ECEF (https://en.wikipedia.org/wiki/ECEF),下面的代码计算你要找的角度:
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)
这将从数据框中读取每一行,计算角度并将其附加到角度列表中。显然,在计算出这些角度后,您可以随心所欲地使用这些角度。
希望对您有所帮助!