Python代码来计算三个点之间的角度(长坐标) [英] Python code to calculate angle between three points (lat long coordinates)

问题描述

有人可以建议如何计算三点(长坐标)之间的角度吗?

Can anybody suggest how to calculate angle between three points (lat long coordinates)

A : (12.92473, 77.6183)
B : (12.92512, 77.61923)
C : (12.92541, 77.61985)

推荐答案

假设您要使用角度ABC(B是该角度的顶点)，我看到了两种主要的解决方法.由于您的三个点彼此靠近(纬度小于0.0007°，经度小于0.002°)，因此我们可以将地球近似为一个平面，并使用二维矢量计算.当我们离开赤道时，经度和纬度的距​​离不是相同的距离，但是我们可以对此进行调整.另一种解决方案是将点视为三维空间，并使用三维矢量计算.在这里，我们只需要将给定的球坐标转换为3D笛卡尔坐标.

I see two main ways to solve your problem, assuming you want angle ABC (B is the vertex of the angle). Since your three points are close to each other (less than 0.0007° latitude and 0.002° longitude apart), we can approximate the earth as a plane and use two-dimensional vector calculations. A degree of longitude and of latitude are not the same distance when we are away from the equator, but we can adjust for that. Another solution is to treat your points as in three-dimensional space and use three-dimensional vector calculations. Here we just need to convert the given spherical coordinates to 3D Cartesian coordinates.

这是您遇到的问题的我的代码.为了方便起见，我在这里使用numpy模块，但是没有它就可以很容易地完成.这段代码比较冗长，因此您可以更好地了解正在执行的操作.

Here is my code for your problem. I use the numpy module here for convenience, but this can be done pretty easily without it. This code is fairly wordy so you can see better just what is being done.

import numpy as np
import math

def latlong_to_3d(latr, lonr):
    """Convert a point given latitude and longitude in radians to
    3-dimensional space, assuming a sphere radius of one."""
    return np.array((
        math.cos(latr) * math.cos(lonr),
        math.cos(latr) * math.sin(lonr),
        math.sin(latr)
    ))

def angle_between_vectors_degrees(u, v):
    """Return the angle between two vectors in any dimension space,
    in degrees."""
    return np.degrees(
        math.acos(np.dot(u, v) / (np.linalg.norm(u) * np.linalg.norm(v))))

# The points in tuple latitude/longitude degrees space
A = (12.92473, 77.6183)
B = (12.92512, 77.61923)
C = (12.92541, 77.61985)

# Convert the points to numpy latitude/longitude radians space
a = np.radians(np.array(A))
b = np.radians(np.array(B))
c = np.radians(np.array(C))

# Vectors in latitude/longitude space
avec = a - b
cvec = c - b

# Adjust vectors for changed longitude scale at given latitude into 2D space
lat = b[0]
avec[1] *= math.cos(lat)
cvec[1] *= math.cos(lat)

# Find the angle between the vectors in 2D space
angle2deg = angle_between_vectors_degrees(avec, cvec)


# The points in 3D space
a3 = latlong_to_3d(*a)
b3 = latlong_to_3d(*b)
c3 = latlong_to_3d(*c)

# Vectors in 3D space
a3vec = a3 - b3
c3vec = c3 - b3

# Find the angle between the vectors in 2D space
angle3deg = angle_between_vectors_degrees(a3vec, c3vec)


# Print the results
print('\nThe angle ABC in 2D space in degrees:', angle2deg)
print('\nThe angle ABC in 3D space in degrees:', angle3deg)

这给出了结果

The angle ABC in 2D space in degrees: 177.64369006

The angle ABC in 3D space in degrees: 177.643487338

请注意，结果非常接近(大约相差了五千分之一度)，这与三个如此接近的点所期望的一样.

Note that the results are very close (off by about one five-thousandth of a degree), as expected for three points so close together.

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