如何使用 NumPy 计算角度？

Question

I have x and y pixel coordinates that I am using to compute angles. Initially I used the math.atan2 approach but it gave me some issues as I was passing my data as an array . So now I am using the code below but the values that are being returned are not in the expected range of degrees. What am I doing wrong?

angle1 = np.arctan2(angle1_x, angle1_y)
angle2 = np.arctan2(angle2_x, angle2_y)
degrees = np.degrees(angle1 - angle2)

Answer 1

It seems you're trying to find the angle between two points (x1, y1) and (x2, y2) . For starters, as mentioned in the comments, you have the arguments the wrong way round. But even then your current solution doesn't really work.

angle1 and angle2 can be anywhere in the range [-pi, pi] so let's assume they are pi and -pi (which although they represent the same value, this is possible due to 0.0 and -0.0 being different floats).

This gives a 'difference' of 2*pi which is not quite what you want. If we switched them around it would be -2*pi . So our values are now in the range [-2*pi, 2*pi] which means that the same answer can be represented by multiple values. How do we fix this? By doing

angle_difference = (angle2 - angle1) % np.pi  # range == [0, 2*pi)

You'd need to do a bit more work to get into the range [-pi, pi].

angle_difference = angle_difference if angle_difference < np.pi else -2*np.pi + angle_difference

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There is another way: https://math.stackexchange.com/questions/227014/find-the-angle-between-two-vectors

import numpy as np
from numpy.linalg import norm

v1 = np.array([x1, y1])
v2 = np.array([x2, y2])

angle_difference = np.arccos((v1 @ v2) / (norm(v1) * norm(v2)))  # in range [0, pi]

However since this is symmetric it will give no indication as to the direction of the angle.