如何有效地计算2d点的三个向量之间的全（2pi）角

Question

I have three numpy arrays of shape [n, 2] containing the a list of points. Let's call these a, b, and c. I want to find the full angle between ab and bc. Using acos nets me only pi radians, but I want the full 2pi scale. I considered using atan2, but am unsure of how to calculate the y and x vectors necessory for atan2 - I tried using vector norms, but these are inherently positive. Is there any way I can do this completely using numpy functions for efficiency?

Answer 1

Using the arccos method alone only gives you the absolute angle between the vectors, not wether it is clock-wise or counter-clockwise. You can augment this by checking if the dot product of a against the perpendicular of b is negative, signifying a counter-clockwise angle.

import numpy as np

def dot(a, b):
  return np.sum(a * b, axis=-1)

def mag(a):
  return np.sqrt(np.sum(a*a, axis=-1))

def angle(a, b):
  cosab = dot(a, b) / (mag(a) * mag(b)) # cosine of angle between vectors
  angle = np.arccos(cosab) # what you currently have (absolute angle)

  b_t = b[:,[1,0]] * [1, -1] # perpendicular of b

  is_cc = dot(a, b_t) < 0

  # invert the angles for counter-clockwise rotations
  angle[is_cc] = 2*np.pi - angle[is_cc]
  return angle

print(angle(
  np.array([[1, 0], [1, 0]]),
  np.array([[0, 1], [0, -1]])
))

Will print the float values of [pi/2, 3pi/2] .

This function outputs in the range [0, 2*pi] .

Answer 2

Unsurprisingly, the angle function can be used here. It takes a complex argument x + yi :

The advantage of this method is that relative angles are easily obtained. With the atan2 this would be a bit trickier.

def get_angle(a,b,yx=False):
    # make sure inputs are contiguous float
    # swap x and  if requested
    a,b = map(np.ascontiguousarray, (a[...,::-1],b[...,::-1]) if yx else (a,b), (float,float))
    # view cast to complex, prune excess dimension
    A,B = (z.view(complex).reshape(z.shape[:-1]) for z in (a,b))
    # to get the relative angle we must either divide 
    # or (probably cheaper) multiply with the conjugate  
    return np.angle(A.conj()*B)

a,b,c = np.random.randn(3,20,2)
# let's look at a roundtrip as a test
get_angle(a,b)+get_angle(b,c)+get_angle(c,a)
# array([ 0.00000000e+00,  1.66533454e-16,  4.44089210e-16, -2.22044605e-16,
#         0.00000000e+00,  0.00000000e+00,  0.00000000e+00, -4.44089210e-16,
#         0.00000000e+00, -1.66533454e-16,  2.22044605e-16,  0.00000000e+00,
#         0.00000000e+00,  2.22044605e-16,  6.28318531e+00,  8.32667268e-17,
#         2.22044605e-16, -6.28318531e+00, -2.22044605e-16,  6.28318531e+00])
# some zeros, some 2pi and some -2pi ==> looks ok

# Let's also check the sum of angles of triangles abc:
get_angle(a-c,b-c)+get_angle(b-a,c-a)+get_angle(c-b,a-b)
# array([-3.14159265, -3.14159265,  3.14159265, -3.14159265, -3.14159265,
#         3.14159265, -3.14159265, -3.14159265,  3.14159265, -3.14159265,
#        -3.14159265,  3.14159265, -3.14159265, -3.14159265,  3.14159265,
#         3.14159265, -3.14159265, -3.14159265,  3.14159265,  3.14159265])