3D 向量的旋转？

Question

I have two vectors as Python lists and an angle. Eg:

v = [3,5,0]
axis = [4,4,1]
theta = 1.2 #radian

What is the best/easiest way to get the resulting vector when rotating the v vector around the axis?

The rotation should appear to be counter clockwise for an observer to whom the axis vector is pointing. This is called the right hand rule

Answer 1

Using theEuler-Rodrigues formula :

import numpy as np
import math

def rotation_matrix(axis, theta):
    """
    Return the rotation matrix associated with counterclockwise rotation about
    the given axis by theta radians.
    """
    axis = np.asarray(axis)
    axis = axis / math.sqrt(np.dot(axis, axis))
    a = math.cos(theta / 2.0)
    b, c, d = -axis * math.sin(theta / 2.0)
    aa, bb, cc, dd = a * a, b * b, c * c, d * d
    bc, ad, ac, ab, bd, cd = b * c, a * d, a * c, a * b, b * d, c * d
    return np.array([[aa + bb - cc - dd, 2 * (bc + ad), 2 * (bd - ac)],
                     [2 * (bc - ad), aa + cc - bb - dd, 2 * (cd + ab)],
                     [2 * (bd + ac), 2 * (cd - ab), aa + dd - bb - cc]])

v = [3, 5, 0]
axis = [4, 4, 1]
theta = 1.2 

print(np.dot(rotation_matrix(axis, theta), v)) 
# [ 2.74911638  4.77180932  1.91629719]

Answer 2

A one-liner, with numpy/scipy functions.

We use the following:

let a be the unit vector along axis , ie a = axis/norm(axis)
and A = I × a be the skew-symmetric matrix associated to a , ie the cross product of the identity matrix with a

then M = exp(θ A) is the rotation matrix.

from numpy import cross, eye, dot
from scipy.linalg import expm, norm

def M(axis, theta):
    return expm(cross(eye(3), axis/norm(axis)*theta))

v, axis, theta = [3,5,0], [4,4,1], 1.2
M0 = M(axis, theta)

print(dot(M0,v))
# [ 2.74911638  4.77180932  1.91629719]

expm (code here) computes the taylor series of the exponential:
\sum_{k=0}^{20} \frac{1}{k!} (θ A)^k , so it's time expensive, but readable and secure. It can be a good way if you have few rotations to do but a lot of vectors.

Answer 3

I just wanted to mention that if speed is required, wrapping unutbu's code in scipy's weave.inline and passing an already existing matrix as a parameter yields a 20-fold decrease in the running time.

The code (in rotation_matrix_test.py):

import numpy as np
import timeit

from math import cos, sin, sqrt
import numpy.random as nr

from scipy import weave

def rotation_matrix_weave(axis, theta, mat = None):
    if mat == None:
        mat = np.eye(3,3)

    support = "#include <math.h>"
    code = """
        double x = sqrt(axis[0] * axis[0] + axis[1] * axis[1] + axis[2] * axis[2]);
        double a = cos(theta / 2.0);
        double b = -(axis[0] / x) * sin(theta / 2.0);
        double c = -(axis[1] / x) * sin(theta / 2.0);
        double d = -(axis[2] / x) * sin(theta / 2.0);

        mat[0] = a*a + b*b - c*c - d*d;
        mat[1] = 2 * (b*c - a*d);
        mat[2] = 2 * (b*d + a*c);

        mat[3*1 + 0] = 2*(b*c+a*d);
        mat[3*1 + 1] = a*a+c*c-b*b-d*d;
        mat[3*1 + 2] = 2*(c*d-a*b);

        mat[3*2 + 0] = 2*(b*d-a*c);
        mat[3*2 + 1] = 2*(c*d+a*b);
        mat[3*2 + 2] = a*a+d*d-b*b-c*c;
    """

    weave.inline(code, ['axis', 'theta', 'mat'], support_code = support, libraries = ['m'])

    return mat

def rotation_matrix_numpy(axis, theta):
    mat = np.eye(3,3)
    axis = axis/sqrt(np.dot(axis, axis))
    a = cos(theta/2.)
    b, c, d = -axis*sin(theta/2.)

    return np.array([[a*a+b*b-c*c-d*d, 2*(b*c-a*d), 2*(b*d+a*c)],
                  [2*(b*c+a*d), a*a+c*c-b*b-d*d, 2*(c*d-a*b)],
                  [2*(b*d-a*c), 2*(c*d+a*b), a*a+d*d-b*b-c*c]])

The timing:

>>> import timeit
>>> 
>>> setup = """
... import numpy as np
... import numpy.random as nr
... 
... from rotation_matrix_test import rotation_matrix_weave
... from rotation_matrix_test import rotation_matrix_numpy
... 
... mat1 = np.eye(3,3)
... theta = nr.random()
... axis = nr.random(3)
... """
>>> 
>>> timeit.repeat("rotation_matrix_weave(axis, theta, mat1)", setup=setup, number=100000)
[0.36641597747802734, 0.34883809089660645, 0.3459300994873047]
>>> timeit.repeat("rotation_matrix_numpy(axis, theta)", setup=setup, number=100000)
[7.180983066558838, 7.172032117843628, 7.180462837219238]

Answer 4

Here is an elegant method using quaternions that are blazingly fast; I can calculate 10 million rotations per second with appropriately vectorised numpy arrays. It relies on the quaternion extension to numpy found here .

Quaternion Theory: A quaternion is a number with one real and 3 imaginary dimensions usually written as q = w + xi + yj + zk where 'i', 'j', 'k' are imaginary dimensions. Just as a unit complex number 'c' can represent all 2d rotations by c=exp(i * theta) , a unit quaternion 'q' can represent all 3d rotations by q=exp(p) , where 'p' is a pure imaginary quaternion set by your axis and angle.

We start by converting your axis and angle to a quaternion whose imaginary dimensions are given by your axis of rotation, and whose magnitude is given by half the angle of rotation in radians. The 4 element vectors (w, x, y, z) are constructed as follows:

import numpy as np
import quaternion as quat

v = [3,5,0]
axis = [4,4,1]
theta = 1.2 #radian

vector = np.array([0.] + v)
rot_axis = np.array([0.] + axis)
axis_angle = (theta*0.5) * rot_axis/np.linalg.norm(rot_axis)

First, a numpy array of 4 elements is constructed with the real component w=0 for both the vector to be rotated vector and the rotation axis rot_axis . The axis angle representation is then constructed by normalizing then multiplying by half the desired angle theta . See here for why half the angle is required.

Now create the quaternions v and qlog using the library, and get the unit rotation quaternion q by taking the exponential.

vec = quat.quaternion(*v)
qlog = quat.quaternion(*axis_angle)
q = np.exp(qlog)

Finally, the rotation of the vector is calculated by the following operation.

v_prime = q * vec * np.conjugate(q)

print(v_prime) # quaternion(0.0, 2.7491163, 4.7718093, 1.9162971)

Now just discard the real element and you have your rotated vector!

v_prime_vec = v_prime.imag # [2.74911638 4.77180932 1.91629719] as a numpy array

Note that this method is particularly efficient if you have to rotate a vector through many sequential rotations, as the quaternion product can just be calculated as q = q1 * q2 * q3 * q4 *... * qn and then the vector is only rotated by 'q' at the very end using v' = q * v * conj(q).

This method gives you a seamless transformation between axis angle <---> 3d rotation operator simply by exp and log functions (yes log(q) just returns the axis-angle representation.). For further clarification of how quaternion multiplication etc, work, see here

Answer 5

Take a look at http://vpython.org/contents/docs/visual/VisualIntro.html .

It provides a vector class which has a method A.rotate(theta,B) . It also provides a helper function rotate(A,theta,B) if you don't want to call the method on A .

http://vpython.org/contents/docs/visual/vector.html

Answer 6

Use scipy's Rotation.from_rotvec() . The argument is the rotation vector (a unit vector) multiplied by the rotation angle in rads.

from scipy.spatial.transform import Rotation
from numpy.linalg import norm


v = [3, 5, 0]
axis = [4, 4, 1]
theta = 1.2

axis = axis / norm(axis)  # normalize the rotation vector first
rot = Rotation.from_rotvec(theta * axis)

new_v = rot.apply(v)  
print(new_v)    # results in [2.74911638 4.77180932 1.91629719]

There are several more ways to use Rotation based on what data you have about the rotation:

• from_quat Initialized from quaternions.

• from_dcm Initialized from direction cosine matrices.

• from_euler Initialized from Euler angles.

---

_{Off-topic note: One line code is not necessarily better code as implied by some users.}

Answer 7

I made a fairly complete library of 3D mathematics for Python{2,3}. It still does not use Cython, but relies heavily on the efficiency of numpy. You can find it here with pip:

python[3] -m pip install math3d

Or have a look at my gitweb http://git.automatics.dyndns.dk/?p=pymath3d.git and now also on github: https://github.com/mortlind/pymath3d .

Once installed, in python you may create the orientation object which can rotate vectors, or be part of transform objects. Eg the following code snippet composes an orientation that represents a rotation of 1 rad around the axis [1,2,3], applies it to the vector [4,5,6], and prints the result:

import math3d as m3d
r = m3d.Orientation.new_axis_angle([1,2,3], 1)
v = m3d.Vector(4,5,6)
print(r * v)

The output would be

<Vector: (2.53727, 6.15234, 5.71935)>

This is more efficient, by a factor of approximately four, as far as I can time it, than the oneliner using scipy posted by BM above. However, it requires installation of my math3d package.

Answer 8

It can also be solved using quaternion theory:

def angle_axis_quat(theta, axis):
    """
    Given an angle and an axis, it returns a quaternion.
    """
    axis = np.array(axis) / np.linalg.norm(axis)
    return np.append([np.cos(theta/2)],np.sin(theta/2) * axis)

def mult_quat(q1, q2):
    """
    Quaternion multiplication.
    """
    q3 = np.copy(q1)
    q3[0] = q1[0]*q2[0] - q1[1]*q2[1] - q1[2]*q2[2] - q1[3]*q2[3]
    q3[1] = q1[0]*q2[1] + q1[1]*q2[0] + q1[2]*q2[3] - q1[3]*q2[2]
    q3[2] = q1[0]*q2[2] - q1[1]*q2[3] + q1[2]*q2[0] + q1[3]*q2[1]
    q3[3] = q1[0]*q2[3] + q1[1]*q2[2] - q1[2]*q2[1] + q1[3]*q2[0]
    return q3

def rotate_quat(quat, vect):
    """
    Rotate a vector with the rotation defined by a quaternion.
    """
    # Transfrom vect into an quaternion 
    vect = np.append([0],vect)
    # Normalize it
    norm_vect = np.linalg.norm(vect)
    vect = vect/norm_vect
    # Computes the conjugate of quat
    quat_ = np.append(quat[0],-quat[1:])
    # The result is given by: quat * vect * quat_
    res = mult_quat(quat, mult_quat(vect,quat_)) * norm_vect
    return res[1:]

v = [3, 5, 0]
axis = [4, 4, 1]
theta = 1.2 

print(rotate_quat(angle_axis_quat(theta, axis), v))
# [2.74911638 4.77180932 1.91629719]

Answer 9

Using pyquaternion is extremely simple; to install it (while still in python), run in your console:

import pip;
pip.main(['install','pyquaternion'])

Once installed:

  from pyquaternion import Quaternion
  v = [3,5,0]
  axis = [4,4,1]
  theta = 1.2 #radian
  rotated_v = Quaternion(axis=axis,angle=theta).rotate(v)

Answer 10

Disclaimer: I am the author of this package

While special classes for rotations can be convenient, in some cases one needs rotation matrices (eg for working with other libraries like the affine_transform functions in scipy). To avoid everyone implementing their own little matrix generating functions, there exists a tiny pure python package which does nothing more than providing convenient rotation matrix generating functions. The package is on github ( mgen ) and can be installed via pip:

pip install mgen

Example usage copied from the readme:

import numpy as np
np.set_printoptions(suppress=True)

from mgen import rotation_around_axis
from mgen import rotation_from_angles
from mgen import rotation_around_x

matrix = rotation_from_angles([np.pi/2, 0, 0], 'XYX')
matrix.dot([0, 1, 0])
# array([0., 0., 1.])

matrix = rotation_around_axis([1, 0, 0], np.pi/2)
matrix.dot([0, 1, 0])
# array([0., 0., 1.])

matrix = rotation_around_x(np.pi/2)
matrix.dot([0, 1, 0])
# array([0., 0., 1.])

Note that the matrices are just regular numpy arrays, so no new data-structures are introduced when using this package.

Answer 11

I needed to rotate a 3D model around one of the three axes {x, y, z} in which that model was embedded and this was the top result for a search of how to do this in numpy. I used the following simple function:

def rotate(X, theta, axis='x'):
  '''Rotate multidimensional array `X` `theta` degrees around axis `axis`'''
  c, s = np.cos(theta), np.sin(theta)
  if axis == 'x': return np.dot(X, np.array([
    [1.,  0,  0],
    [0 ,  c, -s],
    [0 ,  s,  c]
  ]))
  elif axis == 'y': return np.dot(X, np.array([
    [c,  0,  -s],
    [0,  1,   0],
    [s,  0,   c]
  ]))
  elif axis == 'z': return np.dot(X, np.array([
    [c, -s,  0 ],
    [s,  c,  0 ],
    [0,  0,  1.],
  ]))