Fastest way to get dot product of a matrix and each point in the given array

I have an array Nx3 of N points, each one has X, Y and Z coordinate. I need to rotate each point, so i have the rotation matrix 3×3. To do this, i need to get dot product of the rotation matrix and each point. The problem is the array of points is quite massive (~1_000_000x3) and therefore it takes noticeable amount of time to compute the rotated points.

The only solution i come up with so far is simple for loop iterating over array of points. Here is the example with a piece of data:

# Given set of points Nx3: np.ndarray
points = np.array([
     [285.679, 219.75,  524.733],
     [285.924, 219.404, 524.812],
     [285.116, 219.217, 524.813],
     [285.839, 219.557, 524.842],
     [285.173, 219.507, 524.606]
        ])
points_t = points.T 
# Array to store results
points_rotated = np.empty((points_t.shape))
# Given rotation matrix 3x3: np.ndarray
rot_matrix = np.array([
        [0.57423549, 0.81862056, -0.01067613],
        [-0.81866133, 0.57405696, -0.01588193],
        [-0.00687256, 0.0178601, 0.99981688]
            ])

for i in range(points.shape[0]):
    points_rotated[:, i] = np.dot(rot_matrix, points_t[:, i])

# Result
points_rotated.T 
# [[ 338.33677093 -116.05910163  526.59831864]
#  [ 338.1933725  -116.45955203  526.6694408 ]
#  [ 337.5762975  -115.90543822  526.67265381]
#  [ 338.26949115 -116.30261156  526.70275207]
#  [ 337.84863885 -115.78233784  526.47047941]]

I dont feel confident using numpy as i quite new to it, so i’m sure there is at least more elegant way to do. I’ve heard about np.einsum() but i don’t understand yet how to implement it here and i’m not sure it will make it quicker. And the main problem is still the computation time, so i want to now how to make it work faster?

Thank you very much in advance!

Answer

You can write the code using numba parallelization in no python mode as:

@nb.njit("float64[:, ::1](float64[:, ::1], float64[:, ::1])", parallel=True)
def dot(a, b):  # dot(points, rot_matrix)

    dot_ = np.zeros((a.shape[0], b.shape[0]))
    for i in nb.prange(a.shape[0]):
        for j in range(b.shape[0]):
            dot_[i, j] = a[i, 0] * b[j, 0] + a[i, 1] * b[j, 1] + a[i, 2] * b[j, 2]
    return dot_

It must be better than ko3 answer due to parallelization and signatures and using algebra instead np.dot. I have tested similar code (that was applied just on upper triangle of an array) instead dot product in another answer which showed that was at least 2 times faster (as far as I can remember).