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 3x3. 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): 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!
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, b.shape)) for i in nb.prange(a.shape): for j in range(b.shape): 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).
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