Say I have a matrix Y
of random float numbers from 0 to 10 with shape (10, 3)
:
import numpy as np
np.random.seed(99)
Y = np.random.uniform(0, 10, (10, 3))
print(Y)
Output:
[[6.72278559 4.88078399 8.25495174]
[0.31446388 8.08049963 5.6561742 ]
[2.97622499 0.46695721 9.90627399]
[0.06825733 7.69793028 7.46767101]
[3.77438936 4.94147452 9.28948392]
[3.95454044 9.73956297 5.24414715]
[0.93613093 8.13308413 2.11686786]
[5.54345785 2.92269116 8.1614236 ]
[8.28042566 2.21577372 6.44834702]
[0.95181622 4.11663239 0.96865261]]
I am now given a matrix X
with same shape that can be seen as obtained by adding small noises to Y
and then shuffling the rows:
X = np.random.normal(Y, scale=0.1)
np.random.shuffle(X)
print(X)
Output:
[[ 4.04067271 9.90959141 5.19126867]
[ 5.59873104 2.84109306 8.11175891]
[ 0.10743952 7.74620162 7.51100441]
[ 3.60396019 4.91708372 9.07551354]
[ 0.9400948 4.15448712 1.04187208]
[ 2.91884302 0.47222752 10.12700505]
[ 0.30995155 8.09263241 5.74876947]
[ 1.11247872 8.02092335 1.99767444]
[ 6.68543696 4.8345869 8.17330513]
[ 8.38904822 2.11830619 6.42013343]]
Now I want to sort the matrix X
based on Y
by row. I already know each pair of column values in each matching pair of rows are not different from each other more than a tolerance of 0.5. I managed to write the following code and it is working fine.
def sort_X_by_Y(X, Y, tol):
idxs = [next(i for i in range(len(X)) if all(abs(X[i] - row) <= tol)) for row in Y]
return X[idxs]
print(sort_X_by_Y(X, Y, tol=0.5))
Output:
[[ 6.68543696 4.8345869 8.17330513]
[ 0.30995155 8.09263241 5.74876947]
[ 2.91884302 0.47222752 10.12700505]
[ 0.10743952 7.74620162 7.51100441]
[ 3.60396019 4.91708372 9.07551354]
[ 4.04067271 9.90959141 5.19126867]
[ 1.11247872 8.02092335 1.99767444]
[ 5.59873104 2.84109306 8.11175891]
[ 8.38904822 2.11830619 6.42013343]
[ 0.9400948 4.15448712 1.04187208]]
However, in reality I am sorting (1000, 3)
matrices and my code is way too slow. I feel like there should be more numpyic way to code this. Any suggestions?
CodePudding user response:
This is a vectorized version of your algorithm. It runs ~26.5x faster than your implementation for 1000 samples. But an additional boolean array with shape (1000,1000,3)
is created. There is a chance that rows will have similar values within the tolerance and a wrong row is selected.
tol = .5
X[(np.abs(Y[:, np.newaxis] - X) <= tol).all(2).argmax(1)]
Output
array([[ 6.68543696, 4.8345869 , 8.17330513],
[ 0.30995155, 8.09263241, 5.74876947],
[ 2.91884302, 0.47222752, 10.12700505],
[ 0.10743952, 7.74620162, 7.51100441],
[ 3.60396019, 4.91708372, 9.07551354],
[ 4.04067271, 9.90959141, 5.19126867],
[ 1.11247872, 8.02092335, 1.99767444],
[ 5.59873104, 2.84109306, 8.11175891],
[ 8.38904822, 2.11830619, 6.42013343],
[ 0.9400948 , 4.15448712, 1.04187208]])
More robust solutions with L1-norm
X[np.abs(Y[:, np.newaxis] - X).sum(2).argmin(1)]
Or L2-norm
X[((Y[:, np.newaxis] - X)**2).sum(2).argmin(1)]