Computing the magnitude

I came across this question and became curious about your method for normalizing. I use a different method to compute the magnitudes. Note: I also typically compute norms across the last index (rows in this case, not columns).

magnitudes = np.sqrt((vectors ** 2).sum(-1))[..., np.newaxis]

Typically, however, I just normalize like so:

vectors /= np.sqrt((vectors ** 2).sum(-1))[..., np.newaxis]

A time comparison

I ran a test to compare the times, and found that my method is faster by quite a bit, but Freddie Witherdon's suggestion is even faster.

import numpy as np    vectors = np.random.rand(100, 25)# OP's%timeit np.apply_along_axis(np.linalg.norm, 1, vectors)# Output: 100 loops, best of 3: 2.39 ms per loop# Mine%timeit np.sqrt((vectors ** 2).sum(-1))[..., np.newaxis]# Output: 10000 loops, best of 3: 13.8 us per loop# Freddie's (from comment below)%timeit np.sqrt(np.einsum('...i,...i', vectors, vectors))# Output: 10000 loops, best of 3: 6.45 us per loop

Beware though, as this StackOverflow answer notes, there are some safety checks not happening with einsum, so you should be sure that the dtype of vectors is sufficient to store the square of the magnitudes accurately enough.

Well, unless I missed something, this does work:

vectors / norms

The problem in your suggestion is the broadcasting rules.

vectors  # shape 2, 10norms  # shape 10

The shape do not have the same length! So the rule is to first extend the small shape by one on the left:

norms  # shape 1,10

You can do that manually by calling:

vectors / norms.reshape(1,-1)  # same as vectors/norms

If you wanted to compute vectors.T/norms, you would have to do the reshaping manually, as follows:

vectors.T / norms.reshape(-1,1)  # this works

Alright: NumPy's array shape broadcast adds dimensions to the left of the array shape, not to its right. NumPy can however be instructed to add a dimension to the right of the norms array:

print vectors.T / norms[:, newaxis]

does work!