NumPy: how to quickly normalize many vectors?

Question

How can a list of vectors be elegantly normalized, in NumPy?

Here is an example that does not work:

from numpy import *

vectors = array([arange(10), arange(10)])  # All x's, then all y's
norms = apply_along_axis(linalg.norm, 0, vectors)

# Now, what I was expecting would work:
print vectors.T / norms  # vectors.T has 10 elements, as does norms, but this does not work

The last operation yields "shape mismatch: objects cannot be broadcast to a single shape".

How can the normalization of the 2D vectors in vectors be elegantly done, with NumPy?

Edit: Why does the above not work while adding a dimension to norms does work (as per my answer below)?

Answer 1

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.