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         


        

Answer 1

My preferred way to normalize vectors is by using numpy's inner1d to calculate their magnitudes. Here's what's been suggested so far compared to inner1d

import numpy as np
from numpy.core.umath_tests import inner1d
COUNT = 10**6 # 1 million points

points = np.random.random_sample((COUNT,3,))
A      = np.sqrt(np.einsum('...i,...i', points, points))
B      = np.apply_along_axis(np.linalg.norm, 1, points)   
C      = np.sqrt((points ** 2).sum(-1))
D      = np.sqrt((points*points).sum(axis=1))
E      = np.sqrt(inner1d(points,points))

print [np.allclose(E,x) for x in [A,B,C,D]] # [True, True, True, True]

Testing performance with cProfile:

import cProfile
cProfile.run("np.sqrt(np.einsum('...i,...i', points, points))**0.5") # 3 function calls in 0.013 seconds
cProfile.run('np.apply_along_axis(np.linalg.norm, 1, points)')       # 9000018 function calls in 10.977 seconds
cProfile.run('np.sqrt((points ** 2).sum(-1))')                       # 5 function calls in 0.028 seconds
cProfile.run('np.sqrt((points*points).sum(axis=1))')                 # 5 function calls in 0.027 seconds
cProfile.run('np.sqrt(inner1d(points,points))')                      # 2 function calls in 0.009 seconds

inner1d computed the magnitudes a hair faster than einsum. So using inner1d to normalize:

n = points/np.sqrt(inner1d(points,points))[:,None]
cProfile.run('points/np.sqrt(inner1d(points,points))[:,None]') # 2 function calls in 0.026 seconds

Testing against scikit:

import sklearn.preprocessing as preprocessing
n_ = preprocessing.normalize(points, norm='l2')
cProfile.run("preprocessing.normalize(points, norm='l2')") # 47 function calls in 0.047 seconds
np.allclose(n,n_) # True

Conclusion: using inner1d seems to be the best option