I have a 2D matrix and I want to take norm of each row. But when I use numpy.linalg.norm(X) directly, it takes the norm of the whole matrix.
I can take norm of each row by using a for loop and then taking norm of each X[i], but it takes a huge time since I have 30k rows.
Any suggestions to find a quicker way? Or is it possible to apply np.linalg.norm to each row of a matrix?
Note that, as perimosocordiae shows, as of NumPy version 1.9, np.linalg.norm(x, axis=1) is the fastest way to compute the L2-norm.
If you are computing an L2-norm, you could compute it directly (using the axis=-1 argument to sum along rows):
np.sum(np.abs(x)**2,axis=-1)**(1./2) Lp-norms can be computed similarly of course.
It is considerably faster than np.apply_along_axis, though perhaps not as convenient:
In [48]: %timeit np.apply_along_axis(np.linalg.norm, 1, x) 1000 loops, best of 3: 208 us per loop In [49]: %timeit np.sum(np.abs(x)**2,axis=-1)**(1./2) 100000 loops, best of 3: 18.3 us per loop Other ord forms of norm can be computed directly too (with similar speedups):
In [55]: %timeit np.apply_along_axis(lambda row:np.linalg.norm(row,ord=1), 1, x) 1000 loops, best of 3: 203 us per loop In [54]: %timeit np.sum(abs(x), axis=-1) 100000 loops, best of 3: 10.9 us per loop Resurrecting an old question due to a numpy update. As of the 1.9 release, numpy.linalg.norm now accepts an axis argument. [code, documentation]
This is the new fastest method in town:
In [10]: x = np.random.random((500,500)) In [11]: %timeit np.apply_along_axis(np.linalg.norm, 1, x) 10 loops, best of 3: 21 ms per loop In [12]: %timeit np.sum(np.abs(x)**2,axis=-1)**(1./2) 100 loops, best of 3: 2.6 ms per loop In [13]: %timeit np.linalg.norm(x, axis=1) 1000 loops, best of 3: 1.4 ms per loop And to prove it's calculating the same thing:
In [14]: np.allclose(np.linalg.norm(x, axis=1), np.sum(np.abs(x)**2,axis=-1)**(1./2)) Out[14]: True Try the following:
In [16]: numpy.apply_along_axis(numpy.linalg.norm, 1, a) Out[16]: array([ 5.38516481, 1.41421356, 5.38516481]) where a is your 2D array.
The above computes the L2 norm. For a different norm, you could use something like:
In [22]: numpy.apply_along_axis(lambda row:numpy.linalg.norm(row,ord=1), 1, a) Out[22]: array([9, 2, 9]) Much faster than the accepted answer is
numpy.sqrt(numpy.einsum('ij,ij->i', a, a)) Note the log-scale:
Code to reproduce the plot:
import numpy import perfplot def sum_sqrt(a): return numpy.sqrt(numpy.sum(numpy.abs(a)**2, axis=-1)) def apply_norm_along_axis(a): return numpy.apply_along_axis(numpy.linalg.norm, 1, a) def norm_axis(a): return numpy.linalg.norm(a, axis=1) def einsum_sqrt(a): return numpy.sqrt(numpy.einsum('ij,ij->i', a, a)) perfplot.show( setup=lambda n: numpy.random.rand(n, 3), kernels=[sum_sqrt, apply_norm_along_axis, norm_axis, einsum_sqrt], n_range=[2**k for k in range(15)], logx=True, logy=True, xlabel='len(a)' )