How to apply numpy.linalg.norm to each row of a matrix?

Question

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?

Asked By: user103021

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Source

Answer 1

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])

Answered By: NPE

Answer 2

For numpy 1.9+

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.

For numpy < 1.9

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

Answered By: unutbu

Answer 3

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

Answered By: perimosocordiae

Answer 4

Much faster than the accepted answer is using NumPy’s einsum,

numpy.sqrt(numpy.einsum('ij,ij->i', a, a))

And even faster than that is arranging the data such that the norms are computed across all columns,

numpy.sqrt(numpy.einsum('ij,ij->j', aT, aT))

Note the log-scale:

Code to reproduce the plot:

import numpy as np
import perfplot

rng = np.random.default_rng(0)


def setup(n):
    x = rng.random((n, 3))
    xt = np.ascontiguousarray(x.T)
    return x, xt


def sum_sqrt(a, _):
    return np.sqrt(np.sum(np.abs(a) ** 2, axis=-1))


def apply_norm_along_axis(a, _):
    return np.apply_along_axis(np.linalg.norm, 1, a)


def norm_axis(a, _):
    return np.linalg.norm(a, axis=1)


def einsum_sqrt(a, _):
    return np.sqrt(np.einsum("ij,ij->i", a, a))


def einsum_sqrt_columns(_, aT):
    return np.sqrt(np.einsum("ij,ij->j", aT, aT))


b = perfplot.bench(
    setup=setup,
    kernels=[
        sum_sqrt,
        apply_norm_along_axis,
        norm_axis,
        einsum_sqrt,
        einsum_sqrt_columns,
    ],
    n_range=[2**k for k in range(20)],
    xlabel="len(a)",
)
b.show()
b.save("out.png")

Answered By: Nico Schlömer

How to apply numpy.linalg.norm to each row of a matrix?

Question:

Answers:

For numpy 1.9+

For numpy < 1.9