如何将numpy.linalg.norm应用于矩阵的每一行? [英] How to apply numpy.linalg.norm to each row of a matrix?

问题描述

我有一个2D矩阵，我想对每一行进行规范.但是，当我直接使用numpy.linalg.norm(X)时，它将采用整个矩阵的范数.

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.

我可以通过使用for循环来确定每一行的范数，然后再对每个X[i]进行准则，但是由于我有3万行，这需要花费大量时间.

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.

有什么建议可以找到更快的方法吗?还是可以将np.linalg.norm应用于矩阵的每一行?

Any suggestions to find a quicker way? Or is it possible to apply np.linalg.norm to each row of a matrix?

推荐答案

请注意，如 perimosocordiae所示， NumPy 1.9版np.linalg.norm(x, axis=1)是计算L2-范数的最快方法.

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.

如果要计算L2范数，则可以直接计算(使用axis=-1参数沿行求和):

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范数当然可以类似地计算.

Lp-norms can be computed similarly of course.

它比np.apply_along_axis快得多，尽管可能不那么方便:

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

norm的其他ord形式也可以直接计算(具有类似的加速比):

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

这篇关于如何将numpy.linalg.norm应用于矩阵的每一行?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！