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

我有一个二维矩阵，我想取每一行的范数.但是当我直接使用 numpy.linalg.norm(X) 时，它需要整个矩阵的范数.

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 循环获取每行的范数，然后获取每个 X[i] 的范数，但是因为我有 30k 行，所以需要很长时间.

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?或者是否可以将 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

对于 numpy 1.9+

请注意，正如 perimosocordiae 所示，从 NumPy 1.9 版开始，np.linalg.norm(x,axis=1) 是计算 L2 范数的最快方法.

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.

如果您正在计算 L2 范数，您可以直接计算它(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，但可能不那么方便:

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

其他 ord 形式的 norm 也可以直接计算(具有类似的加速):

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

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