NumPy:如何快速归一化许多向量? [英] NumPy: how to quickly normalize many vectors?

问题描述

如何在NumPy中对向量列表进行优雅的标准化?

How can a list of vectors be elegantly normalized, in NumPy?

这是一个不起作用的示例:

Here is an example that does not work:

from numpy import *

vectors = array([arange(10), arange(10)])  # All x's, then all y's
norms = apply_along_axis(linalg.norm, 0, vectors)

# Now, what I was expecting would work:
print vectors.T / norms  # vectors.T has 10 elements, as does norms, but this does not work

最后一个操作会产生形状不匹配:对象无法广播为单个形状".

The last operation yields "shape mismatch: objects cannot be broadcast to a single shape".

如何使用NumPy优雅地完成vectors中2D向量的归一化?

How can the normalization of the 2D vectors in vectors be elegantly done, with NumPy?

编辑:为什么在向norms添加尺寸时上述方法不起作用(根据我在下面的回答)?

Edit: Why does the above not work while adding a dimension to norms does work (as per my answer below)?

推荐答案

好吧，除非我错过了什么，否则确实可以:

Well, unless I missed something, this does work:

vectors / norms

您的建议中的问题是广播规则.

The problem in your suggestion is the broadcasting rules.

vectors  # shape 2, 10
norms  # shape 10

形状不一样长！因此，规则是先在左侧上将小形状扩展一个:

The shape do not have the same length! So the rule is to first extend the small shape by one on the left:

norms  # shape 1,10

您可以通过以下方式手动进行操作:

You can do that manually by calling:

vectors / norms.reshape(1,-1)  # same as vectors/norms

如果要计算vectors.T/norms，则必须手动进行重塑，如下所示:

If you wanted to compute vectors.T/norms, you would have to do the reshaping manually, as follows:

vectors.T / norms.reshape(-1,1)  # this works

这篇关于NumPy:如何快速归一化许多向量?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！