擅长:python、mysql、java
<p>比公认的答案快得多</p>
<pre><code>numpy.sqrt(numpy.einsum('ij,ij->i', a, a))
</code></pre>
<p>注意对数刻度:</p>
<p><a href="https://i.stack.imgur.com/Rpk4D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rpk4D.png" alt="enter image description here"/></a></p>
<hr/>
<p>复制绘图的代码:</p>
<pre><code>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(20)],
logx=True,
logy=True,
xlabel='len(a)'
)
</code></pre>