你如何在Numpy中获得向量的大小？

Question

为了与"只有一种明显的方法"相符,你如何在Numpy中获得向量(1D数组)的大小？

def mag(x): 
    return math.sqrt(sum(i**2 for i in x))

上面的作品,但我不敢相信我必须自己指定这样一个琐碎的核心功能.

Answer 1

你所追求的功能是numpy.linalg.norm.(我认为它应该在基础numpy中作为数组的属性 - 比如说x.norm()- 但是很好).

import numpy as np
x = np.array([1,2,3,4,5])
np.linalg.norm(x)

您也可以ord为您想要的第n个订单提供可选项.说你想要1-norm:

np.linalg.norm(x,ord=1)

等等.

Answer 2

如果你对速度感到担忧,你应该使用:

mag = np.sqrt(x.dot(x))

以下是一些基准测试:

>>> import timeit
>>> timeit.timeit('np.linalg.norm(x)', setup='import numpy as np; x = np.arange(100)', number=1000)
0.0450878
>>> timeit.timeit('np.sqrt(x.dot(x))', setup='import numpy as np; x = np.arange(100)', number=1000)
0.0181372

编辑:真正的速度改进来自你必须采取许多向量的规范.使用纯numpy函数不需要任何for循环.例如:

In [1]: import numpy as np

In [2]: a = np.arange(1200.0).reshape((-1,3))

In [3]: %timeit [np.linalg.norm(x) for x in a]
100 loops, best of 3: 4.23 ms per loop

In [4]: %timeit np.sqrt((a*a).sum(axis=1))
100000 loops, best of 3: 18.9 us per loop

In [5]: np.allclose([np.linalg.norm(x) for x in a],np.sqrt((a*a).sum(axis=1)))
Out[5]: True

Answer 3

另一种方法是einsum在numpy中使用这两个数组的函数:

In [1]: import numpy as np

In [2]: a = np.arange(1200.0).reshape((-1,3))

In [3]: %timeit [np.linalg.norm(x) for x in a]
100 loops, best of 3: 3.86 ms per loop

In [4]: %timeit np.sqrt((a*a).sum(axis=1))
100000 loops, best of 3: 15.6 µs per loop

In [5]: %timeit np.sqrt(np.einsum('ij,ij->i',a,a))
100000 loops, best of 3: 8.71 µs per loop

或矢量:

In [5]: a = np.arange(100000)

In [6]: %timeit np.sqrt(a.dot(a))
10000 loops, best of 3: 80.8 µs per loop

In [7]: %timeit np.sqrt(np.einsum('i,i', a, a))
10000 loops, best of 3: 60.6 µs per loop

但是,似乎有一些与调用它相关的开销可能会因为小输入而变慢:

In [2]: a = np.arange(100)

In [3]: %timeit np.sqrt(a.dot(a))
100000 loops, best of 3: 3.73 µs per loop

In [4]: %timeit np.sqrt(np.einsum('i,i', a, a))
100000 loops, best of 3: 4.68 µs per loop

Answer 4

我发现最快的方法是通过inner1d.以下是它与其他numpy方法的比较:

import numpy as np
from numpy.core.umath_tests import inner1d

V = np.random.random_sample((10**6,3,)) # 1 million vectors
A = np.sqrt(np.einsum('...i,...i', V, V))
B = np.linalg.norm(V,axis=1)   
C = np.sqrt((V ** 2).sum(-1))
D = np.sqrt((V*V).sum(axis=1))
E = np.sqrt(inner1d(V,V))

print [np.allclose(E,x) for x in [A,B,C,D]] # [True, True, True, True]

import cProfile
cProfile.run("np.sqrt(np.einsum('...i,...i', V, V))") # 3 function calls in 0.013 seconds
cProfile.run('np.linalg.norm(V,axis=1)')              # 9 function calls in 0.029 seconds
cProfile.run('np.sqrt((V ** 2).sum(-1))')             # 5 function calls in 0.028 seconds
cProfile.run('np.sqrt((V*V).sum(axis=1))')            # 5 function calls in 0.027 seconds
cProfile.run('np.sqrt(inner1d(V,V))')                 # 2 function calls in 0.009 seconds

inner1d比linalg.norm快3倍,头发比einsum快