5

看起来显式计算向量数组的叉积比使用np.cross. 我已经尝试过vector-first和vector-last,它似乎没有什么区别,尽管这是在对类似问题的回答中提出的。是我用错了,还是只是慢了点?

在笔记本电脑上,每个交叉产品的显式计算似乎需要大约 60ns。这〜大致〜尽可能快吗?在这种情况下,似乎没有任何理由去 Cython 或 PyPy 或编写一个特别的ufunc

我也看到了使用 einsum 的参考资料,但我并不真正了解如何使用它,并且怀疑它并没有更快。

a = np.random.random(size=300000).reshape(100000,3) # vector last
b = np.random.random(size=300000).reshape(100000,3)
c, d = a.swapaxes(0, 1),  b.swapaxes(0, 1)          # vector first

def npcross_vlast():        return np.cross(a, b)
def npcross_vfirst():       return np.cross(c, d, axisa=0, axisb=0)
def npcross_vfirst_axisc(): return np.cross(c, d, axisa=0, axisb=0, axisc=0)
def explicitcross_vlast():
    e = np.zeros_like(a)
    e[:,0] = a[:,1]*b[:,2] - a[:,2]*b[:,1]
    e[:,1] = a[:,2]*b[:,0] - a[:,0]*b[:,2]
    e[:,2] = a[:,0]*b[:,1] - a[:,1]*b[:,0]
    return e
def explicitcross_vfirst():
    e = np.zeros_like(c)
    e[0,:] = c[1,:]*d[2,:] - c[2,:]*d[1,:]
    e[1,:] = c[2,:]*d[0,:] - c[0,:]*d[2,:]
    e[2,:] = c[0,:]*d[1,:] - c[1,:]*d[0,:]
    return e
print "explicit"
print timeit.timeit(explicitcross_vlast,  number=10)
print timeit.timeit(explicitcross_vfirst, number=10)
print "np.cross"
print timeit.timeit(npcross_vlast,        number=10)
print timeit.timeit(npcross_vfirst,       number=10)
print timeit.timeit(npcross_vfirst_axisc, number=10)
print all([npcross_vlast()[7,i] == npcross_vfirst()[7,i] ==
           npcross_vfirst_axisc()[i,7] == explicitcross_vlast()[7,i] ==
           explicitcross_vfirst()[i,7] for i in range(3)]) # check one

explicit
0.0582590103149
0.0560920238495
np.cross
0.399816989899
0.412983894348
0.411231040955
True
4

3 回答 3

4

在numpy 的发布中性能np.cross显着提升。1.9.x

%timeit explicitcross_vlast()
%timeit explicitcross_vfirst()
%timeit npcross_vlast()
%timeit npcross_vfirst()
%timeit npcross_vfirst_axisc() 

这些是我得到的时间1.8.0

100 loops, best of 3: 4.47 ms per loop
100 loops, best of 3: 4.41 ms per loop
10 loops, best of 3: 29.1 ms per loop
10 loops, best of 3: 29.3 ms per loop
10 loops, best of 3: 30.6 ms per loop

这些时间是1.9.0

100 loops, best of 3: 4.62 ms per loop
100 loops, best of 3: 4.19 ms per loop
100 loops, best of 3: 4.05 ms per loop
100 loops, best of 3: 4.09 ms per loop
100 loops, best of 3: 4.24 ms per loop

我怀疑加速是由合并请求#4338引入的。

于 2015-04-18T10:23:33.990 回答
2

首先,如果您希望加快代码速度,您可能应该尝试完全摆脱交叉产品。这在许多情况下都是可能的,例如,当与点积结合使用时<a x b, c x d> = <a, c><b, d> - <a, d><b, c>

无论如何,如果您真的需要明确的交叉产品,请查看

eijk = np.zeros((3, 3, 3))
eijk[0, 1, 2] = eijk[1, 2, 0] = eijk[2, 0, 1] = 1
eijk[0, 2, 1] = eijk[2, 1, 0] = eijk[1, 0, 2] = -1

np.einsum('ijk,aj,ak->ai', eijk, a, b)
np.einsum('iak,ak->ai', np.einsum('ijk,aj->iak', eijk, a), b)

这两个等效于np.cross,其中第二个使用两个 einsum,每个都有两个参数,这是在类似问题中提出的一种技术。

但是,结果令人失望:这两个变体都比np.cross(除了 tiny n)慢:

在此处输入图像描述

该情节是用

import numpy as np
import perfplot

eijk = np.zeros((3, 3, 3))
eijk[0, 1, 2] = eijk[1, 2, 0] = eijk[2, 0, 1] = 1
eijk[0, 2, 1] = eijk[2, 1, 0] = eijk[1, 0, 2] = -1


b = perfplot.bench(
    setup=lambda n: np.random.rand(2, n, 3),
    n_range=[2 ** k for k in range(23)],
    kernels=[
        lambda X: np.cross(X[0], X[1]),
        lambda X: np.einsum("ijk,aj,ak->ai", eijk, X[0], X[1]),
        lambda X: np.einsum("iak,ak->ai", np.einsum("ijk,aj->iak", eijk, X[0]), X[1]),
    ],
    labels=["np.cross", "einsum", "double einsum"],
    xlabel="len(a)",
)

b.save("out.png")
于 2016-09-23T13:45:43.813 回答
1

只需将您的更改vlast

def stacked_vlast(a,b):
        x = a[:,1]*b[:,2] - a[:,2]*b[:,1]
        y = a[:,2]*b[:,0] - a[:,0]*b[:,2]
        z = a[:,0]*b[:,1] - a[:,1]*b[:,0]
        return np.array([x,y,z]).T

即用堆叠替换列分配,就像(旧)cross一样,将速度减慢5倍。

当我使用开发cross功能的本地副本时,我的速度比您的explicit_vlast. 它cross使用该out参数来尝试减少临时数组,但我的粗略测试表明它对速度没有太大影响。

https://github.com/numpy/numpy/blob/master/numpy/core/numeric.py

如果您的显式版本有效,我不会numpy仅仅为了获得这个新的cross.

于 2015-04-18T15:26:55.053 回答