numpy.cross()表现不佳

Question

我一直在做一些性能测试,以提高我正在写的宠物项目的性能.这是一个非常数字化的密集型应用程序,因此我一直在使用Numpy作为提高计算性能的方法.

但是,以下性能测试的结果非常令人惊讶....

测试源代码 (更新了提升和批量提交的测试用例)

import timeit

numpySetup = """
import numpy
left = numpy.array([1.0,0.0,0.0])
right = numpy.array([0.0,1.0,0.0])
"""

hoistSetup = numpySetup +'hoist = numpy.cross\n'

pythonSetup = """
left = [1.0,0.0,0.0]
right = [0.0,1.0,0.0]
"""

numpyBatchSetup = """
import numpy

l = numpy.array([1.0,0.0,0.0])
left = numpy.array([l]*10000)

r = numpy.array([0.0,1.0,0.0])
right = numpy.array([r]*10000)
"""

pythonCrossCode = """
x = ((left[1] * right[2]) - (left[2] * right[1]))
y = ((left[2] * right[0]) - (left[0] * right[2]))
z = ((left[0] * right[1]) - (left[1] * right[0]))
"""

pythonCross = timeit.Timer(pythonCrossCode, pythonSetup)
numpyCross = timeit.Timer ('numpy.cross(left, right)' , numpySetup)
hybridCross = timeit.Timer(pythonCrossCode, numpySetup)
hoistCross = timeit.Timer('hoist(left, right)', hoistSetup)
batchCross = timeit.Timer('numpy.cross(left, right)', numpyBatchSetup) 

print 'Python Cross Product : %4.6f ' % pythonCross.timeit(1000000)
print 'Numpy Cross Product  : %4.6f ' % numpyCross.timeit(1000000) 
print 'Hybrid Cross Product : %4.6f ' % hybridCross.timeit(1000000) 
print 'Hoist Cross Product  : %4.6f ' % hoistCross.timeit(1000000) 
# 100 batches of 10000 each is equivalent to 1000000
print 'Batch Cross Product  : %4.6f ' % batchCross.timeit(100) 

原始结果

Python Cross Product : 0.754945 
Numpy Cross Product  : 20.752983 
Hybrid Cross Product : 4.467417 

最终结果

Python Cross Product : 0.894334 
Numpy Cross Product  : 21.099040 
Hybrid Cross Product : 4.467194 
Hoist Cross Product  : 20.896225 
Batch Cross Product  : 0.262964 

不用说,这不是我预期的结果.纯Python版本的执行速度比Numpy快30倍.其他测试中的Numpy性能优于Python等效(这是预期的结果).

所以,我有两个相关的问题:

• 任何人都能解释为什么NumPy在这种情况下表现如此糟糕吗？
• 有什么我可以做的来解决它吗？

Answer 1

尝试使用更大的数组.我认为只是调用numpy这里的方法的成本超过了Python版本所需的简单几个列表访问.如果你处理更大的阵列,我认为你会看到大的胜利numpy.

Answer 2

您可以在此处自行查看源代码:http://www.google.com/codesearch/p？hl = zh_CN&5mAq98l -MUw/turunk / dunumpy/numpy/core / nuumeric.py&q = cros%20package:numpy& sa = N&rd = 1&CT = RC

numpy.cross只处理大量案例并做一些额外的副本.

一般来说,numpy对于像矩阵乘法或反演这样的慢速事件来说足够快 - 但对像这样的小向量的操作有很多开销.