对于基本求和计算,Cython并不比Python快得多

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我试图按照 Continuum Analytics blog基准测试Python,Cython,Numba上给出的一个例子来计算使用for循环计算的总和.不幸的是,我发现Cython比Python慢​​!

这是我的Python函数定义:

@H_502_4@def python_sum(y): N = len(y) x = y[0] for i in xrange(1,N): x += y[i] return x

现在我的Cython功能

@H_502_4@def cython_sum(int[:] y): cdef int N = y.shape[0] cdef int x = y[0] cdef int i for i in xrange(1,N): x += y[i] return x

现在我有一个脚本可以提取两个函数和基准:

@H_502_4@import timeit import numpy as np import cython_sum import python_sum b = np.ones(10000) timer = timeit.Timer(stmt='python_sum.python_sum(b)',setup='from __main__ import python_sum,b') print "Python Sum (ms): %g" % (timer.timeit(1)*1000) timer = timeit.Timer(stmt='cython_sum.cython_sum(b)',setup='from __main__ import cython_sum,b') print "Cython (ms): %g" % (timer.timeit(1)*1000)

现在我的输出是:

@H_502_4@Python Sum (ms): 9.44624 Cython (ms): 8.54868

基于上面链接博客文章中的图表,我期待速度提高100倍-1000倍,但我所看到的只是Cython比vanilla Python略快.

我在这里做错了吗?这似乎是一个非常基本的问题,具有简单的函数定义,很明显很多人使用Cython取得了巨大的成功,所以很明显错误必须在于我.任何人都可以对此有所了解并告诉我我做错了什么?
谢谢!

解决方法

我不确定你为什么会得到那个结果.正如一位评论者所说,你的代码原样不应该工作,因为你要将浮点数传递给一个期望整数的函数.也许你在同一目录中留下了一个cython_sum.py文件

我做了以下.我创建了一个python_sum.py,其中包含您对python_sum的确切定义.然后我稍微修改了你的Cython代码

cython_sum.pyx:

@H_502_4@def cython_sum(long[:] y): #changed `int` to `long` cdef int N = y.shape[0] cdef int x = y[0] cdef int i for i in xrange(1,N): x += y[i] return x

我创建了一个安装文件来构建Cython模块:

setup.py:

@H_502_4@from distutils.core import setup from Cython.Build import cythonize setup( name = 'Cython sum test',ext_modules = cythonize("cython_sum.pyx"),)

我使用python setup.py build_ext –inplace构建了模块.接下来,我对您的测试代码进行了一些修改

test.py:

@H_502_4@import timeit import numpy as np import cython_sum import python_sum # ** added dtype=np.int to create integers ** b = np.ones(10000,dtype=np.int) # ** changed .timeit(1) to .timeit(1000) for each one ** timer = timeit.Timer(stmt='python_sum.python_sum(b)',b') print "Python Sum (ms): %g" % (timer.timeit(1000)*1000) timer = timeit.Timer(stmt='cython_sum.cython_sum(b)',b') print "Cython (ms): %g" % (timer.timeit(1000)*1000)

我得到了以下结果:

@H_502_4@$python test.py Python Sum (ms): 4111.74 Cython (ms): 7.06697

现在这是一个很好的加速!

此外,按照here所述的指导原则,我可以获得额外的(小)加速:

cython_fast_sum.pyx:

@H_502_4@import numpy as np cimport numpy as np DTYPE = np.int ctypedef np.int_t DTYPE_t def cython_sum(np.ndarray[DTYPE_t,ndim=1] y): cdef int N = y.shape[0] cdef int x = y[0] cdef int i for i in xrange(1,N): x += y[i] return x

setup_fast.py:

@H_502_4@from distutils.core import setup from Cython.Build import cythonize import numpy as np setup( name = 'Cython fast sum test',ext_modules = cythonize("cython_fast_sum.pyx"),include_dirs = [np.get_include()],)

test.py:

@H_502_4@import timeit import numpy as np import cython_sum import cython_fast_sum b = np.ones(10000,dtype=np.int) # ** note 100000 runs,not 1000 ** timer = timeit.Timer(stmt='cython_sum.cython_sum(b)',b') print "Cython naive (ms): %g" % (timer.timeit(100000)*1000) timer = timeit.Timer(stmt='cython_fast_sum.cython_sum(b)',setup='from __main__ import cython_fast_sum,b') print "Cython fast (ms): %g" % (timer.timeit(100000)*1000)

结果:

@H_502_4@$python test.py Cython naive (ms): 676.437 Cython fast (ms): 645.797

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