首先我知道有很多类似主题的问题,但是经过一天的搜索、阅读和测试,我还是找不到答案。在
我有一个python函数,它计算numpy-ndarray(mxn)的成对相关性。我最初只是用numpy来做这个,但是这个函数也计算了倒数对(也就是说,除了计算矩阵的A行和B行之间的相关性,它还计算了B行和A行之间的相关性),所以我采用了一种稍微不同的方法,对于大m的矩阵(实际大小为我的问题是m~8000)。在
这是伟大的,但仍然有点慢,因为将有许多这样的矩阵,并且要完成所有这些将需要很长的时间。所以我开始调查cython作为一种加快速度的方法。我从我读到的书中了解到,cython不会真正加速numpy。这是真的,还是我遗漏了什么?在
我认为下面的瓶颈是np.sqrt
、np.dot
、对ndarray的.T
方法的调用和np.absolute
。我见过有人用sqrt
中的libc.math
来代替np.sqrt公司,所以我想我的第一个问题是,libc.math
中其他方法的类似函数是否可以使用?恐怕我完全不熟悉C/C++和C语言或任何C族语言,所以这个类型和Cython业务对我来说是一个新的领域,如果原因/解决方案显而易见的话,道歉。在
如果做不到这一点,我能做些什么来提高性能吗?在
下面是我的pyx代码、设置代码和对pyx函数的调用。我不知道它是否重要,但当我调用python setup build_ext --inplace
时,它起作用了,但是有很多警告我并不真正理解。这是否也是我没有看到速度提高的原因吗?在
非常感谢任何帮助,并为超长的帖子感到抱歉。在
设置.py
from distutils.core import setup
from distutils.extension import Extension
import numpy
from Cython.Distutils import build_ext
setup(
cmdclass = {'build_ext': build_ext},
ext_modules = [Extension("calcBrownCombinedP",
["calcBrownCombinedP.pyx"],
include_dirs=[numpy.get_include()])]
)
以及设置的输出:
^{pr2}$pyx代码-'calcBrownCombinedP.pyx'
import numpy as np
cimport numpy as np
from scipy import stats
DTYPE = np.int
ctypedef np.int_t DTYPE_t
def calcBrownCombinedP(np.ndarray genotypeArray):
cdef int nSNPs, i
cdef np.ndarray ms, datam, datass, d, rs, temp
cdef float runningSum, sigmaSq, E, df
nSNPs = genotypeArray.shape[0]
ms = genotypeArray.mean(axis=1)[(slice(None,None,None),None)]
datam = genotypeArray - ms
datass = np.sqrt(stats.ss(datam,axis=1))
runningSum = 0
for i in xrange(nSNPs):
temp = np.dot(datam[i:],datam[i].T)
d = (datass[i:]*datass[i])
rs = temp / d
rs = np.absolute(rs)[1:]
runningSum += sum(rs*(3.25+(0.75*rs)))
sigmaSq = 4*nSNPs+2*runningSum
E = 2*nSNPs
df = (2*(E*E))/sigmaSq
runningSum = sigmaSq/(2*E)
return runningSum
针对纯python测试上述内容的代码-'测试.py'
import numpy as np
from scipy import stats
import random
import time
from calcBrownCombinedP import calcBrownCombinedP
from PycalcBrownCombinedP import PycalcBrownCombinedP
ms = [10,50,100,500,1000,5000]
for m in ms:
print '---testing implentation with m = {0}---'.format(m)
genotypeArray = np.empty((m,20),dtype=int)
for i in xrange(m):
genotypeArray[i] = [random.randint(0,2) for j in xrange(20)]
print genotypeArray.shape
start = time.time()
print calcBrownCombinedP(genotypeArray)
print 'cython implementation took {0}'.format(time.time() - start)
start = time.time()
print PycalcBrownCombinedP(genotypeArray)
print 'python implementation took {0}'.format(time.time() - start)
代码的输出是:
---testing implentation with m = 10---
(10L, 20L)
2.13660168648
cython implementation took 0.000999927520752
2.13660167749
python implementation took 0.000999927520752
---testing implentation with m = 50---
(50L, 20L)
8.82721138
cython implementation took 0.00399994850159
8.82721130234
python implementation took 0.00500011444092
---testing implentation with m = 100---
(100L, 20L)
16.7438983917
cython implementation took 0.0139999389648
16.7438965333
python implementation took 0.0120000839233
---testing implentation with m = 500---
(500L, 20L)
80.5343856812
cython implementation took 0.183000087738
80.5343694046
python implementation took 0.161000013351
---testing implentation with m = 1000---
(1000L, 20L)
160.122573853
cython implementation took 0.615000009537
160.122491308
python implementation took 0.598000049591
---testing implentation with m = 5000---
(5000L, 20L)
799.813842773
cython implementation took 10.7159998417
799.813880445
python implementation took 11.2510001659
最后,纯python实现'PycalCbDownCombinedP.py'
import numpy as np
from scipy import stats
def PycalcBrownCombinedP(genotypeArray):
nSNPs = genotypeArray.shape[0]
ms = genotypeArray.mean(axis=1)[(slice(None,None,None),None)]
datam = genotypeArray - ms
datass = np.sqrt(stats.ss(datam,axis=1))
runningSum = 0
for i in xrange(nSNPs):
temp = np.dot(datam[i:],datam[i].T)
d = (datass[i:]*datass[i])
rs = temp / d
rs = np.absolute(rs)[1:]
runningSum += sum(rs*(3.25+(0.75*rs)))
sigmaSq = 4*nSNPs+2*runningSum
E = 2*nSNPs
df = (2*(E*E))/sigmaSq
runningSum = sigmaSq/(2*E)
return runningSum
使用^{} 进行分析显示瓶颈是循环的最后一行:
这并不奇怪,因为在Python和Cython版本中都使用了Python内置函数
sum
。当输入数组的形状为(5000, 20)
时,切换到np.sum
会将代码速度提高4.5倍。在如果精度有一点损失是可以的,那么您可以利用线性代数进一步加快最后一行的速度:
^{pr2}$实际上是向量点积,即
这仍然是次优的,因为它在
rs
上循环了三次,并构造了两个rs
大小的临时数组。使用初等代数,这个表达式可以重写为它不仅给出了原始结果而没有上一个版本中的舍入错误,而且只在
rs
上循环两次并使用常量内存。(*)现在的瓶颈是
np.dot
,所以安装一个更好的BLAS库比用Cython重写整个程序要多。在(*)或最新NumPy中的对数内存,它有一个递归的
np.sum
的重新实现,比旧的迭代实现更快。在相关问题 更多 >
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