有没有办法提高cython代码的速度

问题描述 投票:0回答:2

我正在寻找加速python numpy代码:

def fun_np(m,data):
a, b, c = data[:,0], data[:,1], data[:,2] 

M = len(data[:,0]) 
n = round((m+1)*(m+2)*(m+3)/6) 
u =np.zeros((M,n))

C = 0
for i in range(0,m+1):
    for j in range(0,i+1):
        for k in range(0,j+1):
            if ((i-j)!=0):
                u[:,C] = (j-k)*(a)**(i-j)*(b)**(j-k-1)*(c)**k

        C=C+1  
return u

相应的cython代码如下:

%%cython 
import numpy as np
cimport numpy as np
from cython import wraparound, boundscheck, nonecheck

@boundscheck(False)
@wraparound(False)
@nonecheck(False)

cpdef fun_cyt(int m,np.ndarray[np.float64_t, ndim=2] data):

cdef:
    np.ndarray[np.float64_t, ndim=1] a = data[:,0]
    np.ndarray[np.float64_t, ndim=1] b = data[:,1]
    np.ndarray[np.float64_t, ndim=1] c = data[:,2]
    int M, n
    Py_ssize_t i, j, k, s
M = len(data[:,0]) 
n = round((m+1)*(m+2)*(m+3)/6)   
cdef np.ndarray[np.float64_t, ndim=2]  u = np.zeros((M,n), dtype=np.float64)

cdef int C = 0
for i in range(m+1): #range(0,m+1):
    for j in range(i+1):
        for k in range(j+1):
            for s in range(M):
                if (i-j)!=0:
                    u[s,C] = (j-k)*(a[s])**(i-j)*(b[s])**(j-k-1)*(c[s])**k

            C=C+1
return u

这是时间

z = np.random.randn(6000, 3); m=20;

%timeit fun_np(m,z);

结果:每循环1.97 s±11.2 ms(平均值±标准偏差,7次运行,每次1次循环)

%timeit fun_cyt(m,z);

结果:每循环1.91 s±12.7 ms(平均值±标准偏差,7次运行,每次1次循环)

正如您所看到的,numpy和cython代码之间没有显着的速度。如果你能帮助优化cython代码,我将不胜感激。

注释的cython代码html的html

python performance numpy optimization cython
2个回答
1
投票

正如评论中已经提到的,你可以尝试使用numba。我建议进一步并行化循环:

from numba import prange, jit

@jit(nopython=True, parallel=True)
def fun_numba(m,data):
    a, b, c = data[:,0], data[:,1], data[:,2] 

    M = len(data[:,0]) 
    n = round((m+1)*(m+2)*(m+3)/6) 
    u = np.zeros((M,n))

    C = 0
    for i in range(0,m+1):
        for j in range(0,i+1):
            for k in prange(0,j+1):
                if ((i-j)!=0):
                    u[:,C] = (j-k)*(a)**(i-j)*(b)**(j-k-1)*(c)**k

            C=C+1  
    return u

在我的机器上给我:

In [11]: %timeit fun_np(m,z)                                                                         
642 ms ± 4.13 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [12]: %timeit fun_numba(m,z)                                                                      
101 ms ± 7.15 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

1
投票

非常有趣的例子!大多数操作都在6000个元素向量上。当涉及到大矢量功率,乘法和加法时,Cython实际上不能比numpy更快。通过在Cython中实现这一点,你可能会像numpy一样快,甚至可以通过消除numpy的一些开销来获得10%到20%。

但是,还有其他方法可以加快计算速度。向量操作是对数据向量的三列进行操作,然后写入输出向量的列。默认情况下,numpy数组具有Row-major排序,即在内存中行在内存中是连续的。对于这里完成的操作,这很糟糕。进一步阅读:https://en.wikipedia.org/wiki/Row-_and_column-major_order

这两个函数基本相同,如果输出向量的创建发生在函数之外,它们将是相同的。

请注意以下内容:我将u [:,C] = ...替换为u [:,C] + =,因为否则结果仅由k = j定义,因此始终为0.我不知道是什么意思这些计算是,但可能不是这样。

import numpy as np
def fun_np(m,data):
    a, b, c = data[:,0], data[:,1], data[:,2] 

    M = len(data[:,0]) 
    n = round((m+1)*(m+2)*(m+3)/6) 
    u = np.zeros((M,n))

    C = 0
    for i in range(0,m+1):
        for j in range(0,i+1):
            for k in range(0,j+1):
                if ((i-j)!=0):
                    u[:,C] += (j-k)*(a)**(i-j)*(b)**(j-k-1)*(c)**k

            C=C+1  
    return u

def fun_npF(m,data):
    a, b, c = data[:,0], data[:,1], data[:,2] 

    M = len(data[:,0]) 
    n = round((m+1)*(m+2)*(m+3)/6) 
    u = np.zeros((M,n),order='F')

    C = 0
    for i in range(0,m+1):
        for j in range(0,i+1):
            for k in range(0,j+1):
                if ((i-j)!=0):
                    u[:,C] += (j-k)*(a)**(i-j)*(b)**(j-k-1)*(c)**k

            C=C+1  
    return u

z = np.random.randn(6000, 3); m=20;
print("Numpy Row-major")
%timeit fun_np(m,z);

# Fortran order, because vector operations on columns
print("Numpy Column-major")
zF = np.asarray(z.copy(),order='F')
%timeit fun_npT(m,zF);

# Check if output the same
diff = (max(np.ravel(abs(fun_np(m,z)-fun_npF(m,zF)))))
max_rm = (max(np.ravel(abs(fun_np(m,z)))))
max_cm = (max(np.ravel(abs(fun_npF(m,zF)))))
print("Dffference: %f, Max value Row-major: %f, Max value Column-major: %f"%(diff, max_rm, max_cm))

这给了我

Numpy Row-major
1.64 s ± 12.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Numpy Column-major
16 ms ± 345 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Dffference: 0.000000, Max value Row-major: 196526643123.792450, Max value Column-major: 196526643123.792450

在考虑将“if”放在哪里并将其与Cython结合时,你可以获得更多,但我只能估计10%到20%。

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