为什么代码加速不适用于 Cython？

Question

我需要将这段代码加速到 4 毫秒。

import numpy as np



def return_call(data):
    num = int(data.shape[0] / 4096)
    buff_spectrum  = np.empty(2048,dtype= np.uint64)
    buff_detect =  np.empty(2048,dtype= np.uint64)
    end_spetrum = np.empty(num*1024,dtype=np.uint64)
    end_detect = np.empty(num*1024,dtype= np.uint64)
    _data = np.reshape(data,(num,4096))

    for _raw_data_spec in _data:
        raw_data_spec = np.reshape(_raw_data_spec,(2048,2))
        for i in range(2048):
            buff_spectrum[i] = (np.int16(raw_data_spec[i][0])<<17)|(np.int16(raw_data_spec[i][1] <<1))>>1
            buff_detect[i] = (np.int16(raw_data_spec[i][0])>>15)
        for i in range (511,-1,-1):
            if buff_spectrum[i+1024] != 0:
                end_spetrum[i]=(np.log10(buff_spectrum[i+1024]))
                end_detect[i]=buff_detect[i+1024]
            else:
                end_spetrum[i] =0
                end_detect[i] = 0
        for i in range(1023, 511, -1):
            if buff_spectrum[i+1024] != 0:
                end_spetrum[i] = (np.log10(buff_spectrum[i + 1024]))
                end_detect[i] = buff_detect[i + 1024]
            else:
                end_spetrum[i] = 0
                end_detect[i] = 0

    return end_spetrum, end_detect

我决定使用 Cython 来完成这项任务。但是我没有得到任何加速。

import numpy as np
cimport numpy


ctypedef signed short DTYPE_t
cpdef return_call(numpy.ndarray[DTYPE_t, ndim=1] data):
    cdef int i
    cdef int num = data.shape[0]/4096
    cdef numpy.ndarray _data

    cdef numpy.ndarray[unsigned long long, ndim=1] buff_spectrum  = np.empty(2048,dtype= np.uint64)
    cdef numpy.ndarray[ unsigned long long, ndim=1] buff_detect =  np.empty(2048,dtype= np.uint64)
    cdef numpy.ndarray[double , ndim=1] end_spetrum = np.empty(num*1024,dtype= np.double)
    cdef numpy.ndarray[double , ndim=1] end_detect = np.empty(num*1024,dtype= np.double)
    _data = np.reshape(data,(num,4096))

    for _raw_data_spec in _data:
        raw_data_spec = np.reshape(_raw_data_spec,(2048,2))
        for i in range(2048):
            buff_spectrum[i] = (np.uint16(raw_data_spec[i][0])<<17)|(np.uint16(raw_data_spec[i][1] <<1))>>1
            buff_detect[i] = (np.uint16(raw_data_spec[i][0])>>15)
        for i in range (511,-1,-1):
            if buff_spectrum[i+1024] != 0:
                end_spetrum[i]=(np.log10(buff_spectrum[i+1024]))
                end_detect[i]=buff_detect[i+1024]
            else:
                end_spetrum[i] =0
                end_detect[i] = 0
        for i in range(1023, 511, -1):
            if buff_spectrum[i+1024] != 0:
                end_spetrum[i] = (np.log10(buff_spectrum[i + 1024]))
                end_detect[i] = buff_detect[i + 1024]
            else:
                end_spetrum[i] = 0
                end_detect[i] = 0

    return end_spetrum, end_detect

我达到的最大速度是 80 毫秒，但我需要更快。由于您需要几乎实时地处理来自 iron 的数据告诉我原因。并且达到预期的结果是否现实。我还附上了测试文件的代码。


import numpy as np
import example_original
import example_cython
data = np.empty(8192*2, dtype=np.int16)
import time
startpy = time.time()


example_original.return_call(data)
finpy = time.time() -startpy
startcy = time.time()
k,r = example_cython.return_call(data)
fincy = time.time() -startcy
print( fincy, finpy)
print('Cython is {}x faster'.format(finpy/fincy))

Answer 1

我认为一个主要原因可能是因为您的 python 代码几乎没有 python 操作并且所有操作都是 numpy 操作。大部分 numpy 代码是用 C 语言编写的。其中一些是用 Fortran 语言编写的。 Python里面写了很多。编写良好的 numpy 代码在速度上可与 C 代码相媲美。

Answer 2

我对 Cython 的经验不多，所以这只是一个例子，Cython 也应该可以实现什么计时。

例子

import numpy as np
import numba as nb

@nb.njit(cache=True)
def return_call(data_in):
    #If the input is not contigous the reshape will fail
    #-> make a c-contigous copy if the array isn't c-contigous
    data=np.ascontiguousarray(data_in)

    num = int(data.shape[0] / 4096)
    buff_spectrum  = np.zeros(2048,dtype= np.uint64)
    buff_detect =  np.zeros(2048,dtype= np.uint64)
    end_spetrum = np.zeros(num*1024,dtype=np.float64)
    end_detect = np.zeros(num*1024,dtype= np.float64)
    _data = np.reshape(data,(num,4096))

    #for _raw_data_spec in _data: is not supported
    #but the followin works
    for x in range(_data.shape[0]):
        raw_data_spec = np.reshape(_data[x],(2048,2))
        for i in range(2048):
            buff_spectrum[i] = (np.int16(raw_data_spec[i][0])<<17)|(np.int16(raw_data_spec[i][1] <<1))>>1
            buff_detect[i] = (np.int16(raw_data_spec[i][0])>>15)
        for i in range (511,-1,-1):
            if buff_spectrum[i+1024] != 0:
                end_spetrum[i]=(np.log10(buff_spectrum[i+1024]))
                end_detect[i]=buff_detect[i+1024]
            else:
                end_spetrum[i] =0
                end_detect[i] = 0
        for i in range(1023, 511, -1):
            if buff_spectrum[i+1024] != 0:
                end_spetrum[i] = (np.log10(buff_spectrum[i + 1024]))
                end_detect[i] = buff_detect[i + 1024]
            else:
                end_spetrum[i] = 0
                end_detect[i] = 0

    return end_spetrum, end_detect

计时

data = np.random.rand(8192*2)*20
data=data.astype(np.int16)

#with compilation
%timeit end_spetrum, end_detect=return_call(data)
#32.7 µs ± 5.61 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)

#without compilation
%timeit end_spetrum, end_detect=return_call_orig(data)
#106 ms ± 448 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Answer 3

raw_data_spec = np.reshape(_raw_data_spec,(2048,2))

raw_data_spec 未输入。在函数的开头为其添加一个定义。我推荐较新的 memoryview 语法（但如果需要，请使用旧的 numpy 语法）：

cdef DTYPE_t[:,:] raw_data_spec

这一行（您已确定为瓶颈）一团糟：

buff_spectrum[i] = (np.int16(raw_data_spec[i][0])<<17)|(np.int16(raw_data_spec[i][1] <<1))>>1

一步完成索引，而不是两步：raw_data_spec[i, 0]（注意很多括号和逗号）。
重新考虑转换为 16 位整数。将 16 位整数移动 17 位 真的有意义吗？
您可能根本不需要强制转换，因为已知数据是 DTYPE_t，但如果您确实需要强制转换，请使用尖括号：<numpy.uint16_t>(raw_data_spec[i, 0])

考虑关闭 boundscheck 和 wraparound。 向自己确认这样做是安全的，并且当您索引超出数组末尾或使用负索引时，您不会依赖异常来告诉您。只有经过深思熟虑后才这样做 - 不会以 "cargo cult" 的方式自动执行。

cimport cython    

@cython.boundscheck(False)
@cython.wraparound(False)
cpdef return_call(numpy.ndarray[DTYPE_t, ndim=1] data):

放弃对 np.log10 的呼叫。这是对单个元素的整个 Python 调用，最终效率低下。您可以改用 C 标准库数学函数：

from libc.math cimport log10

然后将 np.log10 替换为 log10。

为什么代码加速不适用于 Cython？

Why doesn't code acceleration work with Cython?

python

numpy

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cython