为什么代码加速不能与Cython一起使用?

问题描述 投票:1回答:2

我需要将此代码加速到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

我决定将Cython用于此任务。但是我没有任何加速。

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

我达到的最大速度为80毫秒,但我需要更快。由于您需要几乎实时处理铁的数据告诉我原因。并且达到期望的结果是否现实。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))
python numpy cython bit
2个回答
1
投票

我认为主要原因可能是因为您的python代码几乎没有python操作,而所有操作都是numpy操作。大部分的numpy代码用C编写。其中一些用Fortran编写。其中很多是用Python编写的。编写良好的numpy代码的速度与C代码相当。


0
投票

我对Cython的经验不足,所以这仅仅是一个示例,也可以使用Cython进行计时。

示例

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

Timings

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)
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