python处理二进制文件的性能

2024-09-26 22:53:38 发布

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我正在尝试转换YCbCr-file 从8个基点到10个基点。在

到目前为止,我最好的方法还是慢一个数量级 而不是最基本的朴素C实现。在

C语言中的朴素方法,大约在8秒内运行 改为在大块上下功夫,把时间缩短到1秒以下

我很好奇它能得到什么样的表演 来自处理二进制文件的标准python。示例文件是 在CIF-resolution中,与1080p中的内容相比“小”。 尽管我主要感兴趣,也可以随意添加一些新的建议 在标准python中。在

测试文件可以从

http://trace.eas.asu.edu/yuv/foreman/foreman_cif.7z

正确的10位输出的sha1sum

^{pr2}$

Python: 在

#!/usr/bin/env python

import array

f_in = 'foreman_cif.yuv'
f_out = 'py_10bpp.yuv'

def bytesfromfile(f):
    while True:
        raw = array.array('B')
        raw.fromstring(f.read(8192))
        if not raw:
            break
        yield raw

with open(f_in, 'rb') as fd_in, \
        open(f_out, 'wb') as fd_out:

    for byte in bytesfromfile(fd_in):
        data = []
        for i in byte:
            i <<= 2
            data.append(i & 0xff)
            data.append((i >> 8) & 0xff)

        fd_out.write(array.array('B', data).tostring())

朴素的C-dito: 在

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char** argv)
{
    int c;
    int d[2];

    FILE* fd_in;
    FILE* fd_out;

    fd_in = fopen("foreman_cif.yuv", "rb");
    fd_out = fopen("c_10bpp.yuv", "wb");

    while((c = fgetc(fd_in)) != EOF) {
        c <<= 2;
        d[0] = c & 0xff;
        d[1] = (c >> 8) & 0xff;

        fwrite(&d[0], 1, 1, fd_out);
        fwrite(&d[1], 1, 1, fd_out);
    }

    fclose(fd_in);
    fclose(fd_out);

    return EXIT_SUCCESS;
}

Tags: 文件方法indata标准rawforemanout
1条回答
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1楼 · 发布于 2024-09-26 22:53:38

问题中的代码在我的机器上需要25秒,numpy0.37秒:

import numpy as np

a_in = np.memmap('foreman_cif.yuv', mode='readonly')
a_out = np.memmap('py_10bpp.yuv', mode='write', shape=2*len(a_in))
a_out[::2] = a_in << 2
a_out[1::2] = a_in >> 6

cython0.20秒:

^{pr2}$

其中bpp8to10.pyx

from cpython.bytes cimport PyBytes_FromStringAndSize

def convert(bytes chunk not None):
    cdef:
        bytes data = PyBytes_FromStringAndSize(NULL, len(chunk)*2)
        char* buf = data # no copy
        Py_ssize_t j = 0
        unsigned char c
    for c in chunk:
        buf[j] = (c << 2) 
        buf[j + 1] = (c >> 6)
        j += 2
    return data

纯CPython版本的主要加速是将代码从模块级移动到函数(main()6.7秒(2个cpu):

from functools import partial
from multiprocessing import Pool

f_in = 'foreman_cif.yuv'
f_out = 'py_10bpp.yuv'

def convert(chunk):
    data = bytearray() # [] -> bytearray(): 17 -> 15 seconds
    data_append = data.append # 15 -> 12  seconds
    for b in bytearray(chunk): # on Python 3: `for b in chunk:`
        data_append((b << 2) & 0xff)
        data_append((b >> 8) & 0xff)
    return data

def main(): # put in main(): # 25 -> 17 seconds
    pool = Pool(processes=2) # 12 -> 6.7 seconds
    with open(f_in, 'rb') as fd_in, open(f_out, 'wb') as fd_out:
        for data in pool.imap(convert, iter(partial(fd_in.read, 8192), b'')):
            fd_out.write(data)
main()

pypy1.6秒:

f_in = 'foreman_cif.yuv'
f_out = 'py_10bpp.yuv'

def convert(chunk):
    data = bytearray() # 1.6 -> 1.5 seconds for preallocated data
    for b in bytearray(chunk): 
        data.append((b << 2) & 0xff)
        data.append((b >> 6) & 0xff)
    return data

with open(f_in, 'rb') as fd_in, open(f_out, 'wb') as fd_out:
    while True:
        chunk = fd_in.read(8192)
        if not chunk:
            break
        fd_out.write(convert(chunk))

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