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我正在尝试将YCbCr 文件 从 8 bpp 转换为 10 bpp。

到目前为止，我最好的方法仍然比最基本的幼稚 C 实现慢一个数量级。

C 中的幼稚方法，运行时间约为 8 秒。使代码改为在块上工作，将时间降低到 1 秒以下。

我很好奇从标准 python 处理二进制文件可以获得什么样的性能。示例文件采用CIF 分辨率，与 1080p 的内容相比“小”。尽管我主要对标准 python 感兴趣，但也可以随意添加 numpy-suggestions。

测试文件可以从

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

sha1sum对于正确的 10 位输出是

c511dabc793383f7fd0ed69b4bb9b9f89ef73b84

Python： 

#!/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())

天真的C-dito： 

#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;
}
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1 回答 1

4

问题中的代码25在我的机器上需要几秒钟，numpy--0.37几秒钟：

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

cython--0.20秒：

from functools import partial

import pyximport; pyximport.install() # pip install cython
from bpp8to10 import convert # bpp8to10.pyx

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

def main():
    with open(f_in, 'rb') as fd_in, open(f_out, 'wb') as fd_out:
        for chunk in iter(partial(fd_in.read, 8192), b''):
            fd_out.write(convert(chunk))
main()

哪里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别移动到函数 ( 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()

pypy--1.6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))
于 2013-04-10T04:09:16.303 回答