python脚本将目录中的所有文件合并为一个文件 [英] python script to concatenate all the files in the directory into one file

问题描述

我编写了以下脚本，将目录中的所有文件连接为一个文件。

I have written the following script to concatenate all the files in the directory into one single file.

可以根据

1. 惯用python进行优化吗？

1. idiomatic python

时间

以下是代码段：

import time, glob

outfilename = 'all_' + str((int(time.time()))) + ".txt"

filenames = glob.glob('*.txt')

with open(outfilename, 'wb') as outfile:
    for fname in filenames:
        with open(fname, 'r') as readfile:
            infile = readfile.read()
            for line in infile:
                outfile.write(line)
            outfile.write("\n\n")

推荐答案

使用 shutil.copyfileobj 复制数据：

Use shutil.copyfileobj to copy data:

import shutil

with open(outfilename, 'wb') as outfile:
    for filename in glob.glob('*.txt'):
        if filename == outfilename:
            # don't want to copy the output into the output
            continue
        with open(filename, 'rb') as readfile:
            shutil.copyfileobj(readfile, outfile)

shutil 从将 readfile 对象成块，将其直接写入 outfile 文件对象。不要使用 readline（）或迭代缓冲区，因为您不需要查找行尾的开销。

shutil reads from the readfile object in chunks, writing them to the outfile fileobject directly. Do not use readline() or a iteration buffer, since you do not need the overhead of finding line endings.

使用相同的模式进行读写；这在使用Python 3时尤其重要；我在这里都使用了二进制模式。

Use the same mode for both reading and writing; this is especially important when using Python 3; I've used binary mode for both here.

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