将文件尽可能方便地组合在一起

Question

将文件尽可能方便地组合起来

假设我有以下markdown文件

    1.md # contains 'foo'
    2.md # contains 'bar'
    3.md # 'zoo'
    4.md # 'zxc'

使用命令cat

很容易合并它们

$ cat {1..4}.md > merged_5.md

尽管如此，Python需要多个步骤才能实现此结果。

1. 创建读写方法

   def read(filename):
       with open(filename) as file:
           content = file.read()
           return content
   def write(filename, content):
       with open(filename, 'w') as file:
           file.write(content)
   

2. 检索合格文件

   import glob
   filenames = glob.glob('*.md')
   In [17]: filenames
   Out[17]: ['1.md', '2.md', '3.md', '4.md']
   

3. 阅读并组合

   def combine(filenames):
       merged_conent = ""
       for filename in filenames:
           content = read(filename)
           merged_content += content
       write('merged.md', merged_content)
   

4. 将数据和方法封装在main模块中，并保存为'combine_files.py'

   def main():
       filenames = glob.glob('*.md')
       combine(filenames)
   if __name__ == '__main__':
       main()
   

5. 在命令行上运行

   python3 combine_files.py

它不方便，因为命令'cat'

如何重构代码尽可能方便？

Answer 1

怎么样？：

with open('merged.md', 'w') as out_f:
    for filename in glob.glob('*.md'):
        with open(filename) as f:
            out_f.write(f.read())

Answer 2

易于轻松：

def cat(out, *src):
    '''Concatenate files'''
    with open(out, 'wb') as f:
        data = b'n'.join(open(i, 'rb').read() for i in src)
        f.write(data)

您现在可以使用cat('merged.md', glob.glob('*.md'))调用它。那方便怎么办？当然比GNU Coreutils的来源容易得多。