Question

我在分开的章节中创建了python_learning_notes，保存为'.md'文件，如：

data-type.md

data-structure.md

high-level-data-structure.md

code-structure.md

object-oriented-programing.md

我复制了每个文件的内容，并将它们粘贴在一个“.md”文件中。

这项工作以丑陋的7个步骤完成，我想更加方便地重构代码。

更改为目标目录
获取并验证'.md'文件名
将'.md'重命名为'.txt'
将内容从“.txt”读取到列表
写一个'.txt'文件名
将单个“.txt”文件重命名为“.md”
将所有'.txt'恢复为'.md'

这是我的代码：

# 1.change to the destination directory
path_name = ' ~/Documents/'
os.chdir(path_name)

# 2.get and validate the '.md' filename
x = os.dirlist(path_name)
qualified_filenames = [i for i in x if '.md' is in x]

# 3.rename '.md' to '.txt'
new_names = []
for i in qualified_filenames:
    name,extension = os.path.splitext(i)
    extension = '.txt'
    new_name = '{}{}'.format(name, extension)
    os.rename(i, new_name)
    new_names.append(new_name)

# 4.read contents from '.txt' to a list
contents_list = []
for i in new_names:
    with open(i) as fp:
        line = fp.read(i)
        contents_list.append(line)

# 5.write to a single '.txt' filename
with open('single-file.txt','w') as fp:
    for i in contents_list:
        fp.write(i + '\n')

# 6.rename the single '.txt' file to '.md'
os.rename('single-file.txt','single-file.md')

# 7.restore all the '.txt' to '.md'
for i, j in zip(new_names, qualified_filenames):
    os.rename(i, j)

Answer 1

这会有点短：

import glob
import fileinput
import os

pat = os.path.expanduser('~/Documents/*.md')
with open('single-file.md', 'w') as fout:
    for line in sorted(fileinput.input(glob.glob(pat))):
        fout.write(line)

如何以优雅的pythonic方式更有效地组合多个'.md'文件？

1 个答案: