python 对文件的扩展名/名称进行排序

Question

我正在为一个学校项目编写脚本:

search_folder = raw_input('Choose Directory')

for root, dirs, files in os.walk(search_folder):
       for file in files:
               pathname = os.path.join(root, file)
               print pathname
               print os.path.getsize(pathname)

如何对扩展上的输出进行排序？

Answer 1

对文件名进行排序

使用sorted()与 os.path.splitext() lambda在 sort-by key 参数中。

示例

import os

search_folder = '/Users/temp/'

results = []
for root, dirs, files in os.walk(search_folder):
       for f in files:
            results.append(os.path.join(root, f))

sorted_results = sorted(results, key=lambda x: os.path.splitext(x)[1])

for path in sorted_results:
    print path
    print os.path.getsize(path)

<小时/>

/Users/temp/.DS_Store
6148
/Users/temp/.localized
0
/Users/temp/profile.png
81168
/Users/temp/IMG_0115.xcf
113212
/Users/temp/profile.xcf
535202

其他注意事项和建议

使用 list comprehension

您还可以使用list comprehension (为了速度和可读性)遍历文件:

results = [os.path.join(root, f) 
               for root, dirs, files in os.walk(search_folder) 
                   for f in files]

内存高效排序

排序而不存储suggestion来自@PM2Ring:

for root, dirs, files in os.walk(search_folder):
    for f in sorted(files, key=lambda x: tuple(reversed(os.path.splitext(x)))):
        pathname = os.path.join(root, f)
        print pathname 
        print os.path.getsize(pathname)