Question

我有一个名为myscript.py的python程序，它会为我提供所提供路径中的文件和文件夹列表。

import os
import sys

def get_files_in_directory(path):
    for root, dirs, files in os.walk(path):
        print(root)
        print(dirs)
        print(files)
path=sys.argv[1]
get_files_in_directory(path)

我提供的路径是D:\Python\TEST，其中有一些文件夹和子文件夹，您可以在下面提供的输出中看到：

C:\Python34>python myscript.py "D:\Python\Test"
D:\Python\Test
['D1', 'D2']
[]
D:\Python\Test\D1
['SD1', 'SD2', 'SD3']
[]
D:\Python\Test\D1\SD1
[]
['f1.bat', 'f2.bat', 'f3.bat']
D:\Python\Test\D1\SD2
[]
['f1.bat']
D:\Python\Test\D1\SD3
[]
['f1.bat', 'f2.bat']
D:\Python\Test\D2
['SD1', 'SD2']
[]
D:\Python\Test\D2\SD1
[]
['f1.bat', 'f2.bat']
D:\Python\Test\D2\SD2
[]
['f1.bat']

我需要以这种方式获得输出：

D1-SD1-f1.bat
D1-SD1-f2.bat
D1-SD1-f3.bat
D1-SD2-f1.bat
D1-SD3-f1.bat
D1-SD3-f2.bat
D2-SD1-f1.bat
D2-SD1-f2.bat
D2-SD2-f1.bat

如何以这种方式获得输出。（请记住，这里的目录结构只是一个例子。程序应该对任何路径都是灵活的）。我该怎么做呢。是否有任何os命令。你能帮我解决一下吗？（附加信息：我正在使用Python3.4）

Answer 1

您可以尝试使用glob模块：

import glob
glob.glob('D:\Python\Test\D1\*\*\*.bat')

或者，只是获取文件名

import os
import glob
[os.path.basename(x) for x in glob.glob('D:\Python\Test\D1\*\*\*.bat')]

Answer 2

要获得所需内容，您可以执行以下操作：

def get_files_in_directory(path):
    # Get the root dir (in your case: test)
    rootDir = path.split('\\')[-1]

    # Walk through all subfolder/files
    for root, subfolder, fileList in os.walk(path):
        for file in fileList:
            # Skip empty dirs
            if file != '':
                # Get the full path of the file
                fullPath = os.path.join(root,file)

                # Split the path and the file (May do this one and the step above in one go
                path, file = os.path.split(fullPath)

                # For each subfolder in the path (in REVERSE order)
                subfolders = []
                for subfolder in path.split('\\')[::-1]:

                    # As long as it isn't the root dir, append it to the subfolders list
                    if subfolder == rootDir:
                        break
                    subfolders.append(subfolder)

                # Print the list of subfolders (joined by '-')
                # + '-' + file
                print('{}-{}'.format( '-'.join(subfolders), file) )

path=sys.argv[1]
get_files_in_directory(path)

我的测试文件夹：

SD1-D1-f1.bat
SD1-D1-f2.bat
SD2-D1-f1.bat
SD3-D1-f1.bat
SD3-D1-f2.bat

这可能不是最好的方法，但它会让你得到你想要的东西。

如何在python

2 个答案: