我想从根目录导航到所有其他目录,并打印相同的内容。

这是我的代码:

#!/usr/bin/python

import os
import fnmatch

for root, dir, files in os.walk("."):
        print root
        print ""
        for items in fnmatch.filter(files, "*"):
                print "..." + items
        print ""

这是我的O/P:

.

...Python_Notes
...pypy.py
...pypy.py.save
...classdemo.py
....goutputstream-J9ZUXW
...latest.py
...pack.py
...classdemo.pyc
...Python_Notes~
...module-demo.py
...filetype.py

./packagedemo

...classdemo.py
...__init__.pyc
...__init__.py
...classdemo.pyc

以上,。和。/packagedemo是目录。

不过,我需要以下列方式列印订单:

A
---a.txt
---b.txt
---B
------c.out

上面,A和B是目录,其余是文件。


当前回答

import os

os.chdir('/your/working/path/')
dir = os.getcwd()
list = sorted(os.listdir(dir))
marks = ""

for s_list in list:
    print marks + s_list
    marks += "---"
    tree_list = sorted(os.listdir(dir + "/" + s_list))
    for i in tree_list:
        print marks + i

其他回答

这是对文件夹名称的处理:

def printFolderName(init_indent, rootFolder):
    fname = rootFolder.split(os.sep)[-1]
    root_levels = rootFolder.count(os.sep)
    # os.walk treats dirs breadth-first, but files depth-first (go figure)
    for root, dirs, files in os.walk(rootFolder):
        # print the directories below the root
        levels = root.count(os.sep) - root_levels
        indent = ' '*(levels*2)
        print init_indent + indent + root.split(os.sep)[-1]

你也可以使用pathlib.Path()递归地遍历一个文件夹并列出它的所有内容

from pathlib import Path


def check_out_path(target_path, level=0):
    """"
    This function recursively prints all contents of a pathlib.Path object
    """
    def print_indented(folder, level):
        print('\t' * level + folder)

    print_indented(target_path.name, level)
    for file in target_path.iterdir():
        if file.is_dir():
            check_out_path(file, level+1)
        else:
            print_indented(file.name, level+1)


my_path = Path(r'C:\example folder')
check_out_path(my_path)

输出:

example folder
    folder
        textfile3.txt
    textfile1.txt
    textfile2.txt

在os包中有更适合的函数。但是如果你必须使用操作系统。走吧,这是我想到的

def walkdir(dirname):
    for cur, _dirs, files in os.walk(dirname):
        pref = ''
        head, tail = os.path.split(cur)
        while head:
            pref += '---'
            head, _tail = os.path.split(head)
        print(pref+tail)
        for f in files:
            print(pref+'---'+f)

输出:

>>> walkdir('.')
.
---file3
---file2
---my.py
---file1
---A
------file2
------file1
---B
------file3
------file2
------file4
------file1
---__pycache__
------my.cpython-33.pyc

这是最好的办法吗

import os

def traverse_dir_recur(directory):
    l = os.listdir(directory)
    for d in l:
        if os.path.isdir(directory + d):
            traverse_dir_recur(directory +  d +"/")
        else:
            print(directory + d)

递归遍历一个目录,你从当前目录的所有dirs中获得所有文件,你从当前目录中获得所有dirs -因为上面的代码没有一个简单性(恕我无礼):

for root, dirs, files in os.walk(rootFolderPath):
    for filename in files:
        doSomethingWithFile(os.path.join(root, filename))
    for dirname in dirs:
        doSomewthingWithDir(os.path.join(root, dirname))