python - Python中使用os.walk()递归遍历目录

Question

我想从根目录导航到其中的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

上面，.和./packagedemo是目录。

但是，我需要按以下方式打印 O/P：

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

上面，A和B是目录，其余的是文件。

score 294 · Accepted Answer

这会给你想要的结果

#!/usr/bin/python

import os

# traverse root directory, and list directories as dirs and files as files
for root, dirs, files in os.walk("."):
    path = root.split(os.sep)
    print((len(path) - 1) * '---', os.path.basename(root))
    for file in files:
        print(len(path) * '---', file)

score 32 · Accepted Answer

尝试这个：

#!/usr/bin/env python
# -*- coding: utf-8 -*-

"""FileTreeMaker.py: ..."""

__author__  = "legendmohe"

import os
import argparse
import time

class FileTreeMaker(object):

    def _recurse(self, parent_path, file_list, prefix, output_buf, level):
        if len(file_list) == 0 \
            or (self.max_level != -1 and self.max_level <= level):
            return
        else:
            file_list.sort(key=lambda f: os.path.isfile(os.path.join(parent_path, f)))
            for idx, sub_path in enumerate(file_list):
                if any(exclude_name in sub_path for exclude_name in self.exn):
                    continue

                full_path = os.path.join(parent_path, sub_path)
                idc = "┣━"
                if idx == len(file_list) - 1:
                    idc = "┗━"

                if os.path.isdir(full_path) and sub_path not in self.exf:
                    output_buf.append("%s%s[%s]" % (prefix, idc, sub_path))
                    if len(file_list) > 1 and idx != len(file_list) - 1:
                        tmp_prefix = prefix + "┃  "
                    else:
                        tmp_prefix = prefix + "    "
                    self._recurse(full_path, os.listdir(full_path), tmp_prefix, output_buf, level + 1)
                elif os.path.isfile(full_path):
                    output_buf.append("%s%s%s" % (prefix, idc, sub_path))

    def make(self, args):
        self.root = args.root
        self.exf = args.exclude_folder
        self.exn = args.exclude_name
        self.max_level = args.max_level

        print("root:%s" % self.root)

        buf = []
        path_parts = self.root.rsplit(os.path.sep, 1)
        buf.append("[%s]" % (path_parts[-1],))
        self._recurse(self.root, os.listdir(self.root), "", buf, 0)

        output_str = "\n".join(buf)
        if len(args.output) != 0:
            with open(args.output, 'w') as of:
                of.write(output_str)
        return output_str

if __name__ == "__main__":
    parser = argparse.ArgumentParser()
    parser.add_argument("-r", "--root", help="root of file tree", default=".")
    parser.add_argument("-o", "--output", help="output file name", default="")
    parser.add_argument("-xf", "--exclude_folder", nargs='*', help="exclude folder", default=[])
    parser.add_argument("-xn", "--exclude_name", nargs='*', help="exclude name", default=[])
    parser.add_argument("-m", "--max_level", help="max level",
                        type=int, default=-1)
    args = parser.parse_args()
    print(FileTreeMaker().make(args))

你会得到这个：

root:.
[.]
┣━[.idea]
┃  ┣━[scopes]
┃  ┃  ┗━scope_settings.xml
┃  ┣━.name
┃  ┣━Demo.iml
┃  ┣━encodings.xml
┃  ┣━misc.xml
┃  ┣━modules.xml
┃  ┣━vcs.xml
┃  ┗━workspace.xml
┣━[test1]
┃  ┗━test1.txt
┣━[test2]
┃  ┣━[test2-2]
┃  ┃  ┗━[test2-3]
┃  ┃      ┣━test2
┃  ┃      ┗━test2-3-1
┃  ┗━test2
┣━folder_tree_maker.py
┗━tree.py

score 26 · Accepted Answer

递归遍历一个目录，您可以从当前目录中的所有目录中获取所有文件，并从当前目录中获取所有目录 - 因为上面的代码不简单（恕我直言）：

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))

score 13 · Accepted Answer

包中有更合适的功能os。但如果你必须使用os.walk，这就是我想出的

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

score 7 · Accepted Answer

您还可以递归遍历文件夹并使用pathlib.Path()列出其所有内容

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

score 5 · Accepted Answer

您可以使用os.walk，这可能是最简单的解决方案，但这是另一个值得探索的想法：

import sys, os

FILES = False

def main():
    if len(sys.argv) > 2 and sys.argv[2].upper() == '/F':
        global FILES; FILES = True
    try:
        tree(sys.argv[1])
    except:
        print('Usage: {} <directory>'.format(os.path.basename(sys.argv[0])))

def tree(path):
    path = os.path.abspath(path)
    dirs, files = listdir(path)[:2]
    print(path)
    walk(path, dirs, files)
    if not dirs:
        print('No subfolders exist')

def walk(root, dirs, files, prefix=''):
    if FILES and files:
        file_prefix = prefix + ('|' if dirs else ' ') + '   '
        for name in files:
            print(file_prefix + name)
        print(file_prefix)
    dir_prefix, walk_prefix = prefix + '+---', prefix + '|   '
    for pos, neg, name in enumerate2(dirs):
        if neg == -1:
            dir_prefix, walk_prefix = prefix + '\\---', prefix + '    '
        print(dir_prefix + name)
        path = os.path.join(root, name)
        try:
            dirs, files = listdir(path)[:2]
        except:
            pass
        else:
            walk(path, dirs, files, walk_prefix)

def listdir(path):
    dirs, files, links = [], [], []
    for name in os.listdir(path):
        path_name = os.path.join(path, name)
        if os.path.isdir(path_name):
            dirs.append(name)
        elif os.path.isfile(path_name):
            files.append(name)
        elif os.path.islink(path_name):
            links.append(name)
    return dirs, files, links

def enumerate2(sequence):
    length = len(sequence)
    for count, value in enumerate(sequence):
        yield count, count - length, value

if __name__ == '__main__':
    main()

您可能会从 Windows 终端中的 TREE 命令中识别出以下文档：

Graphically displays the folder structure of a drive or path.

TREE [drive:][path] [/F] [/A]

   /F   Display the names of the files in each folder.
   /A   Use ASCII instead of extended characters.

score 4 · Accepted Answer

这适用于文件夹名称：

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]

score 3 · Accepted Answer

#!/usr/bin/python

import os 

def tracing(a):
    global i>
    for item in os.listdir(a):
        if os.path.isfile(item):
            print i + item 
        else:
            print i + item 
            i+=i
            tracing(item)

i = "---"
tracing(".")

score 3 · Accepted Answer

将是最好的方式

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

score 2 · Accepted Answer

给定一个文件夹名称，递归遍历其整个层次结构。

#! /usr/local/bin/python3
# findLargeFiles.py - given a folder name, walk through its entire hierarchy
#                   - print folders and files within each folder

import os

def recursive_walk(folder):
    for folderName, subfolders, filenames in os.walk(folder):
        if subfolders:
            for subfolder in subfolders:
                recursive_walk(subfolder)
        print('\nFolder: ' + folderName + '\n')
        for filename in filenames:
            print(filename + '\n')

recursive_walk('/name/of/folder')

score 1 · Accepted Answer

尝试这个：

import os
root_name = next(os.walk("."))[0]
dir_names = next(os.walk("."))[1]
file_names = next(os.walk("."))[2]

在这里，我假设您的路径为“。” 其中 root_file 和其他目录在那里。所以，基本上我们只是通过使用 next() 调用来迭代整个树，因为我们的 os.walk 只是生成函数。通过这样做，我们可以将所有目录和文件名分别保存在 dir_names 和 file_names 中。

score 0 · Accepted Answer

试试这个；容易的

 #!/usr/bin/python
 import os
 # Creating an empty list that will contain the already traversed paths
 donePaths = []
 def direct(path):
       for paths,dirs,files in os.walk(path):
             if paths not in donePaths:
                    count = paths.count('/')
                    if files:
                          for ele1 in files:
                                print '---------' * (count), ele1
                    if dirs:
                          for ele2 in dirs:
                                print '---------' * (count), ele2
                                absPath = os.path.join(paths,ele2)
              # recursively calling the direct function on each directory
                                direct(absPath)
                   # adding the paths to the list that got traversed 
                                donePaths.append(absPath)

 path = raw_input("Enter any path to get the following Dir Tree ...\n")
 direct(path)

========下面的输出========

 /home/test
 ------------------ b.txt
 ------------------ a.txt
 ------------------ a
 --------------------------- a1.txt
 ------------------ b
 --------------------------- b1.txt
 --------------------------- b2.txt
 --------------------------- cde
 ------------------------------------ cde.txt
 ------------------------------------ cdeDir
 --------------------------------------------- cdeDir.txt
 ------------------ c
 --------------------------- c.txt
 --------------------------- c1
 ------------------------------------ c1.txt
 ------------------------------------ c2.txt

score -3 · Accepted Answer

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

python - Python中使用os.walk()递归遍历目录

13 回答 13

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