使用os.walk()遞歸遍歷Python中的目錄

Question

我想從根目錄導航到其中的所有其他目錄並打印相同的內容。

這是我的代碼：

#!/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 ""

這是我的 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是目錄。

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是目錄，rest 是文件。

Answer 1

這會給你想要的結果

#!/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)

Answer 2

嘗試這個：

#!/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))

你會得到這個：

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

Answer 3

遞歸遍歷一個目錄，您可以從當前目錄中的所有目錄中獲取所有文件，並從當前目錄中獲取所有目錄 - 因為上面的代碼並不簡單（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))

Answer 4

在os包中有更合適的函數。 但是如果你必須使用os.walk ，這就是我想出的

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)

輸出：

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

Answer 5

你也可以遞歸遍歷一個文件夾並使用pathlib.Path()列出它的所有內容

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)

輸出：

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

Answer 6

您可以使用os.walk ，這可能是最簡單的解決方案，但這里有另一個想法可以探索：

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()

---

您可能會從 Windows 終端中的 TREE 命令中識別出以下文檔：

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.

Answer 7

這對文件夾名稱執行此操作：

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]

Answer 8

#!/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(".")

Answer 9

將是最好的方法

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)

Answer 10

給定文件夾名稱，遞歸遍歷其整個層次結構。

#! /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')

Answer 11

嘗試這個：

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

在這里，我假設您的路徑為“。” 其中 root_file 和其他目錄在那里。 所以，基本上我們只是使用 next() 調用遍歷整個樹，因為我們的 os.walk 只是生成函數。 通過這樣做，我們可以將所有目錄和文件名分別保存在 dir_names 和 file_names 中。

Answer 12

試試這個； 簡單的

 #!/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)

========輸出如下========

 /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

Answer 13

顯然，現在使用 python 3.x 有更簡單的方法可以做到這一點，但我碰巧正在處理舊的 2.6 東西，所以到底是什么——遲到總比沒有好。

這個遞歸例程將返回給定一組目錄的所有文件（默認 ['.']）：

def find(folders=['.'],result=[]):
    for _folder in [os.path.join(f,'') for f in folders]:
        _,_folders,_files = next(os.walk(_folder))
        for _file in _files:
            result.append(os.path.join(_folder,_file))
        if bool(_folders):
            find(folders=[os.path.join(_folder,f) for f in _folders],result=result)
    return result

files = find(['/opt'])

注意：這可能比使用操作系統的 find 命令慢得多。 如果需要，您可以添加過濾和級別檢查。

更新它運行但沒有通過所有級別 - 已更正。

Answer 14

假設您有一個帶有子目錄的任意父目錄，如下所示：

/home/parent_dir
├── 0_N
├── 1_M
├── 2_P
├── 3_R
└── 4_T

以下是您可以做的估計每個子目錄中的近似百分比分布 #files 相對於父目錄中的總 #files：

from os import listdir as osl
from os import walk as osw
from os.path import join as osj

def subdir_summary(parent_dir):
    parent_dir_len = sum([len(files) for _, _, files in osw(parent_dir)])
    print(f"Total files in parent: {parent_dir_len}")
    for subdir in sorted(osl(parent_dir)):
        subdir_files_len = len(osl(osj(parent_dir, subdir)))
        print(subdir, subdir_files_len, f"{int(100*(subdir_files_len / parent_dir_len))}%")

subdir_summary(image_dataset_dir) ## define folder here

它將在終端打印如下：

Total files in parent: 5876
0_N 3254 55%
1_M 509 8%
2_P 1187 20%
3_R 594 10%
4_T 332 5%

Answer 15

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