在python中列出目录树结构？

Question

我知道我们可以使用os.walk（）来列出目录中的所有子目录或所有文件。不过，我想列出完整的目录树的内容：

• 子目录1：
  • FILE11
  • file12
  • 子子目录11：
    • file111
    • file112
• 子目录2：
  • file21
  • 子子目录21
  • 子子目录22
    
    • 分次子目录221
      • 文件2211

如何在Python中达到最佳效果？

Answer 1

这里是这样做与格式化函数：

import os 

def list_files(startpath): 
    for root, dirs, files in os.walk(startpath): 
     level = root.replace(startpath, '').count(os.sep) 
     indent = ' ' * 4 * (level) 
     print('{}{}/'.format(indent, os.path.basename(root))) 
     subindent = ' ' * 4 * (level + 1) 
     for f in files: 
      print('{}{}'.format(subindent, f)) 

Answer 2

没有你的缩进一个解决方案：

for path, dirs, files in os.walk(path): 
    print path 
    for f in files: 
    print f 

os.walk已做了你正在寻找自上而下，深度优先的步行路程。

忽略目录列表可防止您提到的重叠。

Answer 3

我来到这里寻找同样的事情，使用dhobbs为我解答。作为感谢社区的一种方式，我添加了一些参数来写入文件，正如akshay所问，并且显示文件是可选的，因此它不是一个输出。也使缩进成为一个可选参数，因此您可以更改它，因为有些喜欢它是2而其他人喜欢4.

使用了不同的循环，因此不显示文件的循环不检查它是否必须在每次迭代中进行检查。

希望它可以帮助别人，因为dhobbs回答帮助了我。非常感谢。

def showFolderTree(path,show_files=False,indentation=2,file_output=False): 
""" 
Shows the content of a folder in a tree structure. 
path -(string)- path of the root folder we want to show. 
show_files -(boolean)- Whether or not we want to see files listed. 
         Defaults to False. 
indentation -(int)- Indentation we want to use, defaults to 2. 
file_output -(string)- Path (including the name) of the file where we want 
         to save the tree. 
""" 


tree = [] 

if not show_files: 
    for root, dirs, files in os.walk(path): 
     level = root.replace(path, '').count(os.sep) 
     indent = ' '*indentation*(level) 
     tree.append('{}{}/'.format(indent,os.path.basename(root))) 

if show_files: 
    for root, dirs, files in os.walk(path): 
     level = root.replace(path, '').count(os.sep) 
     indent = ' '*indentation*(level) 
     tree.append('{}{}/'.format(indent,os.path.basename(root)))  
     for f in files: 
      subindent=' ' * indentation * (level+1) 
      tree.append('{}{}'.format(subindent,f)) 

if file_output: 
    output_file = open(file_output,'w') 
    for line in tree: 
     output_file.write(line) 
     output_file.write('\n') 
else: 
    # Default behaviour: print on screen. 
    for line in tree: 
     print line 

Answer 4

基于这个梦幻般的后

http://code.activestate.com/recipes/217212-treepy-graphically-displays-the-directory-structur/

这里上课细化到完全一样

http://linux.die.net/man/1/tree

 
#!/usr/bin/env python2 
# -*- coding: utf-8 -*- 

# tree.py 
# 
# Written by Doug Dahms 
# 
# Prints the tree structure for the path specified on the command line 

from os import listdir, sep 
from os.path import abspath, basename, isdir 
from sys import argv 

def tree(dir, padding, print_files=False, isLast=False, isFirst=False): 
    if isFirst: 
     print padding.decode('utf8')[:-1].encode('utf8') + dir 
    else: 
     if isLast: 
      print padding.decode('utf8')[:-1].encode('utf8') + '└── ' + basename(abspath(dir)) 
     else: 
      print padding.decode('utf8')[:-1].encode('utf8') + '├── ' + basename(abspath(dir)) 
    files = [] 
    if print_files: 
     files = listdir(dir) 
    else: 
     files = [x for x in listdir(dir) if isdir(dir + sep + x)] 
    if not isFirst: 
     padding = padding + ' ' 
    files = sorted(files, key=lambda s: s.lower()) 
    count = 0 
    last = len(files) - 1 
    for i, file in enumerate(files): 
     count += 1 
     path = dir + sep + file 
     isLast = i == last 
     if isdir(path): 
      if count == len(files): 
       if isFirst: 
        tree(path, padding, print_files, isLast, False) 
       else: 
        tree(path, padding + ' ', print_files, isLast, False) 
      else: 
       tree(path, padding + '│', print_files, isLast, False) 
     else: 
      if isLast: 
       print padding + '└── ' + file 
      else: 
       print padding + '├── ' + file 

def usage(): 
    return '''Usage: %s [-f] 
Print tree structure of path specified. 
Options: 
-f  Print files as well as directories 
PATH Path to process''' % basename(argv[0]) 

def main(): 
    if len(argv) == 1: 
     print usage() 
    elif len(argv) == 2: 
     # print just directories 
     path = argv[1] 
     if isdir(path): 
      tree(path, '', False, False, True) 
     else: 
      print 'ERROR: \'' + path + '\' is not a directory' 
    elif len(argv) == 3 and argv[1] == '-f': 
     # print directories and files 
     path = argv[2] 
     if isdir(path): 
      tree(path, '', True, False, True) 
     else: 
      print 'ERROR: \'' + path + '\' is not a directory' 
    else: 
     print usage() 

if __name__ == '__main__': 
    main() 

Answer 5

在dhobbs的顶端回答以上（https://stackoverflow.com/a/9728478/624597） ，这里是一个前例结果存储到文件的TRA功能（我个人用它来复制并粘贴到FreeMind有结构的一个很好的概述，所以我使用的标签，而不是空格缩进）：

import os 

def list_files(startpath): 

    with open("folder_structure.txt", "w") as f_output: 
     for root, dirs, files in os.walk(startpath): 
      level = root.replace(startpath, '').count(os.sep) 
      indent = '\t' * 1 * (level) 
      output_string = '{}{}/'.format(indent, os.path.basename(root)) 
      print(output_string) 
      f_output.write(output_string + '\n') 
      subindent = '\t' * 1 * (level + 1) 
      for f in files: 
       output_string = '{}{}'.format(subindent, f) 
       print(output_string) 
       f_output.write(output_string + '\n') 

list_files(".") 

Answer 6

也许比@快ellockie（也许）

 
import os 
def file_writer(text): 
    with open("folder_structure.txt","a") as f_output: 
     f_output.write(text) 
def list_files(startpath): 


    for root, dirs, files in os.walk(startpath): 
     level = root.replace(startpath, '').count(os.sep) 
     indent = '\t' * 1 * (level) 
     output_string = '{}{}/ \n'.format(indent, os.path.basename(root)) 
     file_writer(output_string) 
     subindent = '\t' * 1 * (level + 1) 
     output_string = '%s %s \n' %(subindent,[f for f in files]) 
     file_writer(''.join(output_string)) 


list_files("/") 

编辑：我测试的截图是：