我知道我们可以使用os.walk()
列出目录中的所有子目录或所有文件。但是,我想列出完整的目录树内容:
- Subdirectory 1:
- file11
- file12
- Sub-sub-directory 11:
- file111
- file112
- Subdirectory 2:
- file21
- sub-sub-directory 21
- sub-sub-directory 22
- sub-sub-sub-directory 221
- file 2211
如何在Python中实现这一目标?
这是一个使用格式化的功能:
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))
只有在系统上安装了tree
时,此解决方案才有效。但是我在这里留下这个解决方案,以防它帮助其他人。
您可以告诉树将树结构输出为XML(tree -X
)或JSON(tree -J
)。当然可以使用python直接解析JSON,并且可以使用lxml
轻松读取XML。
以以下目录结构为例:
[sri@localhost Projects]$ tree --charset=ascii bands
bands
|-- DreamTroll
| |-- MattBaldwinson
| |-- members.txt
| |-- PaulCarter
| |-- SimonBlakelock
| `-- Rob Stringer
|-- KingsX
| |-- DougPinnick
| |-- JerryGaskill
| |-- members.txt
| `-- TyTabor
|-- Megadeth
| |-- DaveMustaine
| |-- DavidEllefson
| |-- DirkVerbeuren
| |-- KikoLoureiro
| `-- members.txt
|-- Nightwish
| |-- EmppuVuorinen
| |-- FloorJansen
| |-- JukkaNevalainen
| |-- MarcoHietala
| |-- members.txt
| |-- TroyDonockley
| `-- TuomasHolopainen
`-- Rush
|-- AlexLifeson
|-- GeddyLee
`-- NeilPeart
5 directories, 25 files
XML
<?xml version="1.0" encoding="UTF-8"?>
<tree>
<directory name="bands">
<directory name="DreamTroll">
<file name="MattBaldwinson"></file>
<file name="members.txt"></file>
<file name="PaulCarter"></file>
<file name="RobStringer"></file>
<file name="SimonBlakelock"></file>
</directory>
<directory name="KingsX">
<file name="DougPinnick"></file>
<file name="JerryGaskill"></file>
<file name="members.txt"></file>
<file name="TyTabor"></file>
</directory>
<directory name="Megadeth">
<file name="DaveMustaine"></file>
<file name="DavidEllefson"></file>
<file name="DirkVerbeuren"></file>
<file name="KikoLoureiro"></file>
<file name="members.txt"></file>
</directory>
<directory name="Nightwish">
<file name="EmppuVuorinen"></file>
<file name="FloorJansen"></file>
<file name="JukkaNevalainen"></file>
<file name="MarcoHietala"></file>
<file name="members.txt"></file>
<file name="TroyDonockley"></file>
<file name="TuomasHolopainen"></file>
</directory>
<directory name="Rush">
<file name="AlexLifeson"></file>
<file name="GeddyLee"></file>
<file name="NeilPeart"></file>
</directory>
</directory>
<report>
<directories>5</directories>
<files>25</files>
</report>
</tree>
JSON
[sri@localhost Projects]$ tree -J bands
[
{"type":"directory","name":"bands","contents":[
{"type":"directory","name":"DreamTroll","contents":[
{"type":"file","name":"MattBaldwinson"},
{"type":"file","name":"members.txt"},
{"type":"file","name":"PaulCarter"},
{"type":"file","name":"RobStringer"},
{"type":"file","name":"SimonBlakelock"}
]},
{"type":"directory","name":"KingsX","contents":[
{"type":"file","name":"DougPinnick"},
{"type":"file","name":"JerryGaskill"},
{"type":"file","name":"members.txt"},
{"type":"file","name":"TyTabor"}
]},
{"type":"directory","name":"Megadeth","contents":[
{"type":"file","name":"DaveMustaine"},
{"type":"file","name":"DavidEllefson"},
{"type":"file","name":"DirkVerbeuren"},
{"type":"file","name":"KikoLoureiro"},
{"type":"file","name":"members.txt"}
]},
{"type":"directory","name":"Nightwish","contents":[
{"type":"file","name":"EmppuVuorinen"},
{"type":"file","name":"FloorJansen"},
{"type":"file","name":"JukkaNevalainen"},
{"type":"file","name":"MarcoHietala"},
{"type":"file","name":"members.txt"},
{"type":"file","name":"TroyDonockley"},
{"type":"file","name":"TuomasHolopainen"}
]},
{"type":"directory","name":"Rush","contents":[
{"type":"file","name":"AlexLifeson"},
{"type":"file","name":"GeddyLee"},
{"type":"file","name":"NeilPeart"}
]}
]},
{"type":"report","directories":5,"files":25}
]
在这里你可以找到像这样的输出代码:https://stackoverflow.com/a/56622847/6671330
V .
|-> V folder1
| |-> V folder2
| | |-> V folder3
| | | |-> file3.txt
| | |-> file2.txt
| |-> V folderX
| |-> file1.txt
|-> 02-hw1_wdwwfm.py
|-> 06-t1-home1.py
|-> 06-t1-home2.py
|-> hw1.py
没有缩进的解决方案:
for path, dirs, files in os.walk(given_path):
print path
for f in files:
print f
os.walk已经完成了你正在寻找的自上而下,深度优先的步行。
忽略dirs列表可以防止您提到的重叠。
我来到这里寻找同样的事情,并为我使用了dhobbs的答案。作为感谢社区的一种方式,我添加了一些写入文件的参数,正如akshay所说的那样,并且显示文件是可选的,所以它不是一个输出。还使缩进成为可选参数,以便您可以更改它,因为有些人喜欢它是2而其他人更喜欢4。
使用不同的循环,因此不显示文件的循环不检查每次迭代是否必须。
希望它帮助别人,因为dhobbs的回答帮助了我。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
与上面的答案类似,但对于python3,可以说是可读的,可以说是可扩展的:
from pathlib import Path
class DisplayablePath(object):
display_filename_prefix_middle = '├──'
display_filename_prefix_last = '└──'
display_parent_prefix_middle = ' '
display_parent_prefix_last = '│ '
def __init__(self, path, parent_path, is_last):
self.path = Path(str(path))
self.parent = parent_path
self.is_last = is_last
if self.parent:
self.depth = self.parent.depth + 1
else:
self.depth = 0
@property
def displayname(self):
if self.path.is_dir():
return self.path.name + '/'
return self.path.name
@classmethod
def make_tree(cls, root, parent=None, is_last=False, criteria=None):
root = Path(str(root))
criteria = criteria or cls._default_criteria
displayable_root = cls(root, parent, is_last)
yield displayable_root
children = sorted(list(path
for path in root.iterdir()
if criteria(path)),
key=lambda s: str(s).lower())
count = 1
for path in children:
is_last = count == len(children)
if path.is_dir():
yield from cls.make_tree(path,
parent=displayable_root,
is_last=is_last,
criteria=criteria)
else:
yield cls(path, displayable_root, is_last)
count += 1
@classmethod
def _default_criteria(cls, path):
return True
@property
def displayname(self):
if self.path.is_dir():
return self.path.name + '/'
return self.path.name
def displayable(self):
if self.parent is None:
return self.displayname
_filename_prefix = (self.display_filename_prefix_last
if self.is_last
else self.display_filename_prefix_middle)
parts = ['{!s} {!s}'.format(_filename_prefix,
self.displayname)]
parent = self.parent
while parent and parent.parent is not None:
parts.append(self.display_parent_prefix_middle
if parent.is_last
else self.display_parent_prefix_last)
parent = parent.parent
return ''.join(reversed(parts))
用法示例:
paths = DisplayablePath.make_tree(Path('doc'))
for path in paths:
print(path.displayable())
示例输出:
doc/
├── _static/
│ ├── embedded/
│ │ ├── deep_file
│ │ └── very/
│ │ └── deep/
│ │ └── folder/
│ │ └── very_deep_file
│ └── less_deep_file
├── about.rst
├── conf.py
└── index.rst
基于这个精彩的帖子
http://code.activestate.com/recipes/217212-treepy-graphically-displays-the-directory-structur/
这里有一个完全像行为的改进
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()
import os
def fs_tree_to_dict(path_):
file_token = ''
for root, dirs, files in os.walk(path_):
tree = {d: fs_tree_to_dict(os.path.join(root, d)) for d in dirs}
tree.update({f: file_token for f in files})
return tree # note we discontinue iteration trough os.walk
如果有人感兴趣 - 递归函数返回嵌套的字典结构。键是file system
名称(目录和文件),值是:
file_token
)在此示例中,字符串指定文件为空。它们也可以是例如给定文件内容或其所有者信息或特权或不同于dict的任何对象。除非它是字典,否则在进一步的操作中可以很容易地与“目录类型”区分开来。
在文件系统中有这样一棵树:
# bash:
$ tree /tmp/ex
/tmp/ex
├── d_a
│ ├── d_a_a
│ ├── d_a_b
│ │ └── f1.txt
│ ├── d_a_c
│ └── fa.txt
├── d_b
│ ├── fb1.txt
│ └── fb2.txt
└── d_c
结果将是:
# python 2 or 3:
>>> fs_tree_to_dict("/tmp/ex")
{
'd_a': {
'd_a_a': {},
'd_a_b': {
'f1.txt': ''
},
'd_a_c': {},
'fa.txt': ''
},
'd_b': {
'fb1.txt': '',
'fb2.txt': ''
},
'd_c': {}
}
如果你喜欢这个,我已经用这个东西创建了一个包(python 2&3)(以及一个很好的pyfakefs
助手):https://pypi.org/project/fsforge/
在上面的dhobbs回答(https://stackoverflow.com/a/9728478/624597)之上,这里有一个额外的功能,可以将结果存储到文件中(我个人使用它来复制并粘贴到FreeMind以便对结构有一个很好的概述,因此我使用制表符代替空格来缩进):
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(".")
你可以执行Linux shell的'tree'命令。
安装:
~$sudo apt install tree
在python中使用
>>> import os
>>> os.system('tree <desired path>')
例:
>>> os.system('tree ~/Desktop/myproject')
这为您提供了更清洁的结构,视觉上更加全面,易于打字。
也许比@ellockie更快(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("/")
测试结果如下截图: