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
以上，.和./packagedemo是目录。
但是，我需要以下列方式打印O / P：
A
---a.txt
---b.txt
---B
------c.out
上面，A和B是目录，其余是文件。
答案 0 :(得分：179)
这将为您提供所需的结果
#!/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)
答案 1 :(得分：17)
试试这个：
#!/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
答案 2 :(得分：11)
os包中有更多合适的功能。但是如果你必须使用os.walk，这就是我想出来的
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
答案 3 :(得分：5)
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终端的TREE命令中识别以下文档：
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.
答案 4 :(得分：4)
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]
答案 5 :(得分：3)
#!/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(".")
答案 6 :(得分：2)
给定文件夹名称，递归遍历整个层次结构。
#! /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')
答案 7 :(得分：2)
您还可以递归地浏览文件夹，并使用pathlib.Path()
列出所有内容。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
答案 8 :(得分：1)
试试这个;很容易
#!/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)
========输出如下========
/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
答案 9 :(得分：1)
将是最好的方法
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)
答案 10 :(得分：0)
递归遍历目录，您可以从当前目录中的所有目录获取所有文件，而从当前目录中获取所有目录-因为上述代码没有简单性（imho）：
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))
答案 11 :(得分：0)
尝试一下：
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“”。其中存在root_file和其他目录。 因此，基本上，我们只是通过使用next（）调用来遍历整个树，因为os.walk只是生成函数。 这样我们可以将所有目录和文件名分别保存在dir_names和file_names中。
答案 12 :(得分：-2)
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