使用 os.walk() 在 Python 中递归遍历目录
- 2024-12-31 08:37:00
- admin 原创
- 130
问题描述:
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 ""
以下是我的 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。
解决方案 1:
这将给你想要的结果
#!/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)
解决方案 2:
递归遍历目录,从当前目录中获取所有目录的所有文件,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:
doSomethingWithDir(os.path.join(root, dirname))
解决方案 3:
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 \n 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 = "
".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
解决方案 4:
包中有更多适合此目的的函数os
。但如果您必须使用os.walk
,以下是我想出的
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
解决方案 5:
您还可以使用pathlib.Path()递归遍历文件夹并列出其所有内容
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(' ' * 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
解决方案 6:
对文件夹名称执行此操作:
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]
解决方案 7:
您可以使用os.walk
,这可能是最简单的解决方案,但这里还有另一个想法可以探索:
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()
您可能会从 Windows 终端中的 TREE 命令中认出以下文档:
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.
解决方案 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(".")
解决方案 9:
是最好的办法
import os
def traverse_dir_recur(directory):
l = os.listdir(directory)
for d in l:
if os.path.isdir(directory + d):
traverse_dir_recur(directory + d +"/")
else:
print(directory + d)
解决方案 10:
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('
Folder: ' + folderName + '
')
for filename in filenames:
print(filename + '
')
recursive_walk('/name/of/folder')
解决方案 11:
尝试一下:
import os
root_name = next(os.walk("."))[0]
dir_names = next(os.walk("."))[1]
file_names = next(os.walk("."))[2]
这里我假设您的路径为“。”,其中有 root_file 和其他目录。因此,基本上我们只是通过使用 next() 调用遍历整个树,因为我们的 os.walk 只是生成函数。通过这样做,我们可以分别将所有目录和文件名保存在 dir_names 和 file_names 中。
解决方案 12:
假设您有一个任意父目录,其中包含如下子目录:
/home/parent_dir
├── 0_N
├── 1_M
├── 2_P
├── 3_R
└── 4_T
您可以按照以下步骤估算每个子目录中文件总数相对于父目录中文件总数的近似百分比分布:
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("/home/parent_dir")
它将在终端中打印如下内容:
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%
解决方案 13:
试试这个;很简单
#!/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 ...
")
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
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