在Python中使用os.walk()递归遍历目录
我想从根目录导航到其中的所有其他目录,并打印相同的目录 这是我的密码:在Python中使用os.walk()递归遍历目录,python,os.walk,Python,Os.walk,我想从根目录导航到其中的所有其他目录,并打印相同的目录 这是我的密码: #!/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 pr
#!/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
上面的
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,其余是文件。在os
软件包中有更适合此功能的功能。但是如果你必须使用os.walk,下面是我的建议
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
您可以使用
os.walk
,这可能是最简单的解决方案,但这里有另一个值得探讨的想法:
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()
这会给你想要的结果
#!/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)
这对文件夹名称不起作用:
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]
给定文件夹名称,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('\nFolder: ' + folderName + '\n')
for filename in filenames:
print(filename + '\n')
recursive_walk('/name/of/folder')
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))
试试这个;简单的
#!/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)
==========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
递归遍历一个目录,从当前目录中的所有目录中获取所有文件,从当前目录中获取所有目录-因为上面的代码并不简单(imho): 试试这个:
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“.”,其中有根文件和其他目录。
因此,基本上我们只是使用next()调用在整个树中进行迭代,因为os.walk只是生成函数。
通过这样做,我们可以将所有目录名和文件名分别保存在目录名和文件名中。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)
您还可以递归地遍历文件夹,并使用
从pathlib导入路径
def签出路径(目标路径,级别=0):
""""
此函数递归打印pathlib.Path对象的所有内容
"""
def打印缩进(文件夹,级别):
打印('\t'*level+文件夹)
缩进打印(目标路径名称、级别)
对于目标路径.iterdir()中的文件:
如果文件.is_dir():
签出路径(文件,级别+1)
其他:
缩进打印(file.name,级别+1)
我的路径=路径(r'C:\示例文件夹')
签出路径(我的路径)
输出:
example folder
folder
textfile3.txt
textfile1.txt
textfile2.txt
我想在这里添加这篇关于python强大功能的帖子:>>>print 2*'------path=os.path.relpath(root,basepath).split(os.sep)@Ajay可能是偏执狂,总是做
os.walk(u)”
因为路径可以是Unicode。更好的是,os.path.curdir
我已经使用了os.path.walk
一段时间了,所以os.walk
对我来说是新的!酷豆。@Semprinibasepath
在你的代码中等于什么?这看起来不像是遍历整个树。不需要递归调用os.walk,因为它已经遍历了lattens递归。这就是它返回folderName参数的原因。检查已遍历路径的目的是什么?如果是检测由链接引起的循环,os.walk显然默认不跟踪链接。还有其他情况吗?您好,我非常喜欢您的脚本,但对于我正在工作的项目来说,它有点太复杂了在上,是否有可能将其作为一个小函数使用,并且只存在-r参数?如何在.txt中打印?我尝试了print(FileTreeMaker().make(args),file=tree)
但它给了我“charmap”编解码器无法对17-21位的字符进行编码:字符映射到
idc代表什么我写了一些类似于os.listdir()的东西
也一样。你的好得多;我无法正确使用递归,它只工作了2到3层。最后我决定用os.walk()
从头开始再试一次,我认为这更合适。我很惊讶你在这里根本没有使用它。那么什么更合适的函数呢?(在3.5中,如果有关系的话)抱歉,没有机会记住我的意思。我的意思可能是os.listdir
,但@ajay的解决方案胜过了这一点。这是最有用的答案。请注意os.path.join(root,filename)
给出了文件的完整路径,即使文件嵌套在多个目录中。在Python3中对我不起作用。我假设错误出现在dir+d
中,这可能会在没有目录分隔符的情况下对它们进行合并。可能最好使用os.path.join
来合并具有文件名的目录,因为它使用了<代码>路径库。谢谢
/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
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))
import os
root_name = next(os.walk("."))[0]
dir_names = next(os.walk("."))[1]
file_names = next(os.walk("."))[2]
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)
example folder
folder
textfile3.txt
textfile1.txt
textfile2.txt