Spe*_*cto 5 python directory-listing
我目前正在使用Here的目录walker
import os
class DirectoryWalker:
# a forward iterator that traverses a directory tree
def __init__(self, directory):
self.stack = [directory]
self.files = []
self.index = 0
def __getitem__(self, index):
while 1:
try:
file = self.files[self.index]
self.index = self.index + 1
except IndexError:
# pop next directory from stack
self.directory = self.stack.pop()
self.files = os.listdir(self.directory)
self.index = 0
else:
# got a filename
fullname = os.path.join(self.directory, file)
if os.path.isdir(fullname) and not os.path.islink(fullname):
self.stack.append(fullname)
return fullname
for file in DirectoryWalker(os.path.abspath('.')):
print file
Run Code Online (Sandbox Code Playgroud)
此次要更改允许您在文件中拥有完整路径.
任何人都可以帮助我如何使用这个来找到文件名?我需要完整的路径,只需要文件名.
Che*_*ery 12
你为什么要自己做这种无聊的事情?
for path, directories, files in os.walk('.'):
print 'ls %r' % path
for directory in directories:
print ' d%r' % directory
for filename in files:
print ' -%r' % filename
Run Code Online (Sandbox Code Playgroud)
输出:
'.'
d'finction'
d'.hg'
-'setup.py'
-'.hgignore'
'./finction'
-'finction'
-'cdg.pyc'
-'util.pyc'
-'cdg.py'
-'util.py'
-'__init__.pyc'
-'__init__.py'
'./.hg'
d'store'
-'hgrc'
-'requires'
-'00changelog.i'
-'undo.branch'
-'dirstate'
-'undo.dirstate'
-'branch'
'./.hg/store'
d'data'
-'undo'
-'00changelog.i'
-'00manifest.i'
'./.hg/store/data'
d'finction'
-'.hgignore.i'
-'setup.py.i'
'./.hg/store/data/finction'
-'util.py.i'
-'cdg.py.i'
-'finction.i'
-'____init____.py.i'
Run Code Online (Sandbox Code Playgroud)
但是如果你坚持,os.path中有路径相关的工具,os.basename就是你正在看的东西.
>>> import os.path
>>> os.path.basename('/hello/world.h')
'world.h'
Run Code Online (Sandbox Code Playgroud)
而不是使用'.' 作为您的目录,请参考其绝对路径:
for file in DirectoryWalker(os.path.abspath('.')):
print file
Run Code Online (Sandbox Code Playgroud)
另外,我建议使用"文件"以外的单词,因为它意味着python语言中的某些东西.不是关键字,但它仍然运行.
顺便说一句,在处理文件名时,我发现os.path模块非常有用 - 我建议你仔细看看,特别是
os.path.normpath
Run Code Online (Sandbox Code Playgroud)
规范化路径(摆脱多余的'.'和'theFolderYouWereJustIn /../')
os.path.join
Run Code Online (Sandbox Code Playgroud)
加入两条路