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Python 中的目录树列表

发布于 2024-07-05 03:05:15 字数 37 浏览 6 评论 0原文

如何获取 Python 给定目录中所有文件（和目录）的列表？

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清风夜微凉 2024-07-12 03:05:16

最简单的方法：

list_output_files = [os.getcwd()+"\\"+f for f in os.listdir(os.getcwd())]

Easiest way:

list_output_files = [os.getcwd()+"\\"+f for f in os.listdir(os.getcwd())]

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緦唸λ蓇 2024-07-12 03:05:16

我知道这是一个老问题。如果你使用的是 liunx 机器，这是我遇到的一个巧妙的方法。

import subprocess
print(subprocess.check_output(["ls", "/"]).decode("utf8"))

I know this is an old question. This is a neat way I came across if you are on a liunx machine.

import subprocess
print(subprocess.check_output(["ls", "/"]).decode("utf8"))

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深海蓝天 2024-07-12 03:05:16

下面是一行 Python 版本：

import os
dir = 'given_directory_name'
filenames = [os.path.join(os.path.dirname(os.path.abspath(__file__)),dir,i) for i in os.listdir(dir)]

此代码列出了给定目录名称中所有文件和目录的完整路径。

Here is a one line Pythonic version:

import os
dir = 'given_directory_name'
filenames = [os.path.join(os.path.dirname(os.path.abspath(__file__)),dir,i) for i in os.listdir(dir)]

This code lists the full path of all files and directories in the given directory name.

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儭儭莪哋寶赑 2024-07-12 03:05:16

如果我想到的话，我会把它扔进去。进行通配符搜索的简单而肮脏的方法。

import re
import os

[a for a in os.listdir(".") if re.search("^.*\.py$",a)]

If figured I'd throw this in. Simple and dirty way to do wildcard searches.

import re
import os

[a for a in os.listdir(".") if re.search("^.*\.py$",a)]

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画▽骨i 2024-07-12 03:05:16

仅供参考添加扩展名或 ext 文件的过滤器
导入操作系统

path = '.'
for dirname, dirnames, filenames in os.walk(path):
    # print path to all filenames with extension py.
    for filename in filenames:
        fname_path = os.path.join(dirname, filename)
        fext = os.path.splitext(fname_path)[1]
        if fext == '.py':
            print fname_path
        else:
            continue

FYI Add a filter of extension or ext file
import os

path = '.'
for dirname, dirnames, filenames in os.walk(path):
    # print path to all filenames with extension py.
    for filename in filenames:
        fname_path = os.path.join(dirname, filename)
        fext = os.path.splitext(fname_path)[1]
        if fext == '.py':
            print fname_path
        else:
            continue

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維他命╮ 2024-07-12 03:05:16

#import modules
import os

_CURRENT_DIR = '.'


def rec_tree_traverse(curr_dir, indent):
    "recurcive function to traverse the directory"
    #print "[traverse_tree]"

    try :
        dfList = [os.path.join(curr_dir, f_or_d) for f_or_d in os.listdir(curr_dir)]
    except:
        print "wrong path name/directory name"
        return

    for file_or_dir in dfList:

        if os.path.isdir(file_or_dir):
            #print "dir  : ",
            print indent, file_or_dir,"\\"
            rec_tree_traverse(file_or_dir, indent*2)

        if os.path.isfile(file_or_dir):
            #print "file : ",
            print indent, file_or_dir

    #end if for loop
#end of traverse_tree()

def main():

    base_dir = _CURRENT_DIR

    rec_tree_traverse(base_dir," ")

    raw_input("enter any key to exit....")
#end of main()


if __name__ == '__main__':
    main()

#import modules
import os

_CURRENT_DIR = '.'


def rec_tree_traverse(curr_dir, indent):
    "recurcive function to traverse the directory"
    #print "[traverse_tree]"

    try :
        dfList = [os.path.join(curr_dir, f_or_d) for f_or_d in os.listdir(curr_dir)]
    except:
        print "wrong path name/directory name"
        return

    for file_or_dir in dfList:

        if os.path.isdir(file_or_dir):
            #print "dir  : ",
            print indent, file_or_dir,"\\"
            rec_tree_traverse(file_or_dir, indent*2)

        if os.path.isfile(file_or_dir):
            #print "file : ",
            print indent, file_or_dir

    #end if for loop
#end of traverse_tree()

def main():

    base_dir = _CURRENT_DIR

    rec_tree_traverse(base_dir," ")

    raw_input("enter any key to exit....")
#end of main()


if __name__ == '__main__':
    main()

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梦归所梦 2024-07-12 03:05:15

def print_directory_contents(sPath):
        import os                                       
        for sChild in os.listdir(sPath):                
            sChildPath = os.path.join(sPath,sChild)
            if os.path.isdir(sChildPath):
                print_directory_contents(sChildPath)
            else:
                print(sChildPath)

Below code will list directories and the files within the dir

def print_directory_contents(sPath):
        import os                                       
        for sChild in os.listdir(sPath):                
            sChildPath = os.path.join(sPath,sChild)
            if os.path.isdir(sChildPath):
                print_directory_contents(sChildPath)
            else:
                print(sChildPath)

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百合的盛世恋 2024-07-12 03:05:15

对于 Python 2

#!/bin/python2

import os

def scan_dir(path):
    print map(os.path.abspath, os.listdir(pwd))

对于 Python 3

对于过滤器和映射，您需要用 list() 包装它们

#!/bin/python3

import os

def scan_dir(path):
    print(list(map(os.path.abspath, os.listdir(pwd))))

现在建议您用生成器表达式或列表推导式替换映射和过滤器的使用：

#!/bin/python

import os

def scan_dir(path):
    print([os.path.abspath(f) for f in os.listdir(path)])

For Python 2

#!/bin/python2

import os

def scan_dir(path):
    print map(os.path.abspath, os.listdir(pwd))

For Python 3

For filter and map, you need wrap them with list()

#!/bin/python3

import os

def scan_dir(path):
    print(list(map(os.path.abspath, os.listdir(pwd))))

The recommendation now is that you replace your usage of map and filter with generators expressions or list comprehensions:

#!/bin/python

import os

def scan_dir(path):
    print([os.path.abspath(f) for f in os.listdir(path)])

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还给你自由 2024-07-12 03:05:15

一个很好的单行代码，仅递归地列出文件。我在 setup.py package_data 指令中使用了这个：

import os

[os.path.join(x[0],y) for x in os.walk('<some_directory>') for y in x[2]]

我知道这不是问题的答案，但可能会派上用场

A nice one liner to list only the files recursively. I used this in my setup.py package_data directive:

import os

[os.path.join(x[0],y) for x in os.walk('<some_directory>') for y in x[2]]

I know it's not the answer to the question, but may come in handy

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莫相离 2024-07-12 03:05:15

与我合作的版本是萨利赫在本页其他地方的答案的修改版本。

代码如下：

dir = 'given_directory_name'
filenames = [os.path.abspath(os.path.join(dir,i)) for i in os.listdir(dir)]

The one worked with me is kind of a modified version from Saleh's answer elsewhere on this page.

The code is as follows:

dir = 'given_directory_name'
filenames = [os.path.abspath(os.path.join(dir,i)) for i in os.listdir(dir)]

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泪冰清 2024-07-12 03:05:15

我写了一个长版本，其中包含我可能需要的所有选项： http://sam.nipl .net/code/python/find.py

我想它也适合这里：

#!/usr/bin/env python

import os
import sys

def ls(dir, hidden=False, relative=True):
    nodes = []
    for nm in os.listdir(dir):
        if not hidden and nm.startswith('.'):
            continue
        if not relative:
            nm = os.path.join(dir, nm)
        nodes.append(nm)
    nodes.sort()
    return nodes

def find(root, files=True, dirs=False, hidden=False, relative=True, topdown=True):
    root = os.path.join(root, '')  # add slash if not there
    for parent, ldirs, lfiles in os.walk(root, topdown=topdown):
        if relative:
            parent = parent[len(root):]
        if dirs and parent:
            yield os.path.join(parent, '')
        if not hidden:
            lfiles   = [nm for nm in lfiles if not nm.startswith('.')]
            ldirs[:] = [nm for nm in ldirs  if not nm.startswith('.')]  # in place
        if files:
            lfiles.sort()
            for nm in lfiles:
                nm = os.path.join(parent, nm)
                yield nm

def test(root):
    print "* directory listing, with hidden files:"
    print ls(root, hidden=True)
    print
    print "* recursive listing, with dirs, but no hidden files:"
    for f in find(root, dirs=True):
        print f
    print

if __name__ == "__main__":
    test(*sys.argv[1:])

I wrote a long version, with all the options I might need: http://sam.nipl.net/code/python/find.py

I guess it will fit here too:

#!/usr/bin/env python

import os
import sys

def ls(dir, hidden=False, relative=True):
    nodes = []
    for nm in os.listdir(dir):
        if not hidden and nm.startswith('.'):
            continue
        if not relative:
            nm = os.path.join(dir, nm)
        nodes.append(nm)
    nodes.sort()
    return nodes

def find(root, files=True, dirs=False, hidden=False, relative=True, topdown=True):
    root = os.path.join(root, '')  # add slash if not there
    for parent, ldirs, lfiles in os.walk(root, topdown=topdown):
        if relative:
            parent = parent[len(root):]
        if dirs and parent:
            yield os.path.join(parent, '')
        if not hidden:
            lfiles   = [nm for nm in lfiles if not nm.startswith('.')]
            ldirs[:] = [nm for nm in ldirs  if not nm.startswith('.')]  # in place
        if files:
            lfiles.sort()
            for nm in lfiles:
                nm = os.path.join(parent, nm)
                yield nm

def test(root):
    print "* directory listing, with hidden files:"
    print ls(root, hidden=True)
    print
    print "* recursive listing, with dirs, but no hidden files:"
    for f in find(root, dirs=True):
        print f
    print

if __name__ == "__main__":
    test(*sys.argv[1:])

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怀念你的温柔 2024-07-12 03:05:15

递归实现

import os

def scan_dir(dir):
    for name in os.listdir(dir):
        path = os.path.join(dir, name)
        if os.path.isfile(path):
            print path
        else:
            scan_dir(path)

A recursive implementation

import os

def scan_dir(dir):
    for name in os.listdir(dir):
        path = os.path.join(dir, name)
        if os.path.isfile(path):
            print path
        else:
            scan_dir(path)

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皇甫轩 2024-07-12 03:05:15

这是另一种选择。

os.scandir(path='.')

它返回与路径给定的目录中的条目（以及文件属性信息）相对应的 os.DirEntry 对象的迭代器。

示例：

with os.scandir(path) as it:
    for entry in it:
        if not entry.name.startswith('.'):
            print(entry.name)

使用 scandir() 代替 listdir() 可以显着提高还需要文件类型或文件属性信息的代码的性能，因为 os.DirEntry 对象公开此信息如果操作系统在扫描目录时提供它。所有 os.DirEntry 方法都可以执行系统调用，但 is_dir() 和 is_file() 通常只需要符号链接的系统调用； os.DirEntry.stat() 在 Unix 上总是需要系统调用，但在 Windows 上只需要符号链接。

Python 文档

Here is another option.

os.scandir(path='.')

It returns an iterator of os.DirEntry objects corresponding to the entries (along with file attribute information) in the directory given by path.

Example:

with os.scandir(path) as it:
    for entry in it:
        if not entry.name.startswith('.'):
            print(entry.name)

Using scandir() instead of listdir() can significantly increase the performance of code that also needs file type or file attribute information, because os.DirEntry objects expose this information if the operating system provides it when scanning a directory. All os.DirEntry methods may perform a system call, but is_dir() and is_file() usually only require a system call for symbolic links; os.DirEntry.stat() always requires a system call on Unix but only requires one for symbolic links on Windows.

Python Docs

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梦醒时光 2024-07-12 03:05:15

尝试这个：

import os
for top, dirs, files in os.walk('./'):
    for nm in files:       
        print os.path.join(top, nm)

Try this:

import os
for top, dirs, files in os.walk('./'):
    for nm in files:       
        print os.path.join(top, nm)

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迷途知返 2024-07-12 03:05:15

虽然 os.listdir() 可以很好地生成文件和目录名称的列表，但通常您希望在获得这些名称后执行更多操作 - 在 Python3 中，pathlib 使其他杂务变得简单。让我们来看看，看看你是否和我一样喜欢它。

要列出目录内容，请构造一个 Path 对象并获取迭代器：

In [16]: Path('/etc').iterdir()
Out[16]: <generator object Path.iterdir at 0x110853fc0>

如果我们只想要一个事物名称列表：

In [17]: [x.name for x in Path('/etc').iterdir()]
Out[17]:
['emond.d',
 'ntp-restrict.conf',
 'periodic',

如果您只想要目录：

In [18]: [x.name for x in Path('/etc').iterdir() if x.is_dir()]
Out[18]:
['emond.d',
 'periodic',
 'mach_init.d',

如果您想要该树中所有conf文件的名称：

In [20]: [x.name for x in Path('/etc').glob('**/*.conf')]
Out[20]:
['ntp-restrict.conf',
 'dnsextd.conf',
 'syslog.conf',

如果您想要一个列表树中的conf文件数>= 1K：

In [23]: [x.name for x in Path('/etc').glob('**/*.conf') if x.stat().st_size > 1024]
Out[23]:
['dnsextd.conf',
 'pf.conf',
 'autofs.conf',

解析相对路径变得容易：

In [32]: Path('../Operational Metrics.md').resolve()
Out[32]: PosixPath('/Users/starver/code/xxxx/Operational Metrics.md')

使用路径导航非常清晰（尽管出乎意料）：

In [10]: p = Path('.')

In [11]: core = p / 'web' / 'core'

In [13]: [x for x in core.iterdir() if x.is_file()]
Out[13]:
[PosixPath('web/core/metrics.py'),
 PosixPath('web/core/services.py'),
 PosixPath('web/core/querysets.py'),

While os.listdir() is fine for generating a list of file and dir names, frequently you want to do more once you have those names - and in Python3, pathlib makes those other chores simple. Let's take a look and see if you like it as much as I do.

To list dir contents, construct a Path object and grab the iterator:

In [16]: Path('/etc').iterdir()
Out[16]: <generator object Path.iterdir at 0x110853fc0>

If we want just a list of names of things:

In [17]: [x.name for x in Path('/etc').iterdir()]
Out[17]:
['emond.d',
 'ntp-restrict.conf',
 'periodic',

If you want just the dirs:

In [18]: [x.name for x in Path('/etc').iterdir() if x.is_dir()]
Out[18]:
['emond.d',
 'periodic',
 'mach_init.d',

If you want the names of all conf files in that tree:

In [20]: [x.name for x in Path('/etc').glob('**/*.conf')]
Out[20]:
['ntp-restrict.conf',
 'dnsextd.conf',
 'syslog.conf',

If you want a list of conf files in the tree >= 1K:

In [23]: [x.name for x in Path('/etc').glob('**/*.conf') if x.stat().st_size > 1024]
Out[23]:
['dnsextd.conf',
 'pf.conf',
 'autofs.conf',

Resolving relative paths become easy:

In [32]: Path('../Operational Metrics.md').resolve()
Out[32]: PosixPath('/Users/starver/code/xxxx/Operational Metrics.md')

Navigating with a Path is pretty clear (although unexpected):

In [10]: p = Path('.')

In [11]: core = p / 'web' / 'core'

In [13]: [x for x in core.iterdir() if x.is_file()]
Out[13]:
[PosixPath('web/core/metrics.py'),
 PosixPath('web/core/services.py'),
 PosixPath('web/core/querysets.py'),

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葬心 2024-07-12 03:05:15

对于当前工作目录中未指定路径的文件

Python 2.7:

import os
os.listdir('.')

Python 3.x:

import os
os.listdir()

For files in current working directory without specifying a path

Python 2.7:

import os
os.listdir('.')

Python 3.x:

import os
os.listdir()

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不即不离 2024-07-12 03:05:15

如果您需要通配能力，也有一个模块可以实现。例如：

import glob
glob.glob('./[0-9].*')

将返回类似以下内容：

['./1.gif', './2.txt']

请参阅此处的文档。

If you need globbing abilities, there's a module for that as well. For example:

import glob
glob.glob('./[0-9].*')

will return something like:

['./1.gif', './2.txt']

See the documentation here.

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是伱的 2024-07-12 03:05:15

import os

for filename in os.listdir("C:\\temp"):
    print  filename

import os

for filename in os.listdir("C:\\temp"):
    print  filename

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热情消退 2024-07-12 03:05:15

这是我经常使用的辅助函数：

import os

def listdir_fullpath(d):
    return [os.path.join(d, f) for f in os.listdir(d)]

Here's a helper function I use quite often:

import os

def listdir_fullpath(d):
    return [os.path.join(d, f) for f in os.listdir(d)]

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人生戏 2024-07-12 03:05:15

您可以使用

os.listdir(path)

参考和更多操作系统功能，请看这里：

Python 2 文档： https ://docs.python.org/2/library/os.html#os.listdir
Python 3 文档：https://docs.python.org/3/library/os.html#os.listdir

You can use

os.listdir(path)

For reference and more os functions look here:

Python 2 docs: https://docs.python.org/2/library/os.html#os.listdir
Python 3 docs: https://docs.python.org/3/library/os.html#os.listdir

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最丧也最甜 2024-07-12 03:05:15

这是一种遍历目录树中每个文件和目录的方法：

import os

for dirname, dirnames, filenames in os.walk('.'):
    # print path to all subdirectories first.
    for subdirname in dirnames:
        print(os.path.join(dirname, subdirname))

    # print path to all filenames.
    for filename in filenames:
        print(os.path.join(dirname, filename))

    # Advanced usage:
    # editing the 'dirnames' list will stop os.walk() from recursing into there.
    if '.git' in dirnames:
        # don't go into any .git directories.
        dirnames.remove('.git')

This is a way to traverse every file and directory in a directory tree:

import os

for dirname, dirnames, filenames in os.walk('.'):
    # print path to all subdirectories first.
    for subdirname in dirnames:
        print(os.path.join(dirname, subdirname))

    # print path to all filenames.
    for filename in filenames:
        print(os.path.join(dirname, filename))

    # Advanced usage:
    # editing the 'dirnames' list will stop os.walk() from recursing into there.
    if '.git' in dirnames:
        # don't go into any .git directories.
        dirnames.remove('.git')

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~没有更多了~