我如何在Python中获得给定目录中的所有文件(和目录)的列表?


当前回答

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)]

这段代码列出给定目录名中所有文件和目录的完整路径。

其他回答

简单的方法:

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

虽然os.listdir()用于生成文件和目录名称的列表很好,但一旦有了这些名称,通常您还想做更多的事情——在Python3中,pathlib使这些其他杂事变得简单。让我们看一看,看看你是否和我一样喜欢它。

要列出目录内容,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的列表

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

如果你只想要dirs:

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

如果你想要该树中所有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',

如果你想在树>= 1K中列出conf文件:

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',

解析相对路径变得简单:

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楚(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'),

这是另一种选择。

os.scandir(path='.')

它返回os的迭代器。对应于path指定目录中的条目(以及文件属性信息)的DirEntry对象。

例子:

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文档

试试这个:

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