Python v3.5 +

在递归函数中使用os.scandir的快速方法。 在文件夹和子文件夹中搜索具有指定扩展名的所有文件。

import os

def findFilesInFolder(path, pathList, extension, subFolders = True):
    """  Recursive function to find all files of an extension type in a folder (and optionally in all subfolders too)

    path:        Base directory to find files
    pathList:    A list that stores all paths
    extension:   File extension to find
    subFolders:  Bool.  If True, find files in all subfolders under path. If False, only searches files in the specified folder
    """

    try:   # Trapping a OSError:  File permissions problem I believe
        for entry in os.scandir(path):
            if entry.is_file() and entry.path.endswith(extension):
                pathList.append(entry.path)
            elif entry.is_dir() and subFolders:   # if its a directory, then repeat process as a nested function
                pathList = findFilesInFolder(entry.path, pathList, extension, subFolders)
    except OSError:
        print('Cannot access ' + path +'. Probably a permissions error')

    return pathList

dir_name = r'J:\myDirectory'
extension = ".txt"

pathList = []
pathList = findFilesInFolder(dir_name, pathList, extension, True)

Python有所有的工具来做到这一点：

import os

the_dir = 'the_dir_that_want_to_search_in'
all_txt_files = filter(lambda x: x.endswith('.txt'), os.listdir(the_dir))

与ghostdog类似的可复制解决方案：

def get_all_filepaths(root_path, ext):
    """
    Search all files which have a given extension within root_path.

    This ignores the case of the extension and searches subdirectories, too.

    Parameters
    ----------
    root_path : str
    ext : str

    Returns
    -------
    list of str

    Examples
    --------
    >>> get_all_filepaths('/run', '.lock')
    ['/run/unattended-upgrades.lock',
     '/run/mlocate.daily.lock',
     '/run/xtables.lock',
     '/run/mysqld/mysqld.sock.lock',
     '/run/postgresql/.s.PGSQL.5432.lock',
     '/run/network/.ifstate.lock',
     '/run/lock/asound.state.lock']
    """
    import os
    all_files = []
    for root, dirs, files in os.walk(root_path):
        for filename in files:
            if filename.lower().endswith(ext):
                all_files.append(os.path.join(root, filename))
    return all_files

以'Pythonic'方式获取'dataPath'文件夹内的所有'.txt'文件名作为列表

from os import listdir
from os.path import isfile, join
path = "/dataPath/"
onlyTxtFiles = [f for f in listdir(path) if isfile(join(path, f)) and  f.endswith(".txt")]
print onlyTxtFiles

你可以简单地使用pathlib s glob ¹ ：

import pathlib

list(pathlib.Path('your_directory').glob('*.txt'))

或者在一个循环中：

for txt_file in pathlib.Path('your_directory').glob('*.txt'):
    # do something with "txt_file"

如果你想递归，你可以使用.glob('**/*.txt)

¹ pathlib模块包含在python 3.4的标准库中。 但是，即使在较早的Python版本（即使用conda或pip ）上，您也可以安装该模块的后端端口： pathlib和pathlib2 。

你可以试试这个代码

import glob
import os
filenames_without_extension = [os.path.basename(c).split('.')[0:1][0] for c in glob.glob('your/files/dir/*.txt')]
filenames_with_extension = [os.path.basename(c) for c in glob.glob('your/files/dir/*.txt')]

使用glob 。

>>> import glob
>>> glob.glob('./*.txt')
['./outline.txt', './pip-log.txt', './test.txt', './testingvim.txt']

使用Python OS模块来查找具有特定扩展名的文件。

简单的例子在这里：

import os

# This is the path where you want to search
path = r'd:'  

# this is extension you want to detect
extension = '.txt'   # this can be : .jpg  .png  .xls  .log .....

for root, dirs_list, files_list in os.walk(path):
    for file_name in files_list:
        if os.path.splitext(file_name)[-1] == extension:
            file_name_path = os.path.join(root, file_name)
            print file_name
            print file_name_path   # This is the full path of the filter file

像这样的东西可以工作：

>>> import os
>>> path = '/usr/share/cups/charmaps'
>>> text_files = [f for f in os.listdir(path) if f.endswith('.txt')]
>>> text_files
['euc-cn.txt', 'euc-jp.txt', 'euc-kr.txt', 'euc-tw.txt', ... 'windows-950.txt']

具有子目录的功能解决方案：

from fnmatch import filter
from functools import partial
from itertools import chain
from os import path, walk

print(*chain(*(map(partial(path.join, root), filter(filenames, "*.txt")) for root, _, filenames in walk("mydir"))))

我做了一个测试（Python 3.6.4，W7x64），以查看哪个解决方案对于一个文件夹（无子目录）最快，以获取具有特定扩展名的文件的完整文件路径列表。

为了简短os.listdir() ，对于这个任务os.listdir()是最快的，并且是下一个最好的pathlib倍： os.walk() （有一个中断！）， pathlib与pathlib一样快，比3.2x快pathlib倍os.scandir()和3.3倍于glob 。
请记住，当您需要递归结果时，这些结果将会改变。 如果您复制/粘贴下面的一种方法，请添加.lower（），否则在搜索.ext时不会找到EXT。

import os
import pathlib
import timeit
import glob

def a():
    path = pathlib.Path().cwd()
    list_sqlite_files = [str(f) for f in path.glob("*.sqlite")]

def b(): 
    path = os.getcwd()
    list_sqlite_files = [f.path for f in os.scandir(path) if os.path.splitext(f)[1] == ".sqlite"]

def c():
    path = os.getcwd()
    list_sqlite_files = [os.path.join(path, f) for f in os.listdir(path) if f.endswith(".sqlite")]

def d():
    path = os.getcwd()
    os.chdir(path)
    list_sqlite_files = [os.path.join(path, f) for f in glob.glob("*.sqlite")]

def e():
    path = os.getcwd()
    list_sqlite_files = [os.path.join(path, f) for f in glob.glob1(str(path), "*.sqlite")]

def f():
    path = os.getcwd()
    list_sqlite_files = []
    for root, dirs, files in os.walk(path):
        for file in files:
            if file.endswith(".sqlite"):
                list_sqlite_files.append( os.path.join(root, file) )
        break



print(timeit.timeit(a, number=1000))
print(timeit.timeit(b, number=1000))
print(timeit.timeit(c, number=1000))
print(timeit.timeit(d, number=1000))
print(timeit.timeit(e, number=1000))
print(timeit.timeit(f, number=1000))

结果：

# Python 3.6.4
0.431
0.515
0.161
0.548
0.537
0.274

我喜欢os.walk() ：

import os, os.path

for root, dirs, files in os.walk(dir):
    for f in files:
        fullpath = os.path.join(root, f)
        if os.path.splitext(fullpath)[1] == '.txt':
            print fullpath

或者与发电机：

import os, os.path

fileiter = (os.path.join(root, f)
    for root, _, files in os.walk(dir)
    for f in files)
txtfileiter = (f for f in fileiter if os.path.splitext(f)[1] == '.txt')
for txt in txtfileiter:
    print txt

类似的东西应该能够完成这项工作

for root, dirs, files in os.walk(directory):
    for file in files:
        if file.endswith('.txt'):
            print file

要从同一个目录中的“data”文件夹中获取“.txt”文件名的数组，我通常使用以下简单的代码行：

import os
fileNames = [fileName for fileName in os.listdir("data") if fileName.endswith(".txt")]

试试这个会发现文件夹或文件夹内的所有文件

import glob, os
os.chdir("H:\\wallpaper")# use whatever you directory 

#double\\ no single \

for file in glob.glob("**/*.psd", recursive = True):#your format
    print(file)

这段代码让我的生活更简单。

import os
fnames = ([file for root, dirs, files in os.walk(dir)
    for file in files
    if file.endswith('.txt') #or file.endswith('.png') or file.endswith('.pdf')
    ])
for fname in fnames: print(fname)

这里的更多版本产生的结果稍有不同：

glob.iglob()

import glob
for f in glob.iglob("/mydir/*/*.txt"): # generator, search immediate subdirectories 
    print f

glob.glob1（）

print glob.glob1("/mydir", "*.tx?")  # literal_directory, basename_pattern

fnmatch.filter()

import fnmatch, os
print fnmatch.filter(os.listdir("/mydir"), "*.tx?") # include dot-files

import glob
import os

path=os.getcwd()

extensions=('*.py','*.cpp')

for i in extensions:
  for files in glob.glob(i):
     print files

import os

path = 'mypath/path' 
files = os.listdir(path)

files_txt = [i for i in files if i.endswith('.txt')]

import os
[x for x in os.listdir() if x.endswith(".txt")]

在DIR和SUBDIRS中有多少文件？

如果你想知道在一个dir和subdirs中有多少个filese：

在这个例子中，我们查找包含在所有目录及其子目录中的文件数量。

import os    

def count(dir, counter=0):
    "returns number of files in dir and subdirs"
    for pack in os.walk(dir):
        for f in pack[2]:
            counter += 1
    return dir + " : " + str(counter) + "files"


print(count("F:\\python"))

产量

'F：\ python'：12057 files'