以python方式获取“dataPath”文件夹中所有的“。txt”文件名:

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

Path.py是另一种选择:https://github.com/jaraco/path.py

from path import path
p = path('/path/to/the/directory')
for f in p.files(pattern='*.txt'):
    print f

我喜欢os.walk():

import os

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

或者使用生成器:

import os

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)

用一团。

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

我做了一个测试(Python 3.6.4, W7x64)，看看对于一个文件夹，没有子目录，哪种解决方案是最快的，以获得具有特定扩展名的文件的完整文件路径列表。

简而言之，对于这个任务，os.listdir()是最快的，比次优os.walk()快1.7倍，比pathlib快2.7倍，比os.scandir()快3.2倍，比glob快3.3倍。 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/粘贴下面的一个方法，请添加.lower()，否则在搜索.ext时将找不到.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

这样做是可行的:

>>> 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']