如何在python中找到扩展名为.txt的目录中的所有文件？
答案 0 :(得分：1950)
您可以使用glob：
import glob, os
os.chdir("/mydir")
for file in glob.glob("*.txt"):
print(file)
或只是os.listdir：
import os
for file in os.listdir("/mydir"):
if file.endswith(".txt"):
print(os.path.join("/mydir", file))
或者如果要遍历目录，请使用os.walk：
import os
for root, dirs, files in os.walk("/mydir"):
for file in files:
if file.endswith(".txt"):
print(os.path.join(root, file))
答案 1 :(得分：207)
使用glob。
>>> import glob
>>> glob.glob('./*.txt')
['./outline.txt', './pip-log.txt', './test.txt', './testingvim.txt']
答案 2 :(得分：125)
这样的事情应该做的工作
for root, dirs, files in os.walk(directory):
for file in files:
if file.endswith('.txt'):
print file
答案 3 :(得分：99)
这样的事情会起作用：
>>> 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']
答案 4 :(得分：27)
import os
path = 'mypath/path'
files = os.listdir(path)
files_txt = [i for i in files if i.endswith('.txt')]
答案 5 :(得分：25)
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
或者使用生成器：
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
答案 6 :(得分：21)
以下是产生略有不同结果的更多版本：
import glob
for f in glob.iglob("/mydir/*/*.txt"): # generator, search immediate subdirectories
print f
print glob.glob1("/mydir", "*.tx?") # literal_directory, basename_pattern
import fnmatch, os
print fnmatch.filter(os.listdir("/mydir"), "*.tx?") # include dot-files
答案 7 :(得分：18)
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
答案 8 :(得分：18)
import pathlib
list(pathlib.Path('your_directory').glob('*.txt'))
或循环：
for txt_file in pathlib.Path('your_directory').glob('*.txt'):
# do something with "txt_file"
如果您想要递归，可以使用.glob('**/*.txt)
1 pathlib模块包含在python 3.4的标准库中。但是，即使在较旧的Python版本上（即使用conda或pip），您也可以安装该模块的后端端口：pathlib和pathlib2。
答案 9 :(得分：11)
Python拥有完成此任务的所有工具：
import os
the_dir = 'the_dir_that_want_to_search_in'
all_txt_files = filter(lambda x: x.endswith('.txt'), os.listdir(the_dir))
答案 10 :(得分：9)
将“dataPath”文件夹中的所有“.txt”文件名称作为Pythonic方式的列表
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
答案 11 :(得分：8)
import os
import sys
if len(sys.argv)==2:
print('no params')
sys.exit(1)
dir = sys.argv[1]
mask= sys.argv[2]
files = os.listdir(dir);
res = filter(lambda x: x.endswith(mask), files);
print res
答案 12 :(得分：7)
在递归函数中使用os.scandir的快速方法。搜索文件夹和子文件夹中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)
如果要搜索包含10,000s文件的目录，则附加到列表会变得效率低下。 “屈服”结果是一个更好的解决方案。我还包括一个将输出转换为Pandas Dataframe的函数。
import os
import re
import pandas as pd
import numpy as np
def findFilesInFolderYield(path, extension, containsTxt='', subFolders = True, excludeText = ''):
""" 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
extension: File extension to find. e.g. 'txt'. Regular expression. Or 'ls\d' to match ls1, ls2, ls3 etc
containsTxt: List of Strings, only finds file if it contains this text. Ignore if '' (or blank)
subFolders: Bool. If True, find files in all subfolders under path. If False, only searches files in the specified folder
excludeText: Text string. Ignore if ''. Will exclude if text string is in path.
"""
if type(containsTxt) == str: # if a string and not in a list
containsTxt = [containsTxt]
myregexobj = re.compile('\.' + extension + '$') # Makes sure the file extension is at the end and is preceded by a .
try: # Trapping a OSError or FileNotFoundError: File permissions problem I believe
for entry in os.scandir(path):
if entry.is_file() and myregexobj.search(entry.path): #
bools = [True for txt in containsTxt if txt in entry.path and (excludeText == '' or excludeText not in entry.path)]
if len(bools)== len(containsTxt):
yield entry.stat().st_size, entry.stat().st_atime_ns, entry.stat().st_mtime_ns, entry.stat().st_ctime_ns, entry.path
elif entry.is_dir() and subFolders: # if its a directory, then repeat process as a nested function
yield from findFilesInFolderYield(entry.path, extension, containsTxt, subFolders)
except OSError as ose:
print('Cannot access ' + path +'. Probably a permissions error ', ose)
except FileNotFoundError as fnf:
print(path +' not found ', fnf)
def findFilesInFolderYieldandGetDf(path, extension, containsTxt, subFolders = True, excludeText = ''):
""" Converts returned data from findFilesInFolderYield and creates and Pandas Dataframe.
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
extension: File extension to find. e.g. 'txt'. Regular expression. Or 'ls\d' to match ls1, ls2, ls3 etc
containsTxt: List of Strings, only finds file if it contains this text. Ignore if '' (or blank)
subFolders: Bool. If True, find files in all subfolders under path. If False, only searches files in the specified folder
excludeText: Text string. Ignore if ''. Will exclude if text string is in path.
"""
fileSizes, accessTimes, modificationTimes, creationTimes , paths = zip(*findFilesInFolderYield(path, extension, containsTxt, subFolders))
df = pd.DataFrame({
'FLS_File_Size':fileSizes,
'FLS_File_Access_Date':accessTimes,
'FLS_File_Modification_Date':np.array(modificationTimes).astype('timedelta64[ns]'),
'FLS_File_Creation_Date':creationTimes,
'FLS_File_PathName':paths,
})
df['FLS_File_Modification_Date'] = pd.to_datetime(df['FLS_File_Modification_Date'],infer_datetime_format=True)
df['FLS_File_Creation_Date'] = pd.to_datetime(df['FLS_File_Creation_Date'],infer_datetime_format=True)
df['FLS_File_Access_Date'] = pd.to_datetime(df['FLS_File_Access_Date'],infer_datetime_format=True)
return df
ext = 'txt' # regular expression
containsTxt=[]
path = 'C:\myFolder'
df = findFilesInFolderYieldandGetDf(path, ext, containsTxt, subFolders = True)
答案 13 :(得分：7)
我做了一个测试（Python 3.6.4，W7x64），看看哪个解决方案对于一个文件夹，没有子目录来说是最快的，以获取具有特定扩展名的文件的完整文件路径列表。
为了简短起见，对于此任务os.listdir()是最快的，并且速度是次佳的1.7倍：os.walk()（休息！），2.7倍于{{1}比pathlib快3.2倍，比os.scandir()快3.3倍
请记住，当您需要递归结果时，这些结果会发生变化。如果你复制/粘贴下面的一个方法，请添加.lower（），否则在搜索.ext时找不到.EXT。
glob结果：
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))
答案 14 :(得分：6)
这段代码让我的生活变得更简单。
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)
答案 15 :(得分：5)
使用fnmatch：https://docs.python.org/2/library/fnmatch.html
import fnmatch
import os
for file in os.listdir('.'):
if fnmatch.fnmatch(file, '*.txt'):
print file
答案 16 :(得分：4)
试试这个会递归找到你的所有文件：
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)
答案 17 :(得分：3)
我建议你使用fnmatch和上面的方法。通过这种方式，您可以找到以下任何内容：
import fnmatch
import os
for file in os.listdir("/Users/Johnny/Desktop/MyTXTfolder"):
if fnmatch.fnmatch(file.upper(), '*.TXT'):
print(file)
答案 18 :(得分：3)
要从同一目录中名为“data”的文件夹中获取一组“.txt”文件名，我通常会使用这一简单的代码行：
import os
fileNames = [fileName for fileName in os.listdir("data") if fileName.endswith(".txt")]
答案 19 :(得分：2)
如果文件夹包含大量文件或内存是约束，请考虑使用生成器：
def yield_files_with_extensions(folder_path, file_extension):
for _, _, files in os.walk(folder_path):
for file in files:
if file.endswith(file_extension):
yield file
选项A：迭代
for f in yield_files_with_extensions('.', '.txt'):
print(f)
选项B：全部获取
files = [f for f in yield_files_with_extensions('.', '.txt')]
答案 20 :(得分：2)
类似于ghostdog的复制可解决方案：
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
答案 21 :(得分：2)
这是一个extend()
types = ('*.jpg', '*.png')
images_list = []
for files in types:
images_list.extend(glob.glob(os.path.join(path, files)))
答案 22 :(得分：2)
使用for循环的简单方法：
import os
dir = ["e","x","e"]
p = os.listdir('E:') #path
for n in range(len(p)):
name = p[n]
myfile = [name[-3],name[-2],name[-1]] #for .txt
if myfile == dir :
print(name)
else:
print("nops")
虽然这可以更加概括。
答案 23 :(得分：2)
带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"))))
答案 24 :(得分：1)
使用Python OS模块查找具有特定扩展名的文件。
这里有一个简单的例子：
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
答案 25 :(得分：0)
许多用户回复了os.walk个答案，其中包括所有文件以及所有目录和子目录及其文件。
import os
def files_in_dir(path, extension=''):
"""
Generator: yields all of the files in <path> ending with
<extension>
\param path Absolute or relative path to inspect,
\param extension [optional] Only yield files matching this,
\yield [filenames]
"""
for _, dirs, files in os.walk(path):
dirs[:] = [] # do not recurse directories.
yield from [f for f in files if f.endswith(extension)]
# Example: print all the .py files in './python'
for filename in files_in_dir('./python', '*.py'):
print("-", filename)
或者对于不需要发电机的一次性接收：
path, ext = "./python", ext = ".py"
for _, _, dirfiles in os.walk(path):
matches = (f for f in dirfiles if f.endswith(ext))
break
for filename in matches:
print("-", filename)
如果您要将匹配用于其他内容，您可能希望将其设为列表而不是生成器表达式：
matches = [f for f in dirfiles if f.endswith(ext)]