在 Python 中查找目录中扩展名为 .txt 的所有文件

问题描述：

.txt如何找到目录中所有具有python扩展名的文件？

解决方案 1：

您可以使用glob：

import glob, os
os.chdir("/mydir")
for file in glob.glob("*.txt"):
    print(file)

或者简单地说os.listdir：

import os
for file in os.listdir("/mydir"):
    if file.endswith(".txt"):
        print(os.path.join("/mydir", file))

或者如果您想遍历目录，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))

解决方案 2：

使用glob。

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

解决方案 3：

您可以简单地使用pathlibs 1：glob

import pathlib

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

或者循环：

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

如果你想要递归的话你可以使用rglob('*.txt')

这就像调用 Path.glob() 并在给定的相对模式前面添加“**/”

1该pathlib模块已包含在 Python 3.4 的标准库中。但即使在较旧的 Python 版本上，您也可以安装该模块的反向移植（即使用conda或pip）：pathlib和pathlib2。

解决方案 4：

类似的东西应该可以完成工作

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

解决方案 5：

类似这样的操作会起作用：

>>> 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']

解决方案 6：

import os

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

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

解决方案 7：

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)

解决方案 8：

以下是相同但产生略有不同的结果的更多版本：

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

解决方案 9：

尝试这个将递归地找到所有文件：

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

#double\\ no single \n
for file in glob.glob("**/*.txt", recursive = True):
    print(file)

解决方案 10：

Python 版本 3.5+

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。即使查找 10,000 个文件，它也很快。

我还添加了一个将输出转换为 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  'lsd' 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  'lsd' 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)

解决方案 11：

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

解决方案 12：

以 Python 方式获取 'dataPath' 文件夹内所有 '.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

解决方案 13：

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))

解决方案 14：

我做了一个测试（Python 3.6.4，W7x64）来查看哪种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。

简而言之，对于这项任务os.listdir()，它是最快的，并且比第二快 1.7 倍：（os.walk()有中断！），比 快 2.7 倍pathlib，比 快 3.2 倍os.scandir()，比 快 3.3 倍glob。

请记住，当您需要递归结果时，这些结果会发生变化。如果您复制/粘贴以下一种方法，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

解决方案 15：

要从同一目录中名为“data”的文件夹中获取“.txt”文件名数组，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")]

解决方案 16：

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

解决方案 17：

这段代码使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)

解决方案 18：

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

解决方案 19：

与 ghostdog 类似的可复制粘贴解决方案：

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

您还可以使用它yield来创建生成器，从而避免组装完整的列表：

def get_all_filepaths(root_path, ext):
    import os
    for root, dirs, files in os.walk(root_path):
        for filename in files:
            if filename.lower().endswith(ext):
                yield os.path.join(root, filename)

解决方案 20：

我建议你使用fnmatch和上面的方法。这样你可以找到以下任何一项：

1. 姓名.txt；

2. 姓名.TXT；

3. 姓名.txt

。

import fnmatch
import os

    for file in os.listdir("/Users/Johnny/Desktop/MyTXTfolder"):
        if fnmatch.fnmatch(file.upper(), '*.TXT'):
            print(file)

解决方案 21：

以下是extend()

types = ('*.jpg', '*.png')
images_list = []
for files in types:
    images_list.extend(glob.glob(os.path.join(path, files)))

解决方案 22：

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"))))

解决方案 23：

如果文件夹包含大量文件或内存有限，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')]

解决方案 24：

使用 Python OS模块查找具有特定扩展名的文件。

简单的例子如下：

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

解决方案 25：

很多用户都回复了os.walk答案，其中包括所有文件，也包括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)

或者对于一次性不需要发电机的情况：

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用途，则可能需要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)]