在 Python 中查找扩展名为 .txt 的目录中的所有文件

Question

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如何在python中找到扩展名为.txt的目录中的所有文件？

Answer 1

您可以使用glob ：

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

或者干脆os.listdir ：

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

或者如果你想遍历目录，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))

Answer 2

使用glob 。

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

Answer 3

这样的事情应该可以完成工作

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

Answer 4

像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']

Answer 5

您可以简单地使用pathlib s glob ¹ ：

import pathlib

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

或在循环中：

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

如果你想要它递归你可以使用.glob('**/*.txt')

---

¹ pathlib模块包含在 python 3.4 的标准库中。 但是您甚至可以在较旧的 Python 版本（即使用conda或pip ）上安装该模块的后向端口： pathlib和pathlib2 。

Answer 6

import os

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

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

Answer 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)

或者使用发电机：

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)

Answer 8

以下是更多版本的相同结果，但结果略有不同：

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

Answer 9

蟒蛇 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。 即使查找 10,000 个文件，它也很快。

我还包含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  '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)

Answer 10

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

Answer 11

试试这个，这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 \

for file in glob.glob("**/*.txt", recursive = True):
    print(file)

Answer 12

要以 Pythonic 方式将“dataPath”文件夹中的所有“.txt”文件名作为列表获取：

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

Answer 13

Python 拥有执行此操作的所有工具：

import os

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

Answer 14

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最快，没有子目录，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，并且是下一个最好的os.walk() pathlib倍（有休息时间！），是pathlib 2.7 倍，比pathlib快pathlib倍os.scandir()并且比glob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。 如果您复制/粘贴下面的一种方法，请添加一个 .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

Answer 15

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

Answer 16

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")]

Answer 17

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)

Answer 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

Answer 19

我建议你使用fnmatch和上层方法。 通过这种方式，您可以找到以下任何一项：

1. 姓名。 .txt ;
2. 姓名。 文本文件；
3. 姓名。 文本

.

import fnmatch
import os

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

Answer 20

一种类似于 ghostdog 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yield创建生成器，从而避免组装完整列表：

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)

Answer 21

这是一个extend()

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

Answer 22

带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"))))

Answer 23

如果文件夹包含大量文件或内存受限，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：获得所有

files = [f for f in yield_files_with_extensions('.', '.txt')]

Answer 24

使用 Python OS模块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

Answer 25

许多用户回复了os.walk答案，其中包括所有文件，但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)

或者对于一个不需要发电机的人：

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)

如果您打算将匹配项用于其他内容，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)]

Answer 26

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")

虽然这可以更概括。