如何在python中找到扩展名为.txt的目录中的所有文件?
当前回答
许多用户都回复了os。Walk回答,其中包括所有文件,还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)
如果你打算为其他东西使用匹配,你可能想让它成为一个列表,而不是一个生成器表达式:
matches = [f for f in dirfiles if f.endswith(ext)]
其他回答
这样的东西就可以了
for root, dirs, files in os.walk(directory):
for file in files:
if file.endswith('.txt'):
print(file)
这里有更多相同的版本,产生略微不同的结果:
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
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']
Python有所有工具可以做到这一点:
import os
the_dir = 'the_dir_that_want_to_search_in'
all_txt_files = filter(lambda x: x.endswith('.txt'), os.listdir(the_dir))
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