如何使用python获取目录中的文件名

Question

There is an mkv file in a folder named " export ". What I want to do is to make a python script which fetches the file name from that export folder. Let's say the folder is at " C:\\Users\\UserName\\Desktop\\New_folder\\export ".

How do I fetch the name?

I tried using this os.path.basename and os.path.splitext .. well.. didn't work out like I expected.

Answer 1

os.path implements some useful functions on pathnames. But it doesn't have access to the contents of the path. For that purpose, you can use os.listdir .

The following command will give you a list of the contents of the given path:

os.listdir("C:\Users\UserName\Desktop\New_folder\export")

Now, if you just want .mkv files you can use fnmatch ( This module provides support for Unix shell-style wildcards ) module to get your expected file names:

import fnmatch
import os

print([f for f in os.listdir("C:\Users\UserName\Desktop\New_folder\export") if fnmatch.fnmatch(f, '*.mkv')])

Also as @Padraic Cunningham mentioned as a more pythonic way for dealing with file names you can use glob module :

map(path.basename,glob.iglob(pth+"*.mkv"))

Answer 2

You can use glob :

from glob import glob

pth ="C:/Users/UserName/Desktop/New_folder/export/"
print(glob(pth+"*.mkv"))

path+"*.mkv" will match all the files ending with .mkv .

To just get the basenames you can use map or a list comp with iglob:

from glob import iglob

print(list(map(path.basename,iglob(pth+"*.mkv"))))


print([path.basename(f) for f in  iglob(pth+"*.mkv")])

iglob returns an iterator so you don't build a list for no reason.

Answer 3

I assume you're basically asking how to list files in a given directory. What you want is:

import os
print os.listdir("""C:\Users\UserName\Desktop\New_folder\export""")

If there's multiple files and you want the one(s) that have a .mkv end you could do:

import os
files = os.listdir("""C:\Users\UserName\Desktop\New_folder\export""")
mkv_files = [_ for _ in files if _[-4:] == ".mkv"]
print mkv_files

Answer 4

If you are searching for recursive folder search, this method will help you to get filename using os.walk , also you can get those file's path and directory using this below code.

import os, fnmatch
for path, dirs, files in os.walk(os.path.abspath(r"C:/Users/UserName/Desktop/New_folder/export/")):
    for filename in fnmatch.filter(files, "*.mkv"):
        print(filename)

Answer 5

You can use glob

import glob
for file in glob.glob('C:\Users\UserName\Desktop\New_folder\export\*.mkv'):
    print(str(file).split('\')[-1])

This will list out all the files having extention .mkv as file.mkv, file2.mkv and so on.

Answer 6

From os.walk you can read file paths as a list

files = [ file_path  for _, _, file_path in os.walk(DIRECTORY_PATH)]
for file_name in files[0]: #note that it has list of lists
    print(file_name)