仅列出目录的文件并忽略 python 的子目录

list only files of a directory and ignore the subdirectories with python

我想只列出目录的文件并跳过子目录。

当前代码还列出了子目录中的文件,但这不是我想要的:


import os 

list_files = []    
for root, dirs, files in os.walk(input_folder):
        for filename in files:
            joined = os.path.join(input_folder, filename)
            list_files.append(joined)

要处理的目录中的文件如下所示:

/headfolder/subfolder/subfile.pdf     #ignore this directory
/headfolder/subfolder2/subfile.pdf    #ignore this directory
/headfolder/file.pdf                  #get this path 
/headfolder/file2.pdf                 #get this path  

期望的输出:

['/headfolder/file.pdf','/headfolder/file2.pdf ']
files=[i for i in os.listdir() if os.path.isfile(i)]

编辑:

用于列出特定目录中的所有文件

import os
from os import listdir
from os.path import isfile, join

input_folder = 'headfolder/'
list_files = []

files = [f for f in listdir(input_folder) if isfile(join(input_folder, f))]
for filename in files:
    joined = os.path.join(input_folder, filename)
    list_files.append(joined)

print(list_files)