python 两个文件夹里的文件名对比

比如需要一个xml对应一个jpg时，有时候会不小心少了其中一个文件，这时可以用以下代码比较缺少的是哪个文件：

# -*- coding: utf-8 -*-
import os

path1 = r'../diff'

def file_name(file_dir):
    jpg_list = []
    xml_list = []
    for root, dirs, files in os.walk(file_dir):
        for file in files:
            if os.path.splitext(file)[1] == '.jpg':
                jpg_list.append(os.path.splitext(file)[0])
            elif os.path.splitext(file)[1] == '.xml':
                xml_list.append(os.path.splitext(file)[0])

    diff = set(xml_list).difference(set(jpg_list))  # 差集，在a中但不在b中的元素
    print(len(diff))
    for name in diff:
        print("no jpg", name + ".xml")

    diff2 = set(jpg_list).difference(set(xml_list))  # 差集，在b中但不在a中的元素
    print(len(diff2))
    for name in diff2:
        print("no xml", name + ".jpg")
    return jpg_list,xml_list

    # 其中os.path.splitext()函数将路径拆分为文件名+扩展名

if __name__ == '__main__':

    file_name(path1)