如何在单独的列表中收集每个子目录中的文件名

import os

def collect_file_names(directory):
    file_names = []
    
    # 获取主目录下的所有子目录
    sub_directories = [os.path.join(directory, sub_directory) for sub_directory in os.listdir(directory) if os.path.isdir(os.path.join(directory, sub_directory))]
    
    # 遍历每个子目录，获取其中的文件名列表
    for sub_directory in sub_directories:
        files = [file for file in os.listdir(sub_directory) if os.path.isfile(os.path.join(sub_directory, file))]
        file_names.extend(files)
    
    return file_names

# 使用示例
main_directory = '/path/to/main/directory'  # 替换为实际的主目录路径
file_names = collect_file_names(main_directory)
print(file_names)