基于路径列表检测目标编号文件夹的video/image子文件夹缺失情况

需求说明 #

给定路径列表all_path和需要检查的子文件夹前缀列表subfolder_startswith（固定为["video","image"]），需判断每个编号文件夹下是否存在这两个子文件夹，输出缺少对应子文件夹的编号及缺失项，格式为编号: 缺失的子文件夹。

输入示例 #

all_path = [
    "my_path/output/12345/video/folder/1",
    "my_path/output/12345/video/folder/2",
    "my_path/output/12345/video/folder/3",
    "my_path/output/98745/video/folder/1",
    "my_path/output/98745/video/folder/2",
    "my_path/output/90000/video/folder/1",
    "my_path/output/12345/image/folder/10",
    "my_path/output/12345/image/folder/9",
    "my_path/output/12345/image/folder/8",
    "my_path/output/98745/image/folder/3",
    "my_path/output/98745/image/folder/9",
    "my_path/output/11145/image/folder/8"
]
subfolder_startswith = ["video","image"]

输出示例 #

11145: video
90000: image

解法实现（Python） #

def find_missing_subfolders(all_path, subfolder_list):
    # 用字典存储每个编号对应的已存在子文件夹集合
    folder_map = {}
    for path in all_path:
        parts = path.split('/')
        # 提取编号（路径中第3段，索引从0开始）
        folder_id = parts[2]
        # 提取子文件夹名称（路径中第4段）
        subfolder = parts[3]
        # 初始化集合，添加子文件夹
        if folder_id not in folder_map:
            folder_map[folder_id] = set()
        folder_map[folder_id].add(subfolder)
    
    # 遍历所有编号，检查缺失的子文件夹
    missing = []
    for folder_id, existing_subfolders in folder_map.items():
        for required_sub in subfolder_list:
            if required_sub not in existing_subfolders:
                missing.append(f"{folder_id}: {required_sub}")
    
    return missing

# 执行函数并输出结果
result = find_missing_subfolders(all_path, subfolder_startswith)
for line in result:
    print(line)

代码逻辑说明 #

1. 路径解析：将每个路径按/分割，提取出编号（路径的第3个分段）和子文件夹名称（第4个分段）。
2. 构建映射关系：用字典记录每个编号下已存在的子文件夹，使用集合避免重复统计。
3. 检查缺失项：对每个编号，对比需要检查的子文件夹列表，找出未在集合中的项，整理成要求的格式。
4. 输出结果：遍历缺失列表，逐行打印。

内容的提问来源于stack exchange，提问作者Beyond