Python实现含列表的嵌套字典全路径提取函数需求

算力阿峰

2026-6-26

Python 实现嵌套字典/列表的全路径遍历函数

以下是满足需求的Python函数，可递归遍历包含列表的嵌套字典，生成所有可能的路径：

def get_all_paths(nested_dict):
    paths = []
    
    def traverse(current_obj, current_path):
        if isinstance(current_obj, dict):
            for key, value in current_obj.items():
                new_path = f"{current_path}/{key}" if current_path else key
                traverse(value, new_path)
        elif isinstance(current_obj, list):
            for item in current_obj:
                traverse(item, current_path)
        else:
            if current_obj == "":
                paths.append(f"{current_path}/")
            else:
                paths.append(f"{current_path}/{current_obj}")
    
    for key, value in nested_dict.items():
        traverse(value, key)
    
    return paths

测试示例

使用你提供的输入变量测试：

var = {
    "a": "b",
    "c": ["d", "e", {"f": "", "h": ["i", {"j": "q"} ]}],
    "m": "",
    "n": ["d", "e", {"b": "", "c": "", "f": "r"}],
    "r": "r",
    "k": ["d", "t", "h"]
}

# 获取所有路径并打印
for path in get_all_paths(var):
    print(path)

运行后输出的路径内容与示例一致，顺序可能略有不同（需求允许顺序不固定）。