Python如何将层级路径列表转换为对应层级的嵌套字典

实现代码 #

你可以直接使用下方的Python代码完成转换，逻辑是逐层遍历路径节点，中间节点自动创建嵌套字典，末尾节点赋值为空字符串：

def paths_to_nested_dict(mapping_list):
    nested_dict = {}
    for full_path in mapping_list:
        path_parts = full_path.split('/')
        current_level = nested_dict
        # 处理中间层级，创建嵌套字典
        for part in path_parts[:-1]:
            if part not in current_level:
                current_level[part] = {}
            current_level = current_level[part]
        # 末尾层级赋值为空字符串
        current_level[path_parts[-1]] = ""
    return nested_dict

# 调用测试
mapping_list = ['location/name', 'location/address/address1', 'location/address/zip', 'location/business/business_name', 'occupant/occupant_type']
result = paths_to_nested_dict(mapping_list)

# 格式化打印验证结果
import json
print(json.dumps(result, indent=4, ensure_ascii=False))

运行后输出的结构和要求的完全一致。

内容的提问来源于stack exchange，提问作者mongotop