修复JSON扁平化递归函数的AttributeError并实现目标结构

问题描述 #

原始JSON结构：

{
    "Teams" : ["Dortmund", "Real Madrid"] ,
    "Countries" : [
        "Colombia",
        {
            "Italia" : ["Milan", "Inter", "Juventus"]
        },
        {
            "France" : ["PSG", "Lille"]
        },
        "China"
    ]
}

期望扁平化后的字典：

{"Teams": ["Dortmund", "Real Madrid"], "Countries":["Colombia", "Italia", "Milan", "Inter", "Juventus", "France", "PSG", "Lille", "China" ]}

编写的递归函数：

def flatten_json(data):
    result = {}
    for key, value in data.items():
        if isinstance(value, dict):
            result.update(flatten_json(value))
        elif isinstance(value, list):
            for i, item in enumerate(value):
                result.update(flatten_json(item))
        else:
            result[key] = value
    return result

运行时出现错误：

Traceback (most recent call last):
  File "prueba.py", line 46, in <module>
    flattened_dict = flatten_json(organigram)
  File "prueba.py", line 36, in flatten_json
    result.update(flatten_json(item))
  File "prueba.py", line 31, in flatten_json
    for key, value in data.items():
AttributeError: 'str' object has no attribute 'items'

错误原因分析 #

函数递归逻辑错误：处理列表元素时，不管元素类型直接调用flatten_json(item)，当元素是字符串（如"Colombia"）时，函数尝试调用data.items()，但字符串没有items方法，因此抛出AttributeError。同时原函数的逻辑不符合需求——它试图合并所有键值对，而非将嵌套的列表结构展开为平级字符串列表。

解决方案代码 #

针对需求调整逻辑，只处理列表中的嵌套结构，保留原始键并生成平级列表：

def flatten_json(data):
    result = {}
    for key, value in data.items():
        if isinstance(value, list):
            flattened_list = []
            for item in value:
                if isinstance(item, str):
                    # 字符串直接加入结果列表
                    flattened_list.append(item)
                elif isinstance(item, dict):
                    # 先添加字典的键（国家名）
                    for country, clubs in item.items():
                        flattened_list.append(country)
                        # 再扩展俱乐部列表的所有元素
                        flattened_list.extend(clubs)
            result[key] = flattened_list
        else:
            # 非列表类型直接保留原内容
            result[key] = value
    return result

# 测试用例
organigram = {
    "Teams": ["Dortmund", "Real Madrid"],
    "Countries": [
        "Colombia",
        {"Italia": ["Milan", "Inter", "Juventus"]},
        {"France": ["PSG", "Lille"]},
        "China"
    ]
}

print(flatten_json(organigram))

运行结果 #

{'Teams': ['Dortmund', 'Real Madrid'], 'Countries': ['Colombia', 'Italia', 'Milan', 'Inter', 'Juventus', 'France', 'PSG', 'Lille', 'China']}

内容的提问来源于stack exchange，提问作者suribe06