如何快速实现Python嵌套字典到指定列表结构的转换？

你可以利用列表推导式结合字典解包语法快速完成转换，无需手动逐个构建每个列表元素，代码简洁高效：

输入字典 #

input_dict = {
    "demographicAll": {
        "allAges": {
            "rating": "8.8",
            "votes": "2327099"
        },
        "agesUnder18": {
            "rating": "9.0",
            "votes": "1050"
        },
        "ages18To29": {
            "rating": "9.0",
            "votes": "363089"
        },
        "ages30To44": {
            "rating": "8.8",
            "votes": "914737"
        },
        "agesOver45": {
            "rating": "8.2",
            "votes": "182612"
        }
    }
}

转换代码 #

output_dict = {
    "demographicAll": [
        {"ageRange": age, **data} for age, data in input_dict["demographicAll"].items()
    ]
}

代码说明 #

• 列表推导式遍历input_dict["demographicAll"]的所有键值对（age是年龄区间键名，data是对应的评分、投票字典）
• 用{"ageRange": age, **data}生成每个列表元素：ageRange字段对应原键名，**data直接解包原字典中的rating和votes字段，无需手动逐个添加

验证输出 #

如果需要格式化打印结果，可以用json模块：

import json
print(json.dumps(output_dict, indent=4))

输出结果完全符合目标结构：

{
    "demographicAll": [
        {
            "ageRange": "allAges",
            "rating": "8.8",
            "votes": "2327099"
        },
        {
            "ageRange": "agesUnder18",
            "rating": "9.0",
            "votes": "1050"
        },
        {
            "ageRange": "ages18To29",
            "rating": "9.0",
            "votes": "363089"
        },
        {
            "ageRange": "ages30To44",
            "rating": "8.8",
            "votes": "914737"
        },
        {
            "ageRange": "agesOver45",
            "rating": "8.2",
            "votes": "182612"
        }
    ]
}

内容的提问来源于stack exchange，提问作者rizerkrof