Python如何合并列表不同类型值为单层字典列表？

给定如下混合了字典、字典子列表的嵌套列表：

data = [
    {"type": "special", "prize": "32220402"},
    {"type": "grand", "prize": "99194290"},
    [
        {"type": "first", "prize": "16525386"},
        {"type": "first", "prize": "28467179"},
        {"type": "first", "prize": "27854976"},
    ],
    [
        {"type": "second", "prize": "6525386"},
        {"type": "second", "prize": "8467179"},
        {"type": "second", "prize": "7854976"},
    ],
]

需要将其展平为仅包含字典的单层列表，期望输出结构为：

[
    {"type": "special", "prize": "32220402"},
    {"type": "grand", "prize": "99194290"},
    {"type": "first", "prize": "16525386"},
    {"type": "first", "prize": "28467179"},
    {"type": "first", "prize": "27854976"},
    {"type": "second", "prize": "6525386"},
    {"type": "second", "prize": "8467179"},
    {"type": "second", "prize": "7854976"},
]

针对当前固定的一层嵌套结构（顶层元素仅为字典或字典组成的子列表，无更深层级嵌套），最简洁优雅的纯原生写法为单行列表推导：

flattened = [item for elem in data for item in (elem if isinstance(elem, list) else [elem])]

• 逻辑说明：遍历原列表每个元素，若元素本身是列表则展开遍历其内部字典，若元素是单个字典则将其包装为单元素列表统一遍历逻辑，最终收集所有字典得到单层结果，无额外依赖、无冗余逻辑，运行结果完全匹配预期。

如果需要兼容任意深度的嵌套列表场景，可以用递归生成器实现，代码可读性和复用性更强：

def flatten_list(input_list):
    for elem in input_list:
        if isinstance(elem, list):
            yield from flatten_list(elem)
        else:
            yield elem

flattened = list(flatten_list(data))

内容的提问来源于stack exchange，提问作者retr0327