如何优化 Python 中嵌套到平面级字典的转换？ [复制]

Question

目标是转换下面的嵌套字典

secondary_citing_paper = [{"paper_href": 'Unique One', 'Paper_year': 1}, \
                          {"paper_href": 'Unique Two', 'Paper_year': 2}]
inner_level = [secondary_citing_paper, secondary_citing_paper]
my_dict_x = [inner_level, inner_level]

到Python中的平级字典（对不起，这里的术语使用得更好！），如下

expected_output = [{"paper_href": 'Unique One', 'Paper_year': 1}, \
                   {"paper_href": 'Unique Two', 'Paper_year': 2}, \
                   {"paper_href": 'Unique One', 'Paper_year': 1}, \
                   {"paper_href": 'Unique Two', 'Paper_year': 2}, \
                   {"paper_href": 'Unique One', 'Paper_year': 1}, \
                   {"paper_href": 'Unique Two', 'Paper_year': 2}, \
                   {"paper_href": 'Unique One', 'Paper_year': 1}, \
                   {"paper_href": 'Unique Two', 'Paper_year': 2}, \
                   ]

以下代码已起草

expected_output = []
for my_dict in my_dict_x:
    for the_ref in my_dict:
        for x_ref in the_ref:
            expected_output.append( x_ref )

虽然代码服务于它的目的，但我想知道是否存在更多 Pythonic 方法？

请注意，我在 SO 上发现了几个问题，但它是关于恰好合并 2 个字典。

编辑： 由于与类似问题相关，该主题已关闭，我无法删除该主题，因为 Vishal Singh 已发布他的建议。

不过，根据OP 的建议，递归转换的一种方法如下

def flatten(container):
    for i in container:
        if isinstance(i, (list,tuple)):
            yield from flatten(i)
        else:
            yield i


expected_output=list(flatten(my_dict_x))

或更快的iteration approach，

def flatten(items, seqtypes=(list, tuple)):
    for i, x in enumerate(items):
        while i < len(items) and isinstance(items[i], seqtypes):
            items[i:i+1] = items[i]
    return items

expected_output = flatten(my_dict_x[:])

Answer 1

您的代码的更 Python 版本将如下所示

expected_output = [
    x_ref
    for my_dict in my_dict_x
    for the_ref in my_dict
    for x_ref in the_ref
]