使用递归生成器：

def hierarchy(d, prefix=None):
    if isinstance(d, dict):
        for k, v in d.items():
            prefix2 = f'{prefix}.{k}' if prefix else k
            yield prefix2
            if isinstance(v, list):
                seen = set()
                for x in v:
                    if isinstance(x, dict):
                        yield from hierarchy({k: v for k, v in x.items()
                                              if k not in seen},
                                             prefix=prefix2)
                        seen.update(x.keys())
                    else:
                        yield from hierarchy(x, prefix=prefix2)
            elif isinstance(v, dict):
                yield from hierarchy(v, prefix=prefix2)
                
out = list(hierarchy(inpt))

# validation
assert out == expected_output

输出：

['boo', 'soo', 'roo', 'qoo', 'fizz', 'buzz', 'yoyo', 'wazzw', 'bnn',
 'cc', 'cc.boo', 'cc.soo', 'cc.roo', 'cc.qoo', 'cc.fizz', 'cc.buzz',
 'cc.yoyo', 'cc.wazzw', 'cc.nummm', 'cc.bsdfff', 'cc.hgdjgkk', 'cc.opu', 'cc.mnb',
 'soos', 'tyu', 'dddd3']

不同示例：

list(hierarchy({'l1': {'l2': {'l3': 'test', 'l4': [['abc'], {'l5': 'def'}]}}}))
# ['l1', 'l1.l2', 'l1.l2.l3', 'l1.l2.l4', 'l1.l2.l4.l5']

莫兹韦回答的修正https://www.mycompiler.io/view/6LB7k4TVOuj

# Includes $ for root node, and [] where access is through an array

def hierarchy(struct, path=None):
    if isinstance(struct, dict):
        path = path if path else '$'
        return set(
            child_path
                for key, obj   in struct.items()
                for child_path in hierarchy(obj, f'{path}.{key}')
        ).union(
            [path]
        )
    elif isinstance(struct, list):
        path = f'{path}[]' if path else '$[]'
        return set(
            child_path
                for obj        in struct
                for child_path in hierarchy(obj, path)
        ).union(
            [path]
        )
    else:
        return [path]

或者...

from itertools import chain

# Excludes those $ and [] markers

def hierarchy2(d):
    if isinstance(d, dict):
        return set(
            f'{k}.{x}' if x else k
                for k,v in d.items()
                for x in chain([''], hierarchy2(v))
        )
    elif isinstance(d, list):
        return set(
            v
                for l in d
                for v in hierarchy2(l)
                    if v
        )
    else:
        return set()

要处理子列表，您可以迭代地检查每个子项是否是一个字典，如果是，则递归地将子字典的键路径追加到当前键：

def get_keys(d):
    keys = []
    for key, value in d.items():
        if isinstance(value, list):
            for obj in value:
                if isinstance(obj, dict):
                    for path in get_keys(obj):
                        keys.append(f'{key}.{path}')
                else:
                    keys.append(key)
        else:
            keys.append(key)
    return keys

因此，对于给定示例input，get_keys(input)将返回：

['boo', 'soo', 'roo', 'qoo', 'fizz', 'buzz', 'yoyo', 'wazzw', 'bnn', 'cc.boo', 'cc.soo', 'cc.roo', 'cc.qoo', 'cc.fizz', 'cc.buzz', 'cc.yoyo', 'cc.wazzw', 'cc.nummm', 'cc.bsdfff', 'cc.hgdjgkk', 'cc.opu', 'cc.mnb', 'cc.boo', 'cc.soo', 'cc.roo', 'cc.qoo', 'cc.fizz', 'cc.buzz', 'cc.yoyo', 'cc.wazzw', 'cc.nummm', 'cc.bsdfff', 'cc.hgdjgkk', 'cc.opu', 'cc.mnb', 'soos', 'tyu', 'dddd3']

演示：https://replit.com/@blhsing/OpenGoldenIntegrationtesting