def register_dict(data: dict, scopes: tuple, v) -> dict:
    temp = data
    for scope in scopes[:-1]:
        if temp.get(scope) is None:
            temp[scope] = {}
        temp = temp[scope]
    a = temp.get(scopes[-1], [])
    a.append(v)
    temp[scopes[-1]] = a
    return data


if __name__ == '__main__':
    ss = [
        ("a1", "b1", "c"),
        ("a1", "b1", "d"),
        ("a1", "b6", "v"),
        ("a2", "b2", "e"),
        ("a2", "b3", "f")
    ]
    d = {}
    for s in ss:
        register_dict(d, s[:-1], s[-1])
    print(d)

谢谢，感觉您的方法比我写的三个循环的更好。

ss = [
        ("a1", "b1", "c"),
        ("a1", "b1", "d"),
        ("a1", "b6", "v"),
        ("a2", "b2", "e"),
        ("a2", "b3", "f")
    ]

result = {}
for (a, b, c) in ss:
    if a not in result:
        result[a] = {b: [c]}
    elif b not in result[a]:
        result[a][b] = [c]
    elif c not in result[a][b]:
        result[a][b].append(c)

print(result)

谢谢，如果是

{"index": 0, "name": a[0], "slice": [1, 2]},
{"index": 1, "name": a[1], "slice": [1, 9]},
{"index": 2, "name": a[2], "slice": [1, 22]}

这种，index 和 name 两个 key 的值有关联，但是另一个 key 的值完全没关系，有没有简易表示的方法呢？
@yiwen985 @yzx200712256