从嵌套字典的底层创建字典的Pythonic方法

我有一个包含嵌套字典的字典，如下所示：

example_dict = {
                "test1": "string here",
                "test2": "another string",
                "test3": {
                        "test4": 25,
                        "test5": {
                                  "test7": "very nested."
                        },
                        "test6": "yep, another string"
                },
}

假设嵌套在其中的所有键都是唯一的，是否有一种Pythonic方法来获取此字典并获得“底部”级别的键和值

我的意思是递归地进入字典并获取所有键：值对，其中值不是字典（但仍然捕获满足标准的键：值对）？因此，对于上述内容，将返回以下内容：

resulting_dict = {
                "test1": "string here",
                "test2": "another string",
                "test4": 25,
                "test7": "very nested.",
                "test6": "yep, another string"
}

---

通过一点递归，它的实现相当简单：

example_dict = {
    "test1": "string here",
    "test2": "another string",
    "test3": {
        "test4": 25,
        "test5": {
            "test7": "very nested."
        },
        "test6": "yep, another string"
    },
}


def flatten(dictionary):
    output = dict()
    for k, v in dictionary.items():
        if isinstance(v, dict):
            output.update(flatten(v))
        else:
            output[k] = v

    return output

resulting_dict = flatten(example_dict)

---

通过一点递归，它的实现相当简单：

example_dict = {
    "test1": "string here",
    "test2": "another string",
    "test3": {
        "test4": 25,
        "test5": {
            "test7": "very nested."
        },
        "test6": "yep, another string"
    },
}


def flatten(dictionary):
    output = dict()
    for k, v in dictionary.items():
        if isinstance(v, dict):
            output.update(flatten(v))
        else:
            output[k] = v

    return output

resulting_dict = flatten(example_dict)

---

我明白了，在函数中引用函数的逻辑是我所缺少的。谢谢。我明白了，在函数中引用函数的逻辑是我所缺少的。谢谢