Python：使用*args作为字典键

我正在尝试找出如何获取字典列表：

some_list = [{'a':1, 'b':{'c':2}}, {'a':3, 'b':{'c':4}}, {'a':5, 'b':{'c':6}}] 

然后使用键的参数获取这种情况下的嵌套值

c

。有什么想法吗？我试着做一些事情，比如：

def compare_something(comparison_list, *args):
    new_list = []
    for something in comparison_list:
        company_list.append(company[arg1?][arg2?])
    return new_list
compare_something(some_list, 'b', 'c')

---

但我不太确定如何指定我想要的参数的具体顺序，在这里我需要它们。有什么想法吗？

如果您确定列表中的每个项目实际上都有必要的嵌套字典

for val in comparison_list:
    for a in args:
        val = val[a]
    # Do something with val

---

在中迭代完成的重复解包/导航也可以通过递归完成：

some_list = [{'a':1, 'b':{'c':2}}, {'a':3, 'b':{'c':4}}, {'a':5, 'b':{'c':6}}]


def extract(nested_dictionary, *keys):
    """
    return the object found by navigating the nested_dictionary along the keys
    """
    if keys:
        # There are keys left to process.

        # Unpack one level by navigating to the first remaining key:
        unpacked_value = nested_dictionary[keys[0]]

        # The rest of the keys won't be handled in this recursion.
        unprocessed_keys = keys[1:]  # Might be empty, we don't care here.

        # Handle yet unprocessed keys in the next recursion:
        return extract(unpacked_value, *unprocessed_keys)
    else:
        # There are no keys left to process. Return the input object (of this recursion) as-is.
        return nested_dictionary  # Might or might not be a dict at this point, so disregard the name.


def compare_something(comparison_list, *keys):
    """
    for each nested dictionary in comparison_list, return the object
    found by navigating the nested_dictionary along the keys

    I'm not really sure what this has to do with comparisons.
    """
    return [extract(d, *keys) for d in comparison_list]


compare_something(some_list, 'b', 'c')    # returns [2, 4, 6]

---

你的预期结果是什么？什么是公司名单，在你的职能范围内新的名单在哪里变化，什么是公司？现在你总是返回一个空列表，还是我遗漏了什么？应该是

new\u list.append（

谢谢你尝试只返回嵌套CY的值你也可以遍历嵌套DICT——不过，它的使用现在有点不受欢迎了。。。