将参数映射到值的Python快捷方式？

我有这个功能：

@staticmethod
def get_curl_params(url=None, headers=None, post=None, curl_proxy=None, curl_success=None, should_use_cookies=False,
                    should_follow_location=False, should_include_referer=False, custom_request='Get',
                    should_verify_ssl=False, head=None, timeout_limit=60, site_rate_limit_seconds=None):

    curl_params = CurlParameters()
    curl_params.url = url
    curl_params.headers = headers
    curl_params.post = post
    curl_params.curl_proxy = curl_proxy
    curl_params.curl_success = curl_success
    curl_params.should_use_cookies = should_use_cookies
    curl_params.should_follow_location = should_follow_location
    curl_params.should_include_referer = should_include_referer
    curl_params.custom_request = custom_request
    curl_params.should_verify_ssl = should_verify_ssl
    curl_params.head = head
    curl_params.timeout_limit = timeout_limit
    curl_params.site_rate_limit_seconds = site_rate_limit_seconds
    return curl_params

---

它将创建一个新的curl_params对象。现在，我传递的所有参数最终都被分配给curl_params.{{VARIABLE NAME}。这样做有捷径吗？以这种方式分配它似乎是重复和明显的

我认为这应该行得通

locals

获取所有局部变量（在本例中为函数中的所有变量）的

dict

---

你能用

locals（）

将所有参数传递给构造函数吗？对不起，没有问题，我怎么做？答案的左边应该有一个复选标记（在向下箭头下方）。点击表示这是您问题的正确答案（然后其他人知道问题已解决）。

def get_curl_params(url=None, headers=None, post=None, curl_proxy=None, curl_success=None, should_use_cookies=False,
                    should_follow_location=False, should_include_referer=False, custom_request='Get',
                    should_verify_ssl=False, head=None, timeout_limit=60, site_rate_limit_seconds=None):

    kwargs = locals()

    # if you can pass everything to init, this would be nice
    curl_params = CurlParameters(**kwargs)

    # otherwise, you can do this:
    for name, val in kwargs.items():
        setattr(curl_params, name, val)

    return curl_params