Question

我有一个看起来像这样的函数：

def request( request = None, account_id = 0, user_id = 0, user = {"ID":0}, headers = None):
    """Send a User request to the server
    Send a request to the server
    """
    url = api_root + "/accounts/" + str(account_id)

    function_dictionary = {'get_user_from_id': (requests.get,                       #Function
                                               (url + "/users/" + str(user_id),),      #Tuple of Arguments
                                               {'headers': headers}),               #Dictionary of keyword args

                               'get_user_from_username': (requests.get,
                                                (url + "/users?username=" + str(user_id),),
                                                {'headers': headers}),

                               'get_user_from_email': (requests.get,
                                             (url + "/users?email=" + str(user_id),),
                                             {'headers': headers}),

                               'delete': (requests.delete,
                                          (url + "/users/" + str(user_id),),
                                          {'headers': headers}),

                               'patch_user':(requests.patch,
                                             (url + "/users/" + str(user["ID"]),),
                                             {'headers': headers, 'data':json.dumps(user)}),

                               'post': (requests.post,
                                        (url + "/users" ,),
                                        {'headers': headers,'data':json.dumps(user)})
                               }

    func, args, kwargs = function_dictionary[request]
    result = func(*args, **kwargs)

    #Throw exception if non-200 response
    result.raise_for_status()

    #Result from query
    print "Query " + request + " result: " + result.text

    return result

是否可以从request函数中删除此字典，以便每次调用请求时都不会重新构建该字典？

主要问题是 - 在调用函数时需要插入变量，即user_id, user, account_id

Answer 1

不容易。

但我不知道你会这样做的原因。在每个请求上构建六项词典的开销是绝对无关紧要的。

从函数中提取包含变量的字典

1 个答案: