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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def find(self, query=None, **kwargs):
 """ You can pass in the appropriate model object from the queries module, or a dictionary with the keys and values for the query, or a set of key=value parameters. """

        url = self.getUrl()
        if query is not None:
            if isinstance(query, queries.SlickQuery):
                url = url + "?" + urlencode(query.to_dict())
            elif isinstance(query, dict):
                url = url + "?" + urlencode(query)
        elif len(kwargs) > 0:
            url = url + "?" + urlencode(kwargs)

        # hopefully when we discover what problems exist in slick to require this, we can take the loop out
        for retry in range(3):
            try:
                self.logger.debug("Making request to slick at url %s", url)
                r = requests.get(url)
                self.logger.debug("Request returned status code %d", r.status_code)
                if r.status_code is 200:
                    retval = []
                    objects = r.json()
                    for dct in objects:
                        retval.append(self.model.from_dict(dct))
                    return retval
                else:
                    self.logger.error("Slick returned an error when trying to access %s: status code %s" % (url, str(r.status_code)))
                    self.logger.error("Slick response: ", pprint.pformat(r))
                    
            except BaseException as error:
                self.logger.warn("Received exception while connecting to slick at %s", url, exc_info=sys.exc_info())
        raise SlickCommunicationError(
            "Tried 3 times to request data from slick at url %s without a successful status code.", url)