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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 query(self, query_data, k=10, queue_size=5.0):
 """Query the training data for the k nearest neighbors Parameters query_data: array-like, last dimension self.dim An array of points to query k: integer (default = 10) The number of nearest neighbors to return queue_size: float (default 5.0) The multiplier of the internal search queue. This controls the speed/accuracy tradeoff. Low values will search faster but with more approximate results. High values will search more accurately, but will require more computation to do so. Values should generally be in the range 1.0 to 10.0. Returns ------- indices, distances: array (n_query_points, k), array (n_query_points, k) The first array, ``indices``, provides the indices of the data points in the training set that are the nearest neighbors of each query point. Thus ``indices[i, j]`` is the index into the training data of the jth nearest neighbor of the ith query points. Similarly ``distances`` provides the distances to the neighbors of the query points such that ``distances[i, j]`` is the distance from the ith query point to its jth nearest neighbor in the training data. """

        # query_data = check_array(query_data, dtype=np.float64, order='C')
        query_data = np.asarray(query_data).astype(np.float32)
        init = initialise_search(
            self._rp_forest,
            self._raw_data,
            query_data,
            int(k * queue_size),
            self._random_init,
            self._tree_init,
            self.rng_state,
        )
        result = self._search(
            self._raw_data,
            self._search_graph.indptr,
            self._search_graph.indices,
            init,
            query_data,
        )

        indices, dists = deheap_sort(result)
        return indices[:, :k], dists[:, :k]