[spark] Make spark model have the same UID with its estimator (#9022)

Signed-off-by: Weichen Xu <weichen.xu@databricks.com>

python-package/xgboost/spark/core.py

@@ -931,7 +931,11 @@ class _SparkXGBEstimator(Estimator, _SparkXGBParams, MLReadable, MLWritable):
         result_xgb_model = self._convert_to_sklearn_model(
             bytearray(booster, "utf-8"), config
         )
-        return self._copyValues(self._create_pyspark_model(result_xgb_model))
+        spark_model = self._create_pyspark_model(result_xgb_model)
+        # According to pyspark ML convention, the model uid should be the same
+        # with estimator uid.
+        spark_model._resetUid(self.uid)
+        return self._copyValues(spark_model)
 
     def write(self):
         """

tests/test_distributed/test_with_spark/test_spark_local.py

@@ -464,6 +464,7 @@ class TestPySparkLocal:
     def test_regressor_basic(self, reg_data: RegData) -> None:
         regressor = SparkXGBRegressor(pred_contrib_col="pred_contribs")
         model = regressor.fit(reg_data.reg_df_train)
+        assert regressor.uid == model.uid
         pred_result = model.transform(reg_data.reg_df_test).collect()
         for row in pred_result:
             np.testing.assert_equal(row.prediction, row.expected_prediction)