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

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Apr 13, 2023
Merged

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

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6 changes: 5 additions & 1 deletion python-package/xgboost/spark/core.py
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Original file line number Diff line number Diff line change
Expand Up @@ -931,7 +931,11 @@ def _run_job():
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):
"""
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1 change: 1 addition & 0 deletions tests/test_distributed/test_with_spark/test_spark_local.py
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Expand Up @@ -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)
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