Snowflake Exam DSA-C02 Topic 2 Question 28 Discussion

Actual exam question for Snowflake's DSA-C02 exam

Question #: 28
Topic #: 2

[All DSA-C02 Questions]

All Snowpark ML modeling and preprocessing classes are in the ________ namespace?

Asnowpark.ml.modeling

Bsnowflake.sklearn.modeling

Csnowflake.scikit.modeling

Dsnowflake.ml.modeling

Show Suggested Answer

Suggested Answer: D

All Snowpark ML modeling and preprocessing classes are in the snowflake.ml.modeling namespace. The Snowpark ML modules have the same name as the corresponding module from the sklearn namespace. For example, the Snowpark ML module corresponding to sklearn.calibration is snow-flake.ml.modeling.calibration.

The xgboost and lightgbm modules correspond to snowflake.ml.modeling.xgboost and snow-flake.ml.modeling.lightgbm, respectively.

Not all of the classes from scikit-learn are supported in Snowpark ML.

by Remedios at Sep 13, 2024, 02:58 AM

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29 days ago

I'm leaning towards D. Snowflake has its own ML capabilities, so the namespace would be snowflake.ml.modeling.

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Jacquelyne

7 days ago

I'm not sure, but I believe it's D) snowflake.ml.modeling.

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Kimberely

16 days ago

I think it's A) snowpark.ml.modeling.

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Essie

1 months ago

I'm pretty sure it's A. Snowpark is Snowflake's own ML platform, so the namespace would be snowpark.ml.modeling.

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Teddy

16 days ago

Yes, you're correct. Snowpark ML modeling classes are in the snowpark.ml.modeling namespace.

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Micaela

30 days ago

I think you're right, it should be A) snowpark.ml.modeling.

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