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Snowflake DSA-C02 Exam - 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?

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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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Jennie
3 months ago
Nope, it's definitely snowpark.ml.modeling.
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Lajuana
3 months ago
I remember it being snowflake.sklearn.modeling.
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Jamal
3 months ago
Wait, is it really that one? I thought it was snowflake.ml.modeling.
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Viki
4 months ago
Agreed, that's the right namespace.
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Chaya
4 months ago
It's definitely snowpark.ml.modeling!
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Alexis
4 months ago
I feel like snowflake.ml.modeling is the correct answer. It aligns with the examples we went through in class.
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Chun
4 months ago
I’m a bit confused. I thought it was snowflake.sklearn.modeling, but now I'm second-guessing myself.
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Marylyn
4 months ago
I remember practicing a question like this, and I think it was snowflake.ml.modeling. That one seems familiar.
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Izetta
5 months ago
I think the namespace is snowpark.ml.modeling, but I'm not entirely sure. It sounds right based on what we covered.
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Shad
5 months ago
I think the answer is A. The question mentions Snowpark, so the namespace should be related to that rather than Snowflake's own ML libraries.
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Nadine
5 months ago
Okay, let me break this down. The question is asking about the namespace for Snowpark ML modeling and preprocessing classes. That means it's likely in the Snowpark namespace, not the Snowflake namespace. I'll go with A.
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Hoa
5 months ago
Hmm, I'm a bit confused on this one. I know Snowpark is a Snowflake feature, but I'm not sure about the exact namespace. I'll have to think this through carefully.
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Jacki
5 months ago
I'm pretty sure it's A. The question specifically mentions Snowpark, so the namespace should be snowpark.ml.modeling.
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Tran
1 year ago
I'm not sure, but I think it's D) snowflake.ml.modeling
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Malcolm
1 year ago
I agree with Nikita, A) snowpark.ml.modeling makes sense
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Nikita
1 year ago
I think it's A) snowpark.ml.modeling
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Vallie
1 year ago
I'm still waiting for Snowflake to come up with a namespace called 'snowlake.ml.modeling' just to keep us on our toes.
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Glen
1 year ago
No, I believe it's D) snowflake.ml.modeling.
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Cecilia
1 year ago
D) snowflake.ml.modeling
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Allene
1 year ago
I think it's A) snowpark.ml.modeling.
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Lucina
1 year ago
A) snowpark.ml.modeling
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Gerald
1 year ago
Hmm, this is a tough one. I'm going to go with A, but I really hope I don't get snow-blindness from all this Snowpark talk.
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Michal
1 year ago
B and C sound like they're trying to trick us. Snowflake doesn't use scikit-learn or sklearn, so those can't be right.
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Felix
1 year ago
I'm leaning towards D. Snowflake has its own ML capabilities, so the namespace would be snowflake.ml.modeling.
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Georgeanna
1 year ago
Let's go with D) snowflake.ml.modeling then.
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Nickolas
1 year ago
I agree with you, D) snowflake.ml.modeling makes sense.
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Jacquelyne
1 year ago
I'm not sure, but I believe it's D) snowflake.ml.modeling.
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Kimberely
1 year ago
I think it's A) snowpark.ml.modeling.
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Essie
1 year 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
1 year ago
Yes, you're correct. Snowpark ML modeling classes are in the snowpark.ml.modeling namespace.
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Micaela
1 year ago
I think you're right, it should be A) snowpark.ml.modeling.
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