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SheffieldML / GPy

Gaussian processes framework in python
BSD 3-Clause "New" or "Revised" License
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Not return/calculate the variance? #1085

Closed Cargonher closed 2 months ago

Cargonher commented 3 months ago

Is there a way to not return (and not compute!) the variance when using predict_noiseless() ? I don't need that information for the moment and I want to save as much computational time as possible

Thank you

MartinBubel commented 3 months ago

Accessing variance may differentiate depending on your setup.

Without more information, I cannot help, I'm afraid. Did you consider looking into the code and trace how predict_noiseless works and where you could acquire the information you won't without additional computations? That might be the best way to get there.

Cargonher commented 3 months ago

Hi Martin,

Maybe I wasn't clear - my question is whether the computation of the variance can be toggled off in predict_noiseless (e.g. there's a boolean variable for this on scikit). I did have a look into the code but couldn't trace back the variance computation nor the corresponding covariance matrices multiplication.

Thank you.

MartinBubel commented 2 months ago

you can set full_covar=False, which onley computes the diagonal entries, surely being less expensive, if that is what you want. I hope that answers your question?

MartinBubel commented 2 months ago

I'm gonna close tihs issue. If you feel like your question has not been answered, feel free to reopen.