Have looked at data on oob but would like to use it as a metric in a grid search on a Random Forest classifier (multiclass) but doesn't seem to be a recognised scorer for the scoring parameter. I do have OoB set to True in the classifier.
Currently using scoring ='accuracy' but would like to change to oob score.
Ideas or comments welcome
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$\begingroup$ See also datascience.stackexchange.com/a/66238/55122 $\endgroup$– Ben Reiniger - on strike ♦Oct 2, 2021 at 13:24
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3 Answers
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Check out the make_scorer function in sklearn: http://scikit-learn.org/stable/modules/generated/sklearn.metrics.make_scorer.html
You may need to code the OoB component yourself, I haven't ever used that metric for scoring before with Sklearn, so I don't know for sure.
Also, consider using random search rather than grid search, it is more practical when dealing with a large hyper parameter space, since it does not need to search the entire dimension of a parameters that have no impact.
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$\begingroup$ Thanks for your response Jinglesting . I did try the make scorer - unsuccessfully but will now try to code the OOB metric . $\endgroup$– RDATAAug 28, 2018 at 14:34
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$\begingroup$ No problem RDATA, if this answer is correct for you, please don't forget to mark it as so! $\endgroup$ Aug 28, 2018 at 15:18
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Sorry - I know this is on old thread but you can try calling model.oob_score_ after running your model fit. It should give the OOB accuracy
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1$\begingroup$ How will you do this with GridSearchCV is the question. $\endgroup$ Jan 4, 2021 at 7:54
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I hope I'm phrasing this correctly.
You can call it as a attribute from the best estimator, i.e.
grid_model.best_estimator_.oob_score_
so you can't really use grid_model.oob_score_ because gridsearchcv does not have such an attribute; but you can call the best instance of the random forest model, then call the attribute.
