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
$\begingroup$
$\endgroup$
1
-
$\begingroup$ See also datascience.stackexchange.com/a/66238/55122 $\endgroup$– Ben ReinigerOct 2 at 13:24
Add a comment
|
$\begingroup$
$\endgroup$
2
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.
-
$\begingroup$ Thanks for your response Jinglesting . I did try the make scorer - unsuccessfully but will now try to code the OOB metric . $\endgroup$– RDATAAug 28 '18 at 14:34
-
$\begingroup$ No problem RDATA, if this answer is correct for you, please don't forget to mark it as so! $\endgroup$ Aug 28 '18 at 15:18
$\begingroup$
$\endgroup$
1
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
-
$\begingroup$ How will you do this with GridSearchCV is the question. $\endgroup$ Jan 4 at 7:54
$\begingroup$
$\endgroup$
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.
