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What will GridsearchCV choose if there are multiple estimators having the same score?


How to graph grid scores from GridSearchCV?GridSeachCV with multiple scoring functions?Difference between GridSearchCV score and error functionSklearn - GridSearchCV with v_measure_score is NOT the sameHow to pass two estimator objects to sklearn's GridSearchCV so that they have the same parameters in each step?Using GridSearchCV and a Random Forest Regressor with the same parameters gives different resultsHow to set own scoring with GridSearchCV from sklearn for regression?Custom scoring function GridsearchCVGridsearchCV Negative Scorewhy GridSearchCv behave different in same code






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1















I'm using RandomForestClassifier in sklearn, and using GridsearchCV for getting best estimator.



I'm wondering when there are many estimators (from simple one to complex one) having the same scores in GridsearchCV, what will be the resulted estimator out of GridsearchCV? The simplest one? or random one?










share|improve this question






























    1















    I'm using RandomForestClassifier in sklearn, and using GridsearchCV for getting best estimator.



    I'm wondering when there are many estimators (from simple one to complex one) having the same scores in GridsearchCV, what will be the resulted estimator out of GridsearchCV? The simplest one? or random one?










    share|improve this question


























      1












      1








      1








      I'm using RandomForestClassifier in sklearn, and using GridsearchCV for getting best estimator.



      I'm wondering when there are many estimators (from simple one to complex one) having the same scores in GridsearchCV, what will be the resulted estimator out of GridsearchCV? The simplest one? or random one?










      share|improve this question














      I'm using RandomForestClassifier in sklearn, and using GridsearchCV for getting best estimator.



      I'm wondering when there are many estimators (from simple one to complex one) having the same scores in GridsearchCV, what will be the resulted estimator out of GridsearchCV? The simplest one? or random one?







      scikit-learn random-forest gridsearchcv






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Mar 27 at 12:02









      GUANGYUAN PIAOGUANGYUAN PIAO

      335 bronze badges




      335 bronze badges

























          1 Answer
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          2














          GridSearchCV does not assess the model complexity (though that would be a neat feature). Neither does it choose among the best models randomly.



          Instead, GridSearchCV simply performs an np.argmin() on the stored errors. See the corresponding line in the source code.



          Now, according to the NumPy docs,




          In case of multiple occurrences of the minimum values, the indices corresponding to the first occurrence are returned.




          That is, GridSearchCV will always select the first among the best models.






          share|improve this answer
























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            1 Answer
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            active

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            active

            oldest

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            active

            oldest

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            2














            GridSearchCV does not assess the model complexity (though that would be a neat feature). Neither does it choose among the best models randomly.



            Instead, GridSearchCV simply performs an np.argmin() on the stored errors. See the corresponding line in the source code.



            Now, according to the NumPy docs,




            In case of multiple occurrences of the minimum values, the indices corresponding to the first occurrence are returned.




            That is, GridSearchCV will always select the first among the best models.






            share|improve this answer





























              2














              GridSearchCV does not assess the model complexity (though that would be a neat feature). Neither does it choose among the best models randomly.



              Instead, GridSearchCV simply performs an np.argmin() on the stored errors. See the corresponding line in the source code.



              Now, according to the NumPy docs,




              In case of multiple occurrences of the minimum values, the indices corresponding to the first occurrence are returned.




              That is, GridSearchCV will always select the first among the best models.






              share|improve this answer



























                2












                2








                2







                GridSearchCV does not assess the model complexity (though that would be a neat feature). Neither does it choose among the best models randomly.



                Instead, GridSearchCV simply performs an np.argmin() on the stored errors. See the corresponding line in the source code.



                Now, according to the NumPy docs,




                In case of multiple occurrences of the minimum values, the indices corresponding to the first occurrence are returned.




                That is, GridSearchCV will always select the first among the best models.






                share|improve this answer













                GridSearchCV does not assess the model complexity (though that would be a neat feature). Neither does it choose among the best models randomly.



                Instead, GridSearchCV simply performs an np.argmin() on the stored errors. See the corresponding line in the source code.



                Now, according to the NumPy docs,




                In case of multiple occurrences of the minimum values, the indices corresponding to the first occurrence are returned.




                That is, GridSearchCV will always select the first among the best models.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Mar 27 at 13:28









                rvfrvf

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