AUC ROC in keras is different when using tensorflow or scikit functions.












1














Two solutions for using AUC-ROC to train keras models, proposed here worked for me. But using tensorflow or scikit rocauc functions I get different results.



def auc(y_true, y_pred):
auc = tf.metrics.auc(y_true, y_pred)[1]
K.get_session().run(tf.local_variables_initializer())
return auc


and



def auc(y_true, y_pred):
return tf.py_func(roc_auc_score, (y_true, y_pred), tf.double)


Based on the history, it looks like both are being applied to train and validation.
When I plot history metrics, tensorflow curve looks very smoothed compared to scikit.



Shouldn't I get about the same results using both functions?










share|improve this question



























    1














    Two solutions for using AUC-ROC to train keras models, proposed here worked for me. But using tensorflow or scikit rocauc functions I get different results.



    def auc(y_true, y_pred):
    auc = tf.metrics.auc(y_true, y_pred)[1]
    K.get_session().run(tf.local_variables_initializer())
    return auc


    and



    def auc(y_true, y_pred):
    return tf.py_func(roc_auc_score, (y_true, y_pred), tf.double)


    Based on the history, it looks like both are being applied to train and validation.
    When I plot history metrics, tensorflow curve looks very smoothed compared to scikit.



    Shouldn't I get about the same results using both functions?










    share|improve this question

























      1












      1








      1







      Two solutions for using AUC-ROC to train keras models, proposed here worked for me. But using tensorflow or scikit rocauc functions I get different results.



      def auc(y_true, y_pred):
      auc = tf.metrics.auc(y_true, y_pred)[1]
      K.get_session().run(tf.local_variables_initializer())
      return auc


      and



      def auc(y_true, y_pred):
      return tf.py_func(roc_auc_score, (y_true, y_pred), tf.double)


      Based on the history, it looks like both are being applied to train and validation.
      When I plot history metrics, tensorflow curve looks very smoothed compared to scikit.



      Shouldn't I get about the same results using both functions?










      share|improve this question













      Two solutions for using AUC-ROC to train keras models, proposed here worked for me. But using tensorflow or scikit rocauc functions I get different results.



      def auc(y_true, y_pred):
      auc = tf.metrics.auc(y_true, y_pred)[1]
      K.get_session().run(tf.local_variables_initializer())
      return auc


      and



      def auc(y_true, y_pred):
      return tf.py_func(roc_auc_score, (y_true, y_pred), tf.double)


      Based on the history, it looks like both are being applied to train and validation.
      When I plot history metrics, tensorflow curve looks very smoothed compared to scikit.



      Shouldn't I get about the same results using both functions?







      deep-learning keras scikit-learn performance






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 2 hours ago









      aerijman

      164




      164






















          1 Answer
          1






          active

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          2














          No, you shouldn't have the same numbers. All depends on the additional parameters:



          tf.metrics.auc(
          labels,
          predictions,
          weights=None,
          num_thresholds=200,
          metrics_collections=None,
          updates_collections=None,
          curve='ROC',
          name=None,
          summation_method='trapezoidal'
          )


          This means that this curve will have 200 points, so very smooth.



          sklearn version doesn't have this kind of parameters:



          roc_auc_score(y_true, y_score, average=’macro’, sample_weight=None, max_fpr=None)


          The number of outputs depends on the curve and the number of points if I remember properly.






          share|improve this answer





















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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

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            2














            No, you shouldn't have the same numbers. All depends on the additional parameters:



            tf.metrics.auc(
            labels,
            predictions,
            weights=None,
            num_thresholds=200,
            metrics_collections=None,
            updates_collections=None,
            curve='ROC',
            name=None,
            summation_method='trapezoidal'
            )


            This means that this curve will have 200 points, so very smooth.



            sklearn version doesn't have this kind of parameters:



            roc_auc_score(y_true, y_score, average=’macro’, sample_weight=None, max_fpr=None)


            The number of outputs depends on the curve and the number of points if I remember properly.






            share|improve this answer


























              2














              No, you shouldn't have the same numbers. All depends on the additional parameters:



              tf.metrics.auc(
              labels,
              predictions,
              weights=None,
              num_thresholds=200,
              metrics_collections=None,
              updates_collections=None,
              curve='ROC',
              name=None,
              summation_method='trapezoidal'
              )


              This means that this curve will have 200 points, so very smooth.



              sklearn version doesn't have this kind of parameters:



              roc_auc_score(y_true, y_score, average=’macro’, sample_weight=None, max_fpr=None)


              The number of outputs depends on the curve and the number of points if I remember properly.






              share|improve this answer
























                2












                2








                2






                No, you shouldn't have the same numbers. All depends on the additional parameters:



                tf.metrics.auc(
                labels,
                predictions,
                weights=None,
                num_thresholds=200,
                metrics_collections=None,
                updates_collections=None,
                curve='ROC',
                name=None,
                summation_method='trapezoidal'
                )


                This means that this curve will have 200 points, so very smooth.



                sklearn version doesn't have this kind of parameters:



                roc_auc_score(y_true, y_score, average=’macro’, sample_weight=None, max_fpr=None)


                The number of outputs depends on the curve and the number of points if I remember properly.






                share|improve this answer












                No, you shouldn't have the same numbers. All depends on the additional parameters:



                tf.metrics.auc(
                labels,
                predictions,
                weights=None,
                num_thresholds=200,
                metrics_collections=None,
                updates_collections=None,
                curve='ROC',
                name=None,
                summation_method='trapezoidal'
                )


                This means that this curve will have 200 points, so very smooth.



                sklearn version doesn't have this kind of parameters:



                roc_auc_score(y_true, y_score, average=’macro’, sample_weight=None, max_fpr=None)


                The number of outputs depends on the curve and the number of points if I remember properly.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 2 hours ago









                Matthieu Brucher

                45813




                45813






























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