Low score in Linear Regression with discrete attributes












0














I'm trying to do a linear regression in my dataframe. The dataframe is about apple applications, and I want to predict the notes of applications. The notes are in following format: 



1.0

1.5

2.0

2.5

...

5.0


My code is:



atributos = ['size_bytes','price','rating_count_tot','cont_rating','sup_devices_num','num_screenshots','num_lang','vpp_lic']

atrib_prev = ['nota']



X = np.array(data_regress.drop(['nota'],1))

y = np.array(data_regress['nota'])



X = preprocessing.scale(X)



X_train, X_test, y_train, y_test = cross_validation.train_test_split(X, y, test_size=0.2)



clf = LinearRegression()

clf.fit(X_train, y_train)



accuracy = clf.score(X_test, y_test)



print(accuracy)


But my accuracy is 0.046295306696438665. I think this occurs because the linear model is predicting real values, while my 'note' is real, but at intervals. I don't know how to round this values before the clf.score.






pandas jupyter-notebook sklearn-pandas







share|improve this question





























    0














    I'm trying to do a linear regression in my dataframe. The dataframe is about apple applications, and I want to predict the notes of applications. The notes are in following format: 



    1.0
    
1.5
    
2.0
    
2.5
    
...
    
5.0


    My code is:



    atributos = ['size_bytes','price','rating_count_tot','cont_rating','sup_devices_num','num_screenshots','num_lang','vpp_lic']
    
atrib_prev = ['nota']
    

    
X = np.array(data_regress.drop(['nota'],1))
    
y = np.array(data_regress['nota'])
    

    
X = preprocessing.scale(X)
    

    
X_train, X_test, y_train, y_test = cross_validation.train_test_split(X, y, test_size=0.2)
    

    
clf = LinearRegression()
    
clf.fit(X_train, y_train)
    

    
accuracy = clf.score(X_test, y_test)
    

    
print(accuracy)


    But my accuracy is 0.046295306696438665. I think this occurs because the linear model is predicting real values, while my 'note' is real, but at intervals. I don't know how to round this values before the clf.score.






    pandas jupyter-notebook sklearn-pandas







    share|improve this question



























      0












      0








      0







      I'm trying to do a linear regression in my dataframe. The dataframe is about apple applications, and I want to predict the notes of applications. The notes are in following format: 



      1.0
      
1.5
      
2.0
      
2.5
      
...
      
5.0


      My code is:



      atributos = ['size_bytes','price','rating_count_tot','cont_rating','sup_devices_num','num_screenshots','num_lang','vpp_lic']
      
atrib_prev = ['nota']
      

      
X = np.array(data_regress.drop(['nota'],1))
      
y = np.array(data_regress['nota'])
      

      
X = preprocessing.scale(X)
      

      
X_train, X_test, y_train, y_test = cross_validation.train_test_split(X, y, test_size=0.2)
      

      
clf = LinearRegression()
      
clf.fit(X_train, y_train)
      

      
accuracy = clf.score(X_test, y_test)
      

      
print(accuracy)


      But my accuracy is 0.046295306696438665. I think this occurs because the linear model is predicting real values, while my 'note' is real, but at intervals. I don't know how to round this values before the clf.score.






      pandas jupyter-notebook sklearn-pandas







      share|improve this question















      I'm trying to do a linear regression in my dataframe. The dataframe is about apple applications, and I want to predict the notes of applications. The notes are in following format: 



      1.0
      
1.5
      
2.0
      
2.5
      
...
      
5.0


      My code is:



      atributos = ['size_bytes','price','rating_count_tot','cont_rating','sup_devices_num','num_screenshots','num_lang','vpp_lic']
      
atrib_prev = ['nota']
      

      
X = np.array(data_regress.drop(['nota'],1))
      
y = np.array(data_regress['nota'])
      

      
X = preprocessing.scale(X)
      

      
X_train, X_test, y_train, y_test = cross_validation.train_test_split(X, y, test_size=0.2)
      

      
clf = LinearRegression()
      
clf.fit(X_train, y_train)
      

      
accuracy = clf.score(X_test, y_test)
      

      
print(accuracy)


      But my accuracy is 0.046295306696438665. I think this occurs because the linear model is predicting real values, while my 'note' is real, but at intervals. I don't know how to round this values before the clf.score.






      pandas jupyter-notebook sklearn-pandas




      pandas jupyter-notebook sklearn-pandas






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 14 '18 at 12:56









      Aqueous Carlos

      293213




      293213










      asked Nov 13 '18 at 0:10









      Giovanni BrogiatoGiovanni Brogiato

      1




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          First, for regression models, clf.score() calculates R-squared value, not accuracy. So you would need to decide if you want to treat this problem as a classification problem (For some fixed number of target labels) or a regression problem (for a real-valued target)



          Secondly, if you insist on using regression models and not classification, you can call clf.predict() to first get the predicted values and then round off as you want to, and then call r2_score() on actual and predicted labels. Something like:



          # Get actual predictions
          
y_pred = clf.predict(X_test)
          

          
# You will need to implement the round function yourself
          
y_pred_rounded = round(y_pred)
          

          
# Call the appropriate scorer
          
score = r2_score(y_test, y_pred_rounded)


          You can look at the sklearn documentation here for available metrics in sklearn.






          share|improve this answer





















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            First, for regression models, clf.score() calculates R-squared value, not accuracy. So you would need to decide if you want to treat this problem as a classification problem (For some fixed number of target labels) or a regression problem (for a real-valued target)



            Secondly, if you insist on using regression models and not classification, you can call clf.predict() to first get the predicted values and then round off as you want to, and then call r2_score() on actual and predicted labels. Something like:



            # Get actual predictions
            
y_pred = clf.predict(X_test)
            

            
# You will need to implement the round function yourself
            
y_pred_rounded = round(y_pred)
            

            
# Call the appropriate scorer
            
score = r2_score(y_test, y_pred_rounded)


            You can look at the sklearn documentation here for available metrics in sklearn.






            share|improve this answer


























              0














              First, for regression models, clf.score() calculates R-squared value, not accuracy. So you would need to decide if you want to treat this problem as a classification problem (For some fixed number of target labels) or a regression problem (for a real-valued target)



              Secondly, if you insist on using regression models and not classification, you can call clf.predict() to first get the predicted values and then round off as you want to, and then call r2_score() on actual and predicted labels. Something like:



              # Get actual predictions
              
y_pred = clf.predict(X_test)
              

              
# You will need to implement the round function yourself
              
y_pred_rounded = round(y_pred)
              

              
# Call the appropriate scorer
              
score = r2_score(y_test, y_pred_rounded)


              You can look at the sklearn documentation here for available metrics in sklearn.






              share|improve this answer
























                0












                0








                0






                First, for regression models, clf.score() calculates R-squared value, not accuracy. So you would need to decide if you want to treat this problem as a classification problem (For some fixed number of target labels) or a regression problem (for a real-valued target)



                Secondly, if you insist on using regression models and not classification, you can call clf.predict() to first get the predicted values and then round off as you want to, and then call r2_score() on actual and predicted labels. Something like:



                # Get actual predictions
                
y_pred = clf.predict(X_test)
                

                
# You will need to implement the round function yourself
                
y_pred_rounded = round(y_pred)
                

                
# Call the appropriate scorer
                
score = r2_score(y_test, y_pred_rounded)


                You can look at the sklearn documentation here for available metrics in sklearn.






                share|improve this answer












                First, for regression models, clf.score() calculates R-squared value, not accuracy. So you would need to decide if you want to treat this problem as a classification problem (For some fixed number of target labels) or a regression problem (for a real-valued target)



                Secondly, if you insist on using regression models and not classification, you can call clf.predict() to first get the predicted values and then round off as you want to, and then call r2_score() on actual and predicted labels. Something like:



                # Get actual predictions
                
y_pred = clf.predict(X_test)
                

                
# You will need to implement the round function yourself
                
y_pred_rounded = round(y_pred)
                

                
# Call the appropriate scorer
                
score = r2_score(y_test, y_pred_rounded)


                You can look at the sklearn documentation here for available metrics in sklearn.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 14 '18 at 14:25









                Vivek KumarVivek Kumar

                15.5k41953




                15.5k41953






























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