normalization/denormalization for linear regression problem












1












$begingroup$


My question is simple actually, I have two features that have big difference in scale. So I used a simple normalization by dividing the scale=np.max(array) for both data and lables. Then after prediction, I mulitiplied this scale value back.



But since I used a DNN, wouldn't the nonlinear change the scale so make the multiply not valid? e.g.



given input data: X, label: y;

y' = y/scale

X' = X/scale

predicted = f(X')

predicted_update = predicted * scale


Anyone could provide some advice on whether I could do this or it's actually not correct? How do we handle this kind of problem?






machine-learning linear-regression preprocessing







share|improve this question











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    1












    $begingroup$


    My question is simple actually, I have two features that have big difference in scale. So I used a simple normalization by dividing the scale=np.max(array) for both data and lables. Then after prediction, I mulitiplied this scale value back.



    But since I used a DNN, wouldn't the nonlinear change the scale so make the multiply not valid? e.g.



    given input data: X, label: y;
    
y' = y/scale
    
X' = X/scale
    
predicted = f(X')
    
predicted_update = predicted * scale


    Anyone could provide some advice on whether I could do this or it's actually not correct? How do we handle this kind of problem?






    machine-learning linear-regression preprocessing







    share|improve this question











    $endgroup$




    bumped to the homepage by Community♦ 2 mins ago


    This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.


















      1












      1








      1


      1



      $begingroup$


      My question is simple actually, I have two features that have big difference in scale. So I used a simple normalization by dividing the scale=np.max(array) for both data and lables. Then after prediction, I mulitiplied this scale value back.



      But since I used a DNN, wouldn't the nonlinear change the scale so make the multiply not valid? e.g.



      given input data: X, label: y;
      
y' = y/scale
      
X' = X/scale
      
predicted = f(X')
      
predicted_update = predicted * scale


      Anyone could provide some advice on whether I could do this or it's actually not correct? How do we handle this kind of problem?






      machine-learning linear-regression preprocessing







      share|improve this question











      $endgroup$




      My question is simple actually, I have two features that have big difference in scale. So I used a simple normalization by dividing the scale=np.max(array) for both data and lables. Then after prediction, I mulitiplied this scale value back.



      But since I used a DNN, wouldn't the nonlinear change the scale so make the multiply not valid? e.g.



      given input data: X, label: y;
      
y' = y/scale
      
X' = X/scale
      
predicted = f(X')
      
predicted_update = predicted * scale


      Anyone could provide some advice on whether I could do this or it's actually not correct? How do we handle this kind of problem?






      machine-learning linear-regression preprocessing




      machine-learning linear-regression preprocessing






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited May 15 '18 at 8:26









      Toros91

      2,0042829




      2,0042829










      asked May 15 '18 at 8:22









      user2189731user2189731

      163




      163





      bumped to the homepage by Community♦ 2 mins ago


      This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.







      bumped to the homepage by Community♦ 2 mins ago


      This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
























          1 Answer
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          $begingroup$

          I think it is ok, as long as your training and test data have the same maximum values for every feature, approximately. The idea is that the scaling has to be done with the training set (remember that using the test set for anything that is not testing is illegal, not even for scaling).



          So, you actually fit $y'$ as a function of $X'$, and you have a model that maps properly $y' = f(X')$. When you get your test data, you just obtain the predictions by doing $f(X_{test}')$. As the paragraph before states, if you have that $scale approx scale_{test}$, then you can just recover $y_{test}$ by doing $y_{test} = scale cdot f(X_{test}')$.



          Edit: Don't worry about nonlinearities



          Even if the function $f$ is highly nonlinear, it is a function capable of mapping $X'$ to $y'$. If you trust this function and trust the fact that $y = y' cdot scale$, then there is no need to worry about the way $f$ acts, as function composition makes sense for all kind of functions, both linear and nonlinear.






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for the comment. For training/test set, I'm splitting the whole data set. And both are scaled before training/testing. And the np.max() is a simplied notation. I use separate scale for different features though I could use one. The question is actually about the bold line you mentioned. Why should i not worry about nonlinearities? I'm doubting the validity of y=y' * scale
            $endgroup$
            – user2189731
            May 16 '18 at 7:11












          • $begingroup$
            If $scale approx scale_{test}$, then $y_{test} = scale cdot y_{test}'$. It is completely valid, scaling and getting back to usual scale are inverse transformations.
            $endgroup$
            – David Masip
            May 16 '18 at 7:29










          • $begingroup$
            Suppose I use a universal scale for all X, and y, regardless of training/test data. Then y=f(X) is my original problem and what I wanted to learn is f. After scaling, I actually is training with y'_train = g(X'_train) and then I get my $y'_{test} = g(X'_{test})$. To make sure y'_test == y_test, we have to make sure g=~f. Am I right?
            $endgroup$
            – user2189731
            May 18 '18 at 3:09












          • $begingroup$
            Correction to last sentence: since $ y_{test} = f(X_{test}) $ and $ y'_{test} = g(X'_{test}) $, to make sure $y_{test} == scale * y'_{test}$, I have to make sure $f(X_{test}) = scale * g(X'_{test}) $ . Which equals to: $f(X_{test}) = scale * g(X_{test}/scale) $ and this is not gauranteed?
            $endgroup$
            – user2189731
            May 18 '18 at 3:21














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          $begingroup$

          I think it is ok, as long as your training and test data have the same maximum values for every feature, approximately. The idea is that the scaling has to be done with the training set (remember that using the test set for anything that is not testing is illegal, not even for scaling).



          So, you actually fit $y'$ as a function of $X'$, and you have a model that maps properly $y' = f(X')$. When you get your test data, you just obtain the predictions by doing $f(X_{test}')$. As the paragraph before states, if you have that $scale approx scale_{test}$, then you can just recover $y_{test}$ by doing $y_{test} = scale cdot f(X_{test}')$.



          Edit: Don't worry about nonlinearities



          Even if the function $f$ is highly nonlinear, it is a function capable of mapping $X'$ to $y'$. If you trust this function and trust the fact that $y = y' cdot scale$, then there is no need to worry about the way $f$ acts, as function composition makes sense for all kind of functions, both linear and nonlinear.






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for the comment. For training/test set, I'm splitting the whole data set. And both are scaled before training/testing. And the np.max() is a simplied notation. I use separate scale for different features though I could use one. The question is actually about the bold line you mentioned. Why should i not worry about nonlinearities? I'm doubting the validity of y=y' * scale
            $endgroup$
            – user2189731
            May 16 '18 at 7:11












          • $begingroup$
            If $scale approx scale_{test}$, then $y_{test} = scale cdot y_{test}'$. It is completely valid, scaling and getting back to usual scale are inverse transformations.
            $endgroup$
            – David Masip
            May 16 '18 at 7:29










          • $begingroup$
            Suppose I use a universal scale for all X, and y, regardless of training/test data. Then y=f(X) is my original problem and what I wanted to learn is f. After scaling, I actually is training with y'_train = g(X'_train) and then I get my $y'_{test} = g(X'_{test})$. To make sure y'_test == y_test, we have to make sure g=~f. Am I right?
            $endgroup$
            – user2189731
            May 18 '18 at 3:09












          • $begingroup$
            Correction to last sentence: since $ y_{test} = f(X_{test}) $ and $ y'_{test} = g(X'_{test}) $, to make sure $y_{test} == scale * y'_{test}$, I have to make sure $f(X_{test}) = scale * g(X'_{test}) $ . Which equals to: $f(X_{test}) = scale * g(X_{test}/scale) $ and this is not gauranteed?
            $endgroup$
            – user2189731
            May 18 '18 at 3:21


















          0












          $begingroup$

          I think it is ok, as long as your training and test data have the same maximum values for every feature, approximately. The idea is that the scaling has to be done with the training set (remember that using the test set for anything that is not testing is illegal, not even for scaling).



          So, you actually fit $y'$ as a function of $X'$, and you have a model that maps properly $y' = f(X')$. When you get your test data, you just obtain the predictions by doing $f(X_{test}')$. As the paragraph before states, if you have that $scale approx scale_{test}$, then you can just recover $y_{test}$ by doing $y_{test} = scale cdot f(X_{test}')$.



          Edit: Don't worry about nonlinearities



          Even if the function $f$ is highly nonlinear, it is a function capable of mapping $X'$ to $y'$. If you trust this function and trust the fact that $y = y' cdot scale$, then there is no need to worry about the way $f$ acts, as function composition makes sense for all kind of functions, both linear and nonlinear.






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for the comment. For training/test set, I'm splitting the whole data set. And both are scaled before training/testing. And the np.max() is a simplied notation. I use separate scale for different features though I could use one. The question is actually about the bold line you mentioned. Why should i not worry about nonlinearities? I'm doubting the validity of y=y' * scale
            $endgroup$
            – user2189731
            May 16 '18 at 7:11












          • $begingroup$
            If $scale approx scale_{test}$, then $y_{test} = scale cdot y_{test}'$. It is completely valid, scaling and getting back to usual scale are inverse transformations.
            $endgroup$
            – David Masip
            May 16 '18 at 7:29










          • $begingroup$
            Suppose I use a universal scale for all X, and y, regardless of training/test data. Then y=f(X) is my original problem and what I wanted to learn is f. After scaling, I actually is training with y'_train = g(X'_train) and then I get my $y'_{test} = g(X'_{test})$. To make sure y'_test == y_test, we have to make sure g=~f. Am I right?
            $endgroup$
            – user2189731
            May 18 '18 at 3:09












          • $begingroup$
            Correction to last sentence: since $ y_{test} = f(X_{test}) $ and $ y'_{test} = g(X'_{test}) $, to make sure $y_{test} == scale * y'_{test}$, I have to make sure $f(X_{test}) = scale * g(X'_{test}) $ . Which equals to: $f(X_{test}) = scale * g(X_{test}/scale) $ and this is not gauranteed?
            $endgroup$
            – user2189731
            May 18 '18 at 3:21
















          0












          0








          0





          $begingroup$

          I think it is ok, as long as your training and test data have the same maximum values for every feature, approximately. The idea is that the scaling has to be done with the training set (remember that using the test set for anything that is not testing is illegal, not even for scaling).



          So, you actually fit $y'$ as a function of $X'$, and you have a model that maps properly $y' = f(X')$. When you get your test data, you just obtain the predictions by doing $f(X_{test}')$. As the paragraph before states, if you have that $scale approx scale_{test}$, then you can just recover $y_{test}$ by doing $y_{test} = scale cdot f(X_{test}')$.



          Edit: Don't worry about nonlinearities



          Even if the function $f$ is highly nonlinear, it is a function capable of mapping $X'$ to $y'$. If you trust this function and trust the fact that $y = y' cdot scale$, then there is no need to worry about the way $f$ acts, as function composition makes sense for all kind of functions, both linear and nonlinear.






          share|improve this answer











          $endgroup$



          I think it is ok, as long as your training and test data have the same maximum values for every feature, approximately. The idea is that the scaling has to be done with the training set (remember that using the test set for anything that is not testing is illegal, not even for scaling).



          So, you actually fit $y'$ as a function of $X'$, and you have a model that maps properly $y' = f(X')$. When you get your test data, you just obtain the predictions by doing $f(X_{test}')$. As the paragraph before states, if you have that $scale approx scale_{test}$, then you can just recover $y_{test}$ by doing $y_{test} = scale cdot f(X_{test}')$.



          Edit: Don't worry about nonlinearities



          Even if the function $f$ is highly nonlinear, it is a function capable of mapping $X'$ to $y'$. If you trust this function and trust the fact that $y = y' cdot scale$, then there is no need to worry about the way $f$ acts, as function composition makes sense for all kind of functions, both linear and nonlinear.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited May 15 '18 at 10:58

























          answered May 15 '18 at 8:59









          David MasipDavid Masip

          2,5711428




          2,5711428












          • $begingroup$
            Thanks for the comment. For training/test set, I'm splitting the whole data set. And both are scaled before training/testing. And the np.max() is a simplied notation. I use separate scale for different features though I could use one. The question is actually about the bold line you mentioned. Why should i not worry about nonlinearities? I'm doubting the validity of y=y' * scale
            $endgroup$
            – user2189731
            May 16 '18 at 7:11












          • $begingroup$
            If $scale approx scale_{test}$, then $y_{test} = scale cdot y_{test}'$. It is completely valid, scaling and getting back to usual scale are inverse transformations.
            $endgroup$
            – David Masip
            May 16 '18 at 7:29










          • $begingroup$
            Suppose I use a universal scale for all X, and y, regardless of training/test data. Then y=f(X) is my original problem and what I wanted to learn is f. After scaling, I actually is training with y'_train = g(X'_train) and then I get my $y'_{test} = g(X'_{test})$. To make sure y'_test == y_test, we have to make sure g=~f. Am I right?
            $endgroup$
            – user2189731
            May 18 '18 at 3:09












          • $begingroup$
            Correction to last sentence: since $ y_{test} = f(X_{test}) $ and $ y'_{test} = g(X'_{test}) $, to make sure $y_{test} == scale * y'_{test}$, I have to make sure $f(X_{test}) = scale * g(X'_{test}) $ . Which equals to: $f(X_{test}) = scale * g(X_{test}/scale) $ and this is not gauranteed?
            $endgroup$
            – user2189731
            May 18 '18 at 3:21




















          • $begingroup$
            Thanks for the comment. For training/test set, I'm splitting the whole data set. And both are scaled before training/testing. And the np.max() is a simplied notation. I use separate scale for different features though I could use one. The question is actually about the bold line you mentioned. Why should i not worry about nonlinearities? I'm doubting the validity of y=y' * scale
            $endgroup$
            – user2189731
            May 16 '18 at 7:11












          • $begingroup$
            If $scale approx scale_{test}$, then $y_{test} = scale cdot y_{test}'$. It is completely valid, scaling and getting back to usual scale are inverse transformations.
            $endgroup$
            – David Masip
            May 16 '18 at 7:29










          • $begingroup$
            Suppose I use a universal scale for all X, and y, regardless of training/test data. Then y=f(X) is my original problem and what I wanted to learn is f. After scaling, I actually is training with y'_train = g(X'_train) and then I get my $y'_{test} = g(X'_{test})$. To make sure y'_test == y_test, we have to make sure g=~f. Am I right?
            $endgroup$
            – user2189731
            May 18 '18 at 3:09












          • $begingroup$
            Correction to last sentence: since $ y_{test} = f(X_{test}) $ and $ y'_{test} = g(X'_{test}) $, to make sure $y_{test} == scale * y'_{test}$, I have to make sure $f(X_{test}) = scale * g(X'_{test}) $ . Which equals to: $f(X_{test}) = scale * g(X_{test}/scale) $ and this is not gauranteed?
            $endgroup$
            – user2189731
            May 18 '18 at 3:21


















          $begingroup$
          Thanks for the comment. For training/test set, I'm splitting the whole data set. And both are scaled before training/testing. And the np.max() is a simplied notation. I use separate scale for different features though I could use one. The question is actually about the bold line you mentioned. Why should i not worry about nonlinearities? I'm doubting the validity of y=y' * scale
          $endgroup$
          – user2189731
          May 16 '18 at 7:11






          $begingroup$
          Thanks for the comment. For training/test set, I'm splitting the whole data set. And both are scaled before training/testing. And the np.max() is a simplied notation. I use separate scale for different features though I could use one. The question is actually about the bold line you mentioned. Why should i not worry about nonlinearities? I'm doubting the validity of y=y' * scale
          $endgroup$
          – user2189731
          May 16 '18 at 7:11














          $begingroup$
          If $scale approx scale_{test}$, then $y_{test} = scale cdot y_{test}'$. It is completely valid, scaling and getting back to usual scale are inverse transformations.
          $endgroup$
          – David Masip
          May 16 '18 at 7:29




          $begingroup$
          If $scale approx scale_{test}$, then $y_{test} = scale cdot y_{test}'$. It is completely valid, scaling and getting back to usual scale are inverse transformations.
          $endgroup$
          – David Masip
          May 16 '18 at 7:29












          $begingroup$
          Suppose I use a universal scale for all X, and y, regardless of training/test data. Then y=f(X) is my original problem and what I wanted to learn is f. After scaling, I actually is training with y'_train = g(X'_train) and then I get my $y'_{test} = g(X'_{test})$. To make sure y'_test == y_test, we have to make sure g=~f. Am I right?
          $endgroup$
          – user2189731
          May 18 '18 at 3:09






          $begingroup$
          Suppose I use a universal scale for all X, and y, regardless of training/test data. Then y=f(X) is my original problem and what I wanted to learn is f. After scaling, I actually is training with y'_train = g(X'_train) and then I get my $y'_{test} = g(X'_{test})$. To make sure y'_test == y_test, we have to make sure g=~f. Am I right?
          $endgroup$
          – user2189731
          May 18 '18 at 3:09














          $begingroup$
          Correction to last sentence: since $ y_{test} = f(X_{test}) $ and $ y'_{test} = g(X'_{test}) $, to make sure $y_{test} == scale * y'_{test}$, I have to make sure $f(X_{test}) = scale * g(X'_{test}) $ . Which equals to: $f(X_{test}) = scale * g(X_{test}/scale) $ and this is not gauranteed?
          $endgroup$
          – user2189731
          May 18 '18 at 3:21






          $begingroup$
          Correction to last sentence: since $ y_{test} = f(X_{test}) $ and $ y'_{test} = g(X'_{test}) $, to make sure $y_{test} == scale * y'_{test}$, I have to make sure $f(X_{test}) = scale * g(X'_{test}) $ . Which equals to: $f(X_{test}) = scale * g(X_{test}/scale) $ and this is not gauranteed?
          $endgroup$
          – user2189731
          May 18 '18 at 3:21




















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          Metroo de Marsejlo

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          Metroo de Marsejlo Logoo Ĝeneralaj informoj Loko Marsejlo  Francio Transportsistema speco Metroo Komenco de trafiko 26-a de novembro 1977 Teknikaj informoj Longeco de sistemo 21,6 km Nombro de linioj 2 Nombro de veturiloj 36 Nombro de stacioj 28 (+ 2 interkonektoj) Aliaj informoj Traklarĝo 1 435 mm Transportisto RTM v   •   d   •   r La Metroo de Marsejlo (Francio) (france  Métro de Marseille ) estas metrosistemo funkcianta per pneŭoj derivita de la teknologio disvolvigita de la RATP por la metroo de Parizo. Ĝi malfermiĝis unuafoje al la publiko en la jaro 1977. Ĝi konsistas el du linioj, parte subteraj, kun longeco de entute 21,6 km , situantaj nur en la municipo Marsejlo. Nuntempe konstruiĝas du plilongigaĵoj. Enhavo 1 Mapo 2 Metrolinioj 2.1 Linio 1 2.2 Linio 2 3 Vidu ankaŭ 4 Eksteraj ligiloj Mapo | Mapo de la metroo de Marsejlo Metrolinioj | ...
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