How do I make LinearModelFit flexible to number of parameters using FromLetterNumber?

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I want to run LinearModelFit where the number of parameters is not fixed. For example, the following



i = 2;
data = RandomReal[20,10,3+i];
lm1 = LinearModelFit[data,FromLetterNumber[Range[2+i]],FromLetterNumber[Range[2+i]];


doesn't work, whereas



lm1 = LinearModelFit[data,a,b,c,d,a,b,c,d]


works as per described, despite FromLetterNumber[Range[2+i]] == a,b,c,d



Can this be made to work using some kind of Hold or do I need to resort to a design matrix, response vector format?




fitting hold




share|improve this question























    up vote
    4
    down vote

    favorite












    I want to run LinearModelFit where the number of parameters is not fixed. For example, the following



    i = 2;
    data = RandomReal[20,10,3+i];
    lm1 = LinearModelFit[data,FromLetterNumber[Range[2+i]],FromLetterNumber[Range[2+i]];


    doesn't work, whereas



    lm1 = LinearModelFit[data,a,b,c,d,a,b,c,d]


    works as per described, despite FromLetterNumber[Range[2+i]] == a,b,c,d



    Can this be made to work using some kind of Hold or do I need to resort to a design matrix, response vector format?




    fitting hold




    share|improve this question





















      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      I want to run LinearModelFit where the number of parameters is not fixed. For example, the following



      i = 2;
      data = RandomReal[20,10,3+i];
      lm1 = LinearModelFit[data,FromLetterNumber[Range[2+i]],FromLetterNumber[Range[2+i]];


      doesn't work, whereas



      lm1 = LinearModelFit[data,a,b,c,d,a,b,c,d]


      works as per described, despite FromLetterNumber[Range[2+i]] == a,b,c,d



      Can this be made to work using some kind of Hold or do I need to resort to a design matrix, response vector format?




      fitting hold




      share|improve this question











      I want to run LinearModelFit where the number of parameters is not fixed. For example, the following



      i = 2;
      data = RandomReal[20,10,3+i];
      lm1 = LinearModelFit[data,FromLetterNumber[Range[2+i]],FromLetterNumber[Range[2+i]];


      doesn't work, whereas



      lm1 = LinearModelFit[data,a,b,c,d,a,b,c,d]


      works as per described, despite FromLetterNumber[Range[2+i]] == a,b,c,d



      Can this be made to work using some kind of Hold or do I need to resort to a design matrix, response vector format?





      fitting hold





      share|improve this question










      share|improve this question




      share|improve this question









      asked Aug 6 at 7:54









      Thadeu Freitas Filho

      866




      866




















          1 Answer
          1






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          FromLetterNumber returns a String:



          Head @ FromLetterNumber @ 3



          String




          You can wrap the strings returned by FromLetterNumber with Symbol or ToExpression to get Symbols:



          lm1 = LinearModelFit[data, Symbol /@ FromLetterNumber[Range[2+i]], 
           Symbol /@ FromLetterNumber[Range[2+i]]];
          Normal @ lm1



          9.60642 + 0.175108 a - 0.723711 b + 0.850788 c - 0.198777 d




          Alternatively,



          lm2 = LinearModelFit[data, Array[a, 2 + i], Array[a, 2 + i] ] ;
          Normal @ lm2



          9.60642 + 0.175108 a[1] - 0.723711 a[2] + 0.850788 a[3] - 0.198777 a[4]







          share|improve this answer



















          • 2




            The Array version should be the canonical choice, IMHO. +1
            – Marius Ladegård Meyer
            Aug 6 at 9:34










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          up vote
          5
          down vote



          accepted










          FromLetterNumber returns a String:



          Head @ FromLetterNumber @ 3



          String




          You can wrap the strings returned by FromLetterNumber with Symbol or ToExpression to get Symbols:



          lm1 = LinearModelFit[data, Symbol /@ FromLetterNumber[Range[2+i]], 
           Symbol /@ FromLetterNumber[Range[2+i]]];
          Normal @ lm1



          9.60642 + 0.175108 a - 0.723711 b + 0.850788 c - 0.198777 d




          Alternatively,



          lm2 = LinearModelFit[data, Array[a, 2 + i], Array[a, 2 + i] ] ;
          Normal @ lm2



          9.60642 + 0.175108 a[1] - 0.723711 a[2] + 0.850788 a[3] - 0.198777 a[4]







          share|improve this answer



















          • 2




            The Array version should be the canonical choice, IMHO. +1
            – Marius Ladegård Meyer
            Aug 6 at 9:34














          up vote
          5
          down vote



          accepted










          FromLetterNumber returns a String:



          Head @ FromLetterNumber @ 3



          String




          You can wrap the strings returned by FromLetterNumber with Symbol or ToExpression to get Symbols:



          lm1 = LinearModelFit[data, Symbol /@ FromLetterNumber[Range[2+i]], 
           Symbol /@ FromLetterNumber[Range[2+i]]];
          Normal @ lm1



          9.60642 + 0.175108 a - 0.723711 b + 0.850788 c - 0.198777 d




          Alternatively,



          lm2 = LinearModelFit[data, Array[a, 2 + i], Array[a, 2 + i] ] ;
          Normal @ lm2



          9.60642 + 0.175108 a[1] - 0.723711 a[2] + 0.850788 a[3] - 0.198777 a[4]







          share|improve this answer



















          • 2




            The Array version should be the canonical choice, IMHO. +1
            – Marius Ladegård Meyer
            Aug 6 at 9:34












          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          FromLetterNumber returns a String:



          Head @ FromLetterNumber @ 3



          String




          You can wrap the strings returned by FromLetterNumber with Symbol or ToExpression to get Symbols:



          lm1 = LinearModelFit[data, Symbol /@ FromLetterNumber[Range[2+i]], 
           Symbol /@ FromLetterNumber[Range[2+i]]];
          Normal @ lm1



          9.60642 + 0.175108 a - 0.723711 b + 0.850788 c - 0.198777 d




          Alternatively,



          lm2 = LinearModelFit[data, Array[a, 2 + i], Array[a, 2 + i] ] ;
          Normal @ lm2



          9.60642 + 0.175108 a[1] - 0.723711 a[2] + 0.850788 a[3] - 0.198777 a[4]







          share|improve this answer















          FromLetterNumber returns a String:



          Head @ FromLetterNumber @ 3



          String




          You can wrap the strings returned by FromLetterNumber with Symbol or ToExpression to get Symbols:



          lm1 = LinearModelFit[data, Symbol /@ FromLetterNumber[Range[2+i]], 
           Symbol /@ FromLetterNumber[Range[2+i]]];
          Normal @ lm1



          9.60642 + 0.175108 a - 0.723711 b + 0.850788 c - 0.198777 d




          Alternatively,



          lm2 = LinearModelFit[data, Array[a, 2 + i], Array[a, 2 + i] ] ;
          Normal @ lm2



          9.60642 + 0.175108 a[1] - 0.723711 a[2] + 0.850788 a[3] - 0.198777 a[4]








          share|improve this answer















          share|improve this answer



          share|improve this answer








          edited Aug 7 at 0:00


























          answered Aug 6 at 8:06









          kglr

          154k8179374




          154k8179374







          • 2




            The Array version should be the canonical choice, IMHO. +1
            – Marius Ladegård Meyer
            Aug 6 at 9:34












          • 2




            The Array version should be the canonical choice, IMHO. +1
            – Marius Ladegård Meyer
            Aug 6 at 9:34







          2




          2




          The Array version should be the canonical choice, IMHO. +1
          – Marius Ladegård Meyer
          Aug 6 at 9:34




          The Array version should be the canonical choice, IMHO. +1
          – Marius Ladegård Meyer
          Aug 6 at 9:34












           

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