How to substitute values from a list into a function?












3












$begingroup$


I was hoping to take pairs of numbers from a list and substitute them into a function. So if my list was 



list = {{1,2}, {3,4}, {5,6}}


and my function was 



function = a x^b


The output I'm hoping to get is 



result =1x^2 + 3x^4 + 5x^6


How would I best do this?






list-manipulation functions







share|improve this question







New contributor




Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    #1 x^#2 & @@@ {{1, 2}, {3, 4}, {5, 6}} // Total
    $endgroup$
    – Okkes Dulgerci
    20 mins ago
















3












$begingroup$


I was hoping to take pairs of numbers from a list and substitute them into a function. So if my list was 



list = {{1,2}, {3,4}, {5,6}}


and my function was 



function = a x^b


The output I'm hoping to get is 



result =1x^2 + 3x^4 + 5x^6


How would I best do this?






list-manipulation functions







share|improve this question







New contributor




Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    #1 x^#2 & @@@ {{1, 2}, {3, 4}, {5, 6}} // Total
    $endgroup$
    – Okkes Dulgerci
    20 mins ago














3












3








3





$begingroup$


I was hoping to take pairs of numbers from a list and substitute them into a function. So if my list was 



list = {{1,2}, {3,4}, {5,6}}


and my function was 



function = a x^b


The output I'm hoping to get is 



result =1x^2 + 3x^4 + 5x^6


How would I best do this?






list-manipulation functions







share|improve this question







New contributor




Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I was hoping to take pairs of numbers from a list and substitute them into a function. So if my list was 



list = {{1,2}, {3,4}, {5,6}}


and my function was 



function = a x^b


The output I'm hoping to get is 



result =1x^2 + 3x^4 + 5x^6


How would I best do this?






list-manipulation functions




list-manipulation functions






share|improve this question







New contributor




Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









PineapplePineapple

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161




New contributor




Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Pineapple is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    #1 x^#2 & @@@ {{1, 2}, {3, 4}, {5, 6}} // Total
    $endgroup$
    – Okkes Dulgerci
    20 mins ago


















  • $begingroup$
    #1 x^#2 & @@@ {{1, 2}, {3, 4}, {5, 6}} // Total
    $endgroup$
    – Okkes Dulgerci
    20 mins ago
















$begingroup$
#1 x^#2 & @@@ {{1, 2}, {3, 4}, {5, 6}} // Total
$endgroup$
– Okkes Dulgerci
20 mins ago




$begingroup$
#1 x^#2 & @@@ {{1, 2}, {3, 4}, {5, 6}} // Total
$endgroup$
– Okkes Dulgerci
20 mins ago










6 Answers
6






active

oldest

votes


















3












$begingroup$

This is not a Function:



function = a x^b


But this is:



function = {a,b} [Function] a x^b


You can Apply it to each element of 



list = {{1,2}, {3,4}, {5,6}}


with



function @@@ list



{x^3, 3 x^5, 5 x^7}




and sum it up with Total:



Total[ function @@@ list ]



x^3 + 3 x^5 + 5 x^7







share|improve this answer









$endgroup$













  • $begingroup$
    Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
    $endgroup$
    – Pineapple
    1 hour ago










  • $begingroup$
    Because [Function] is easily entered with the escape sequence esc f n esc . Also Function defines a pure function while function[a_, b_, x_] = a*x^b defines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
    $endgroup$
    – Henrik Schumacher
    1 hour ago












  • $begingroup$
    @Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference between Set (=) and SetDelayed (:=).
    $endgroup$
    – MarcoB
    1 hour ago





















2












$begingroup$

Total[#*x^#2&@@@list]



x^2 + 3 x^4 + 5 x^6







share|improve this answer









$endgroup$





















    1












    $begingroup$

    Total[(#[[1]] x^#[[2]]) & /@ list]





    share|improve this answer









    $endgroup$













    • $begingroup$
      Amazing, thank you! Sorry - I'm very new to Mathematica.
      $endgroup$
      – Pineapple
      1 hour ago



















    0












    $begingroup$

    Another way, perhaps easier on the eyes. Use a pattern to deconstruct the pairs in a function definition.



    term[{a_, b_}] := a x^b


    Then, Map it to the list.



    Total[term /@ list]
    
(* x^2 + 3 x^4 + 5 x^6 *)





    share|improve this answer









    $endgroup$





















      0












      $begingroup$

      Inner[#1 x^#2 &, Sequence @@ Transpose@list, Plus]



      x^2 + 3 x^4 + 5 x^6




      One could expand this a bit to allow for different variables:



      Clear[f]
      
f[coefflist_][var_] := Inner[#1 var^#2 &, Sequence @@ Transpose@coefflist, Plus]


      so that 



      f[list][x]



      x^2 + 3 x^4 + 5 x^6




      but then:



      f[list][t]



      t^2 + 3 t^4 + 5 t^6







      share|improve this answer











      $endgroup$





















        0












        $begingroup$

        ClearAll[fa, fb]
        
fa = FromCoefficientRules[Thread[#[[All, 2;;]] -> #[[All, 1]]], #2] &;
        
fb = Internal`FromCoefficientList[Normal@SparseArray[1 + #[[All, 2;;]]->#[[All, 1]]], #2] &;


        Examples:



        list1 = {{1, 2}, {3, 4}, {5, 6}};
        
{fa[list1, x], fb[list1, x]}



        {x^2 + 3 x^4 + 5 x^6, x^2 + 3 x^4 + 5 x^6}




        list2 = {{1, 3, 0}, {3, 2, 1}, {3, 1, 2}, {1, 0, 3}};
        
{fa[list2, {x, y}], fb[list2, {x, y}]}



        {x^3 + 3 x^2 y + 3 x y^2 + y^3, x^3 + 3 x^2 y + 3 x y^2 + y^3}







        share|improve this answer











        $endgroup$













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          6 Answers
          6






          active

          oldest

          votes








          6 Answers
          6






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          This is not a Function:



          function = a x^b


          But this is:



          function = {a,b} [Function] a x^b


          You can Apply it to each element of 



          list = {{1,2}, {3,4}, {5,6}}


          with



          function @@@ list



          {x^3, 3 x^5, 5 x^7}




          and sum it up with Total:



          Total[ function @@@ list ]



          x^3 + 3 x^5 + 5 x^7







          share|improve this answer









          $endgroup$













          • $begingroup$
            Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
            $endgroup$
            – Pineapple
            1 hour ago










          • $begingroup$
            Because [Function] is easily entered with the escape sequence esc f n esc . Also Function defines a pure function while function[a_, b_, x_] = a*x^b defines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
            $endgroup$
            – Henrik Schumacher
            1 hour ago












          • $begingroup$
            @Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference between Set (=) and SetDelayed (:=).
            $endgroup$
            – MarcoB
            1 hour ago


















          3












          $begingroup$

          This is not a Function:



          function = a x^b


          But this is:



          function = {a,b} [Function] a x^b


          You can Apply it to each element of 



          list = {{1,2}, {3,4}, {5,6}}


          with



          function @@@ list



          {x^3, 3 x^5, 5 x^7}




          and sum it up with Total:



          Total[ function @@@ list ]



          x^3 + 3 x^5 + 5 x^7







          share|improve this answer









          $endgroup$













          • $begingroup$
            Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
            $endgroup$
            – Pineapple
            1 hour ago










          • $begingroup$
            Because [Function] is easily entered with the escape sequence esc f n esc . Also Function defines a pure function while function[a_, b_, x_] = a*x^b defines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
            $endgroup$
            – Henrik Schumacher
            1 hour ago












          • $begingroup$
            @Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference between Set (=) and SetDelayed (:=).
            $endgroup$
            – MarcoB
            1 hour ago
















          3












          3








          3





          $begingroup$

          This is not a Function:



          function = a x^b


          But this is:



          function = {a,b} [Function] a x^b


          You can Apply it to each element of 



          list = {{1,2}, {3,4}, {5,6}}


          with



          function @@@ list



          {x^3, 3 x^5, 5 x^7}




          and sum it up with Total:



          Total[ function @@@ list ]



          x^3 + 3 x^5 + 5 x^7







          share|improve this answer









          $endgroup$



          This is not a Function:



          function = a x^b


          But this is:



          function = {a,b} [Function] a x^b


          You can Apply it to each element of 



          list = {{1,2}, {3,4}, {5,6}}


          with



          function @@@ list



          {x^3, 3 x^5, 5 x^7}




          and sum it up with Total:



          Total[ function @@@ list ]



          x^3 + 3 x^5 + 5 x^7








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 1 hour ago









          Henrik SchumacherHenrik Schumacher

          55.4k576154




          55.4k576154












          • $begingroup$
            Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
            $endgroup$
            – Pineapple
            1 hour ago










          • $begingroup$
            Because [Function] is easily entered with the escape sequence esc f n esc . Also Function defines a pure function while function[a_, b_, x_] = a*x^b defines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
            $endgroup$
            – Henrik Schumacher
            1 hour ago












          • $begingroup$
            @Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference between Set (=) and SetDelayed (:=).
            $endgroup$
            – MarcoB
            1 hour ago




















          • $begingroup$
            Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
            $endgroup$
            – Pineapple
            1 hour ago










          • $begingroup$
            Because [Function] is easily entered with the escape sequence esc f n esc . Also Function defines a pure function while function[a_, b_, x_] = a*x^b defines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
            $endgroup$
            – Henrik Schumacher
            1 hour ago












          • $begingroup$
            @Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference between Set (=) and SetDelayed (:=).
            $endgroup$
            – MarcoB
            1 hour ago


















          $begingroup$
          Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
          $endgroup$
          – Pineapple
          1 hour ago




          $begingroup$
          Why would you do 'function = {a,b} [Function] a x^b' instead of function[a_, b_, x_] = a*x^b?
          $endgroup$
          – Pineapple
          1 hour ago












          $begingroup$
          Because [Function] is easily entered with the escape sequence esc f n esc . Also Function defines a pure function while function[a_, b_, x_] = a*x^b defines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
          $endgroup$
          – Henrik Schumacher
          1 hour ago






          $begingroup$
          Because [Function] is easily entered with the escape sequence esc f n esc . Also Function defines a pure function while function[a_, b_, x_] = a*x^b defines a replacement rule. However, they act the same way - most of the time. So it is a matter of taste.
          $endgroup$
          – Henrik Schumacher
          1 hour ago














          $begingroup$
          @Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference between Set (=) and SetDelayed (:=).
          $endgroup$
          – MarcoB
          1 hour ago






          $begingroup$
          @Pineapple Take a look at this tutorial on Defining Functions. Another good resource is the tutorial on Immediate and Delayed Definitions for the difference between Set (=) and SetDelayed (:=).
          $endgroup$
          – MarcoB
          1 hour ago













          2












          $begingroup$

          Total[#*x^#2&@@@list]



          x^2 + 3 x^4 + 5 x^6







          share|improve this answer









          $endgroup$


















            2












            $begingroup$

            Total[#*x^#2&@@@list]



            x^2 + 3 x^4 + 5 x^6







            share|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              Total[#*x^#2&@@@list]



              x^2 + 3 x^4 + 5 x^6







              share|improve this answer









              $endgroup$



              Total[#*x^#2&@@@list]



              x^2 + 3 x^4 + 5 x^6








              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 1 hour ago









              J42161217J42161217

              3,935322




              3,935322























                  1












                  $begingroup$

                  Total[(#[[1]] x^#[[2]]) & /@ list]





                  share|improve this answer









                  $endgroup$













                  • $begingroup$
                    Amazing, thank you! Sorry - I'm very new to Mathematica.
                    $endgroup$
                    – Pineapple
                    1 hour ago
















                  1












                  $begingroup$

                  Total[(#[[1]] x^#[[2]]) & /@ list]





                  share|improve this answer









                  $endgroup$













                  • $begingroup$
                    Amazing, thank you! Sorry - I'm very new to Mathematica.
                    $endgroup$
                    – Pineapple
                    1 hour ago














                  1












                  1








                  1





                  $begingroup$

                  Total[(#[[1]] x^#[[2]]) & /@ list]





                  share|improve this answer









                  $endgroup$



                  Total[(#[[1]] x^#[[2]]) & /@ list]






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 1 hour ago









                  David G. StorkDavid G. Stork

                  24.5k22153




                  24.5k22153












                  • $begingroup$
                    Amazing, thank you! Sorry - I'm very new to Mathematica.
                    $endgroup$
                    – Pineapple
                    1 hour ago


















                  • $begingroup$
                    Amazing, thank you! Sorry - I'm very new to Mathematica.
                    $endgroup$
                    – Pineapple
                    1 hour ago
















                  $begingroup$
                  Amazing, thank you! Sorry - I'm very new to Mathematica.
                  $endgroup$
                  – Pineapple
                  1 hour ago




                  $begingroup$
                  Amazing, thank you! Sorry - I'm very new to Mathematica.
                  $endgroup$
                  – Pineapple
                  1 hour ago











                  0












                  $begingroup$

                  Another way, perhaps easier on the eyes. Use a pattern to deconstruct the pairs in a function definition.



                  term[{a_, b_}] := a x^b


                  Then, Map it to the list.



                  Total[term /@ list]
                  
(* x^2 + 3 x^4 + 5 x^6 *)





                  share|improve this answer









                  $endgroup$


















                    0












                    $begingroup$

                    Another way, perhaps easier on the eyes. Use a pattern to deconstruct the pairs in a function definition.



                    term[{a_, b_}] := a x^b


                    Then, Map it to the list.



                    Total[term /@ list]
                    
(* x^2 + 3 x^4 + 5 x^6 *)





                    share|improve this answer









                    $endgroup$
















                      0












                      0








                      0





                      $begingroup$

                      Another way, perhaps easier on the eyes. Use a pattern to deconstruct the pairs in a function definition.



                      term[{a_, b_}] := a x^b


                      Then, Map it to the list.



                      Total[term /@ list]
                      
(* x^2 + 3 x^4 + 5 x^6 *)





                      share|improve this answer









                      $endgroup$



                      Another way, perhaps easier on the eyes. Use a pattern to deconstruct the pairs in a function definition.



                      term[{a_, b_}] := a x^b


                      Then, Map it to the list.



                      Total[term /@ list]
                      
(* x^2 + 3 x^4 + 5 x^6 *)






                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 51 mins ago









                      John DotyJohn Doty

                      7,17311024




                      7,17311024























                          0












                          $begingroup$

                          Inner[#1 x^#2 &, Sequence @@ Transpose@list, Plus]



                          x^2 + 3 x^4 + 5 x^6




                          One could expand this a bit to allow for different variables:



                          Clear[f]
                          
f[coefflist_][var_] := Inner[#1 var^#2 &, Sequence @@ Transpose@coefflist, Plus]


                          so that 



                          f[list][x]



                          x^2 + 3 x^4 + 5 x^6




                          but then:



                          f[list][t]



                          t^2 + 3 t^4 + 5 t^6







                          share|improve this answer











                          $endgroup$


















                            0












                            $begingroup$

                            Inner[#1 x^#2 &, Sequence @@ Transpose@list, Plus]



                            x^2 + 3 x^4 + 5 x^6




                            One could expand this a bit to allow for different variables:



                            Clear[f]
                            
f[coefflist_][var_] := Inner[#1 var^#2 &, Sequence @@ Transpose@coefflist, Plus]


                            so that 



                            f[list][x]



                            x^2 + 3 x^4 + 5 x^6




                            but then:



                            f[list][t]



                            t^2 + 3 t^4 + 5 t^6







                            share|improve this answer











                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Inner[#1 x^#2 &, Sequence @@ Transpose@list, Plus]



                              x^2 + 3 x^4 + 5 x^6




                              One could expand this a bit to allow for different variables:



                              Clear[f]
                              
f[coefflist_][var_] := Inner[#1 var^#2 &, Sequence @@ Transpose@coefflist, Plus]


                              so that 



                              f[list][x]



                              x^2 + 3 x^4 + 5 x^6




                              but then:



                              f[list][t]



                              t^2 + 3 t^4 + 5 t^6







                              share|improve this answer











                              $endgroup$



                              Inner[#1 x^#2 &, Sequence @@ Transpose@list, Plus]



                              x^2 + 3 x^4 + 5 x^6




                              One could expand this a bit to allow for different variables:



                              Clear[f]
                              
f[coefflist_][var_] := Inner[#1 var^#2 &, Sequence @@ Transpose@coefflist, Plus]


                              so that 



                              f[list][x]



                              x^2 + 3 x^4 + 5 x^6




                              but then:



                              f[list][t]



                              t^2 + 3 t^4 + 5 t^6








                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited 46 mins ago

























                              answered 52 mins ago









                              MarcoBMarcoB

                              36.7k556113




                              36.7k556113























                                  0












                                  $begingroup$

                                  ClearAll[fa, fb]
                                  
fa = FromCoefficientRules[Thread[#[[All, 2;;]] -> #[[All, 1]]], #2] &;
                                  
fb = Internal`FromCoefficientList[Normal@SparseArray[1 + #[[All, 2;;]]->#[[All, 1]]], #2] &;


                                  Examples:



                                  list1 = {{1, 2}, {3, 4}, {5, 6}};
                                  
{fa[list1, x], fb[list1, x]}



                                  {x^2 + 3 x^4 + 5 x^6, x^2 + 3 x^4 + 5 x^6}




                                  list2 = {{1, 3, 0}, {3, 2, 1}, {3, 1, 2}, {1, 0, 3}};
                                  
{fa[list2, {x, y}], fb[list2, {x, y}]}



                                  {x^3 + 3 x^2 y + 3 x y^2 + y^3, x^3 + 3 x^2 y + 3 x y^2 + y^3}







                                  share|improve this answer











                                  $endgroup$


















                                    0












                                    $begingroup$

                                    ClearAll[fa, fb]
                                    
fa = FromCoefficientRules[Thread[#[[All, 2;;]] -> #[[All, 1]]], #2] &;
                                    
fb = Internal`FromCoefficientList[Normal@SparseArray[1 + #[[All, 2;;]]->#[[All, 1]]], #2] &;


                                    Examples:



                                    list1 = {{1, 2}, {3, 4}, {5, 6}};
                                    
{fa[list1, x], fb[list1, x]}



                                    {x^2 + 3 x^4 + 5 x^6, x^2 + 3 x^4 + 5 x^6}




                                    list2 = {{1, 3, 0}, {3, 2, 1}, {3, 1, 2}, {1, 0, 3}};
                                    
{fa[list2, {x, y}], fb[list2, {x, y}]}



                                    {x^3 + 3 x^2 y + 3 x y^2 + y^3, x^3 + 3 x^2 y + 3 x y^2 + y^3}







                                    share|improve this answer











                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      ClearAll[fa, fb]
                                      
fa = FromCoefficientRules[Thread[#[[All, 2;;]] -> #[[All, 1]]], #2] &;
                                      
fb = Internal`FromCoefficientList[Normal@SparseArray[1 + #[[All, 2;;]]->#[[All, 1]]], #2] &;


                                      Examples:



                                      list1 = {{1, 2}, {3, 4}, {5, 6}};
                                      
{fa[list1, x], fb[list1, x]}



                                      {x^2 + 3 x^4 + 5 x^6, x^2 + 3 x^4 + 5 x^6}




                                      list2 = {{1, 3, 0}, {3, 2, 1}, {3, 1, 2}, {1, 0, 3}};
                                      
{fa[list2, {x, y}], fb[list2, {x, y}]}



                                      {x^3 + 3 x^2 y + 3 x y^2 + y^3, x^3 + 3 x^2 y + 3 x y^2 + y^3}







                                      share|improve this answer











                                      $endgroup$



                                      ClearAll[fa, fb]
                                      
fa = FromCoefficientRules[Thread[#[[All, 2;;]] -> #[[All, 1]]], #2] &;
                                      
fb = Internal`FromCoefficientList[Normal@SparseArray[1 + #[[All, 2;;]]->#[[All, 1]]], #2] &;


                                      Examples:



                                      list1 = {{1, 2}, {3, 4}, {5, 6}};
                                      
{fa[list1, x], fb[list1, x]}



                                      {x^2 + 3 x^4 + 5 x^6, x^2 + 3 x^4 + 5 x^6}




                                      list2 = {{1, 3, 0}, {3, 2, 1}, {3, 1, 2}, {1, 0, 3}};
                                      
{fa[list2, {x, y}], fb[list2, {x, y}]}



                                      {x^3 + 3 x^2 y + 3 x y^2 + y^3, x^3 + 3 x^2 y + 3 x y^2 + y^3}








                                      share|improve this answer














                                      share|improve this answer



                                      share|improve this answer








                                      edited 42 mins ago

























                                      answered 54 mins ago









                                      kglrkglr

                                      187k10203422




                                      187k10203422






















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