Change arguments of a product of functions












2












$begingroup$


I am trying to do the following with rule-based pattern matching



(f1[t] f2[t] f3[t]) /. {x__ -> n[

     Product[x[[i, 0]][om[i]], {i, 3}]]} // Quiet


which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages



"Part specification x[[1,0]] is longer than depth of object"


I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?






pattern-matching







share|improve this question









$endgroup$

















    2












    $begingroup$


    I am trying to do the following with rule-based pattern matching



    (f1[t] f2[t] f3[t]) /. {x__ -> n[
    
     Product[x[[i, 0]][om[i]], {i, 3}]]} // Quiet


    which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages



    "Part specification x[[1,0]] is longer than depth of object"


    I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?






    pattern-matching







    share|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I am trying to do the following with rule-based pattern matching



      (f1[t] f2[t] f3[t]) /. {x__ -> n[
      
     Product[x[[i, 0]][om[i]], {i, 3}]]} // Quiet


      which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages



      "Part specification x[[1,0]] is longer than depth of object"


      I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?






      pattern-matching







      share|improve this question









      $endgroup$




      I am trying to do the following with rule-based pattern matching



      (f1[t] f2[t] f3[t]) /. {x__ -> n[
      
     Product[x[[i, 0]][om[i]], {i, 3}]]} // Quiet


      which gives me the output n[f1[om[1]] f2[om[2]] f3[om[3]]]. However, it seems as if one cannot specify parts of a pattern MMA gives the error messages



      "Part specification x[[1,0]] is longer than depth of object"


      I would like to do the operatoration for a product of n arbitrary functions f1[t]...fn[t] where the number of functions is automatically detected (I cannot use Length as well). How would one do it correctly?






      pattern-matching




      pattern-matching






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 3 hours ago









      Display NameDisplay Name

      56938




      56938






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Also



          Module[{i = 1}, f1[t] f2[t] f3[t] /. {t :> om[i++]} // n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Update: "If all the functions are equal":



          Inactivate[f1[t] f1[t] f1[t]] /. 
          
 Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[{i = 1}, Inactivate[f1[t] f1[t] f1[t]] /. {t :> om[i++]} // n]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]







          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
            $endgroup$
            – Display Name
            58 mins ago












          • $begingroup$
            @DisplayName, please see the update.
            $endgroup$
            – kglr
            48 mins ago










          • $begingroup$
            Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
            $endgroup$
            – Display Name
            36 mins ago





















          0












          $begingroup$

          A simple function can do it.



          changeArg[p : Times[__], var_Symbol, head_Symbol] := 
          
  head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]
          

          
changeArg[f1[t] f2[t] f3[t], om, n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]







          share|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
            $endgroup$
            – Display Name
            22 mins ago











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Also



          Module[{i = 1}, f1[t] f2[t] f3[t] /. {t :> om[i++]} // n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Update: "If all the functions are equal":



          Inactivate[f1[t] f1[t] f1[t]] /. 
          
 Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[{i = 1}, Inactivate[f1[t] f1[t] f1[t]] /. {t :> om[i++]} // n]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]







          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
            $endgroup$
            – Display Name
            58 mins ago












          • $begingroup$
            @DisplayName, please see the update.
            $endgroup$
            – kglr
            48 mins ago










          • $begingroup$
            Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
            $endgroup$
            – Display Name
            36 mins ago


















          2












          $begingroup$

          f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Also



          Module[{i = 1}, f1[t] f2[t] f3[t] /. {t :> om[i++]} // n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Update: "If all the functions are equal":



          Inactivate[f1[t] f1[t] f1[t]] /. 
          
 Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[{i = 1}, Inactivate[f1[t] f1[t] f1[t]] /. {t :> om[i++]} // n]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]







          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
            $endgroup$
            – Display Name
            58 mins ago












          • $begingroup$
            @DisplayName, please see the update.
            $endgroup$
            – kglr
            48 mins ago










          • $begingroup$
            Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
            $endgroup$
            – Display Name
            36 mins ago
















          2












          2








          2





          $begingroup$

          f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Also



          Module[{i = 1}, f1[t] f2[t] f3[t] /. {t :> om[i++]} // n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Update: "If all the functions are equal":



          Inactivate[f1[t] f1[t] f1[t]] /. 
          
 Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[{i = 1}, Inactivate[f1[t] f1[t] f1[t]] /. {t :> om[i++]} // n]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]







          share|improve this answer











          $endgroup$



          f1[t] f2[t] f3[t] /. Times[x_, y__] :> n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Also



          Module[{i = 1}, f1[t] f2[t] f3[t] /. {t :> om[i++]} // n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]




          Update: "If all the functions are equal":



          Inactivate[f1[t] f1[t] f1[t]] /. 
          
 Inactive[Times][x_, y__] :> Activate@n[Times @@ MapIndexed[#[[0]][om[#2[[1]]]] &, {x, y}]]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]




          Activate@Module[{i = 1}, Inactivate[f1[t] f1[t] f1[t]] /. {t :> om[i++]} // n]



          n[f1[om[1]] f1[om[2]] f1[om[3]]]








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 49 mins ago

























          answered 1 hour ago









          kglrkglr

          179k9198410




          179k9198410












          • $begingroup$
            Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
            $endgroup$
            – Display Name
            58 mins ago












          • $begingroup$
            @DisplayName, please see the update.
            $endgroup$
            – kglr
            48 mins ago










          • $begingroup$
            Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
            $endgroup$
            – Display Name
            36 mins ago




















          • $begingroup$
            Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
            $endgroup$
            – Display Name
            58 mins ago












          • $begingroup$
            @DisplayName, please see the update.
            $endgroup$
            – kglr
            48 mins ago










          • $begingroup$
            Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
            $endgroup$
            – Display Name
            36 mins ago


















          $begingroup$
          Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
          $endgroup$
          – Display Name
          58 mins ago






          $begingroup$
          Thank you for your answer. It works for f1 != f2 != f3 but stops working if all the functions are equal.
          $endgroup$
          – Display Name
          58 mins ago














          $begingroup$
          @DisplayName, please see the update.
          $endgroup$
          – kglr
          48 mins ago




          $begingroup$
          @DisplayName, please see the update.
          $endgroup$
          – kglr
          48 mins ago












          $begingroup$
          Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
          $endgroup$
          – Display Name
          36 mins ago






          $begingroup$
          Thanks again. I think I lack a little bit of knowledge to come up with sth like this. I just read into pattern matching but it needs some time to settle. But I also like the approach with the RuleDelayed. The mathematical background of this question is to write a product in time domain as a product in Fourier space via the convolution theorem. n is then simply a functional involving integrations over the oms.
          $endgroup$
          – Display Name
          36 mins ago













          0












          $begingroup$

          A simple function can do it.



          changeArg[p : Times[__], var_Symbol, head_Symbol] := 
          
  head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]
          

          
changeArg[f1[t] f2[t] f3[t], om, n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]







          share|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
            $endgroup$
            – Display Name
            22 mins ago
















          0












          $begingroup$

          A simple function can do it.



          changeArg[p : Times[__], var_Symbol, head_Symbol] := 
          
  head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]
          

          
changeArg[f1[t] f2[t] f3[t], om, n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]







          share|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
            $endgroup$
            – Display Name
            22 mins ago














          0












          0








          0





          $begingroup$

          A simple function can do it.



          changeArg[p : Times[__], var_Symbol, head_Symbol] := 
          
  head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]
          

          
changeArg[f1[t] f2[t] f3[t], om, n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]







          share|improve this answer









          $endgroup$



          A simple function can do it.



          changeArg[p : Times[__], var_Symbol, head_Symbol] := 
          
  head[MapIndexed[#1[var[#2[[1]]]] &, Head /@ p]]
          

          
changeArg[f1[t] f2[t] f3[t], om, n]



          n[f1[om[1]] f2[om[2]] f3[om[3]]]








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 36 mins ago









          m_goldbergm_goldberg

          84.7k872196




          84.7k872196












          • $begingroup$
            Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
            $endgroup$
            – Display Name
            22 mins ago


















          • $begingroup$
            Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
            $endgroup$
            – Display Name
            22 mins ago
















          $begingroup$
          Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
          $endgroup$
          – Display Name
          22 mins ago




          $begingroup$
          Thank you for another possible answer. I think in the current form it doesn't cover the case of equal functions.
          $endgroup$
          – Display Name
          22 mins ago


















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