Cleanest way to take a[b[c]] to a[b][c]












2












$begingroup$


As indicated in the title I'm looking for the fastest way to transform a[b[c]] into a[b][c], and the natural generalization to an arbitrary chaining of arguments. I'm sure there's got to be a convenient way that I've overlooked.



In my cases a, b, and c can be any expression with any complicated internal structure they like.






function-construction







share|improve this question











$endgroup$








  • 1




    $begingroup$
    The solution likely would use Operate.
    $endgroup$
    – QuantumDot
    1 hour ago










  • $begingroup$
    f = Curry[Replace][a_[b_[c_]] :> a[b][c]] also works, so that a[b[c]] // f gives the desired result.
    $endgroup$
    – Shredderroy
    57 mins ago










  • $begingroup$
    How about generalizing the question to taking a[b[c[d[...]]]] to a[b][c][d]...?
    $endgroup$
    – David G. Stork
    46 mins ago












  • $begingroup$
    @DavidG.Stork sorry I thought that was implicit
    $endgroup$
    – b3m2a1
    40 mins ago










  • $begingroup$
    @b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
    $endgroup$
    – David G. Stork
    37 mins ago
















2












$begingroup$


As indicated in the title I'm looking for the fastest way to transform a[b[c]] into a[b][c], and the natural generalization to an arbitrary chaining of arguments. I'm sure there's got to be a convenient way that I've overlooked.



In my cases a, b, and c can be any expression with any complicated internal structure they like.






function-construction







share|improve this question











$endgroup$








  • 1




    $begingroup$
    The solution likely would use Operate.
    $endgroup$
    – QuantumDot
    1 hour ago










  • $begingroup$
    f = Curry[Replace][a_[b_[c_]] :> a[b][c]] also works, so that a[b[c]] // f gives the desired result.
    $endgroup$
    – Shredderroy
    57 mins ago










  • $begingroup$
    How about generalizing the question to taking a[b[c[d[...]]]] to a[b][c][d]...?
    $endgroup$
    – David G. Stork
    46 mins ago












  • $begingroup$
    @DavidG.Stork sorry I thought that was implicit
    $endgroup$
    – b3m2a1
    40 mins ago










  • $begingroup$
    @b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
    $endgroup$
    – David G. Stork
    37 mins ago














2












2








2





$begingroup$


As indicated in the title I'm looking for the fastest way to transform a[b[c]] into a[b][c], and the natural generalization to an arbitrary chaining of arguments. I'm sure there's got to be a convenient way that I've overlooked.



In my cases a, b, and c can be any expression with any complicated internal structure they like.






function-construction







share|improve this question











$endgroup$




As indicated in the title I'm looking for the fastest way to transform a[b[c]] into a[b][c], and the natural generalization to an arbitrary chaining of arguments. I'm sure there's got to be a convenient way that I've overlooked.



In my cases a, b, and c can be any expression with any complicated internal structure they like.






function-construction




function-construction






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 32 mins ago









David G. Stork

24.1k22153




24.1k22153










asked 1 hour ago









b3m2a1b3m2a1

27.2k257156




27.2k257156








  • 1




    $begingroup$
    The solution likely would use Operate.
    $endgroup$
    – QuantumDot
    1 hour ago










  • $begingroup$
    f = Curry[Replace][a_[b_[c_]] :> a[b][c]] also works, so that a[b[c]] // f gives the desired result.
    $endgroup$
    – Shredderroy
    57 mins ago










  • $begingroup$
    How about generalizing the question to taking a[b[c[d[...]]]] to a[b][c][d]...?
    $endgroup$
    – David G. Stork
    46 mins ago












  • $begingroup$
    @DavidG.Stork sorry I thought that was implicit
    $endgroup$
    – b3m2a1
    40 mins ago










  • $begingroup$
    @b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
    $endgroup$
    – David G. Stork
    37 mins ago














  • 1




    $begingroup$
    The solution likely would use Operate.
    $endgroup$
    – QuantumDot
    1 hour ago










  • $begingroup$
    f = Curry[Replace][a_[b_[c_]] :> a[b][c]] also works, so that a[b[c]] // f gives the desired result.
    $endgroup$
    – Shredderroy
    57 mins ago










  • $begingroup$
    How about generalizing the question to taking a[b[c[d[...]]]] to a[b][c][d]...?
    $endgroup$
    – David G. Stork
    46 mins ago












  • $begingroup$
    @DavidG.Stork sorry I thought that was implicit
    $endgroup$
    – b3m2a1
    40 mins ago










  • $begingroup$
    @b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
    $endgroup$
    – David G. Stork
    37 mins ago








1




1




$begingroup$
The solution likely would use Operate.
$endgroup$
– QuantumDot
1 hour ago




$begingroup$
The solution likely would use Operate.
$endgroup$
– QuantumDot
1 hour ago












$begingroup$
f = Curry[Replace][a_[b_[c_]] :> a[b][c]] also works, so that a[b[c]] // f gives the desired result.
$endgroup$
– Shredderroy
57 mins ago




$begingroup$
f = Curry[Replace][a_[b_[c_]] :> a[b][c]] also works, so that a[b[c]] // f gives the desired result.
$endgroup$
– Shredderroy
57 mins ago












$begingroup$
How about generalizing the question to taking a[b[c[d[...]]]] to a[b][c][d]...?
$endgroup$
– David G. Stork
46 mins ago






$begingroup$
How about generalizing the question to taking a[b[c[d[...]]]] to a[b][c][d]...?
$endgroup$
– David G. Stork
46 mins ago














$begingroup$
@DavidG.Stork sorry I thought that was implicit
$endgroup$
– b3m2a1
40 mins ago




$begingroup$
@DavidG.Stork sorry I thought that was implicit
$endgroup$
– b3m2a1
40 mins ago












$begingroup$
@b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
$endgroup$
– David G. Stork
37 mins ago




$begingroup$
@b3m2a1: Oh.... well I recommend you alter the question... and seem my new solution.
$endgroup$
– David G. Stork
37 mins ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Operate[#[[0]], First@#] &[a[b[c]]]



a[b][c]




ClearAll[deCompose]

deCompose = Nest[Operate[#[[0]], First@#] &, #, Depth[#] - 2] &;



deCompose@a[b[c]]



a[b][c]




exp = Compose[a, b, c, d, e, f, g]



a[b[c[d[e[f[g]]]]]]




deCompose @ exp



a[b][c][d][e][f][g]







share|improve this answer











$endgroup$





















    2












    $begingroup$

    Not particularly "clean," but it works:



    Operate[Head[#], Level[#, 2][[2]]] & @ a[b[c]]


    For the full generalization:



    Nest[Operate[Head[#], Level[#, 2][[2]]] & , #, Depth[#] -2] & @
    
 a[b[c[d[e]]]]





    share|improve this answer











    $endgroup$













      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












      $begingroup$

      Operate[#[[0]], First@#] &[a[b[c]]]



      a[b][c]




      ClearAll[deCompose]
      
deCompose = Nest[Operate[#[[0]], First@#] &, #, Depth[#] - 2] &;
      

      
deCompose@a[b[c]]



      a[b][c]




      exp = Compose[a, b, c, d, e, f, g]



      a[b[c[d[e[f[g]]]]]]




      deCompose @ exp



      a[b][c][d][e][f][g]







      share|improve this answer











      $endgroup$


















        2












        $begingroup$

        Operate[#[[0]], First@#] &[a[b[c]]]



        a[b][c]




        ClearAll[deCompose]
        
deCompose = Nest[Operate[#[[0]], First@#] &, #, Depth[#] - 2] &;
        

        
deCompose@a[b[c]]



        a[b][c]




        exp = Compose[a, b, c, d, e, f, g]



        a[b[c[d[e[f[g]]]]]]




        deCompose @ exp



        a[b][c][d][e][f][g]







        share|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Operate[#[[0]], First@#] &[a[b[c]]]



          a[b][c]




          ClearAll[deCompose]
          
deCompose = Nest[Operate[#[[0]], First@#] &, #, Depth[#] - 2] &;
          

          
deCompose@a[b[c]]



          a[b][c]




          exp = Compose[a, b, c, d, e, f, g]



          a[b[c[d[e[f[g]]]]]]




          deCompose @ exp



          a[b][c][d][e][f][g]







          share|improve this answer











          $endgroup$



          Operate[#[[0]], First@#] &[a[b[c]]]



          a[b][c]




          ClearAll[deCompose]
          
deCompose = Nest[Operate[#[[0]], First@#] &, #, Depth[#] - 2] &;
          

          
deCompose@a[b[c]]



          a[b][c]




          exp = Compose[a, b, c, d, e, f, g]



          a[b[c[d[e[f[g]]]]]]




          deCompose @ exp



          a[b][c][d][e][f][g]








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 34 mins ago

























          answered 41 mins ago









          kglrkglr

          180k9200413




          180k9200413























              2












              $begingroup$

              Not particularly "clean," but it works:



              Operate[Head[#], Level[#, 2][[2]]] & @ a[b[c]]


              For the full generalization:



              Nest[Operate[Head[#], Level[#, 2][[2]]] & , #, Depth[#] -2] & @
              
 a[b[c[d[e]]]]





              share|improve this answer











              $endgroup$


















                2












                $begingroup$

                Not particularly "clean," but it works:



                Operate[Head[#], Level[#, 2][[2]]] & @ a[b[c]]


                For the full generalization:



                Nest[Operate[Head[#], Level[#, 2][[2]]] & , #, Depth[#] -2] & @
                
 a[b[c[d[e]]]]





                share|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Not particularly "clean," but it works:



                  Operate[Head[#], Level[#, 2][[2]]] & @ a[b[c]]


                  For the full generalization:



                  Nest[Operate[Head[#], Level[#, 2][[2]]] & , #, Depth[#] -2] & @
                  
 a[b[c[d[e]]]]





                  share|improve this answer











                  $endgroup$



                  Not particularly "clean," but it works:



                  Operate[Head[#], Level[#, 2][[2]]] & @ a[b[c]]


                  For the full generalization:



                  Nest[Operate[Head[#], Level[#, 2][[2]]] & , #, Depth[#] -2] & @
                  
 a[b[c[d[e]]]]






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 33 mins ago

























                  answered 58 mins ago









                  David G. StorkDavid G. Stork

                  24.1k22153




                  24.1k22153






























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