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Applying a function to a nested list


Nested List ProductExtract function argumentsCreating a simple function to compute the average of the difference between pairs of elements in an arraySorting non-numerical listsExtract part of list by reading it in a cyclic mannerApplying a function along the desired dimensions of a n-dimensional arrayManipulating Elements in a Triple-Nested ListApplying function to all elements of the list of listApplying a function to a list of symbolic ratiosApplying a function to a list (compounding)













1












$begingroup$


Say I have a list:



l = {{{a, b}, c}, d}


I now want to apply a function, call it F to that list in a way that I go from the lowest to highest level, i.e. 



F[{F[{F[{a,b}], c}] , d}]


Is there a function in Mathematica which does exactly that?






list-manipulation functions







share|improve this question











$endgroup$








  • 4




    $begingroup$
    Why does the order for d and c change, and not for a and b?
    $endgroup$
    – Carl Woll
    2 hours ago










  • $begingroup$
    My mistake, thank you for spotting that. I edited it.
    $endgroup$
    – amator2357
    2 hours ago
















1












$begingroup$


Say I have a list:



l = {{{a, b}, c}, d}


I now want to apply a function, call it F to that list in a way that I go from the lowest to highest level, i.e. 



F[{F[{F[{a,b}], c}] , d}]


Is there a function in Mathematica which does exactly that?






list-manipulation functions







share|improve this question











$endgroup$








  • 4




    $begingroup$
    Why does the order for d and c change, and not for a and b?
    $endgroup$
    – Carl Woll
    2 hours ago










  • $begingroup$
    My mistake, thank you for spotting that. I edited it.
    $endgroup$
    – amator2357
    2 hours ago














1












1








1





$begingroup$


Say I have a list:



l = {{{a, b}, c}, d}


I now want to apply a function, call it F to that list in a way that I go from the lowest to highest level, i.e. 



F[{F[{F[{a,b}], c}] , d}]


Is there a function in Mathematica which does exactly that?






list-manipulation functions







share|improve this question











$endgroup$




Say I have a list:



l = {{{a, b}, c}, d}


I now want to apply a function, call it F to that list in a way that I go from the lowest to highest level, i.e. 



F[{F[{F[{a,b}], c}] , d}]


Is there a function in Mathematica which does exactly that?






list-manipulation functions




list-manipulation functions






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 hours ago







amator2357

















asked 3 hours ago









amator2357amator2357

4538




4538








  • 4




    $begingroup$
    Why does the order for d and c change, and not for a and b?
    $endgroup$
    – Carl Woll
    2 hours ago










  • $begingroup$
    My mistake, thank you for spotting that. I edited it.
    $endgroup$
    – amator2357
    2 hours ago














  • 4




    $begingroup$
    Why does the order for d and c change, and not for a and b?
    $endgroup$
    – Carl Woll
    2 hours ago










  • $begingroup$
    My mistake, thank you for spotting that. I edited it.
    $endgroup$
    – amator2357
    2 hours ago








4




4




$begingroup$
Why does the order for d and c change, and not for a and b?
$endgroup$
– Carl Woll
2 hours ago




$begingroup$
Why does the order for d and c change, and not for a and b?
$endgroup$
– Carl Woll
2 hours ago












$begingroup$
My mistake, thank you for spotting that. I edited it.
$endgroup$
– amator2357
2 hours ago




$begingroup$
My mistake, thank you for spotting that. I edited it.
$endgroup$
– amator2357
2 hours ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

Replace[l, x_List :> F[x], All]



F[{F[{F[{a, b}], c}], d}]




Also



ClearAll[f]

f[Except[_List, x_]] := x;

MapAll[f, l]



f[{f[{f[{a, b}], c}], d}]







share|improve this answer









$endgroup$





















    1












    $begingroup$

    F@*Reverse@Map[F@*Reverse, l, -2]



    F[{d, F[{c, F[{b, a}]}]}]




    Fold[F[{#2, #1}] &, Flatten[l]]



    F[{d, F[{c, F[{b, a}]}]}]







    share|improve this answer









    $endgroup$













    • $begingroup$
      I think that Flatten works in this trivial case but it will not "find" the levels in a more complex situation
      $endgroup$
      – J42161217
      2 hours ago



















    1












    $begingroup$

    Another possibility, if you want just lists to acquire the F wrapper:



    l /. List -> F@*List



    F[{F[{F[{a, b}], c}], d}]







    share|improve this answer









    $endgroup$





















      0












      $begingroup$

      This works:



      {{{a, b}, c}, d} //. {{s_?ListQ, t_?(Not@*ListQ)} :> {f[s], t}} // f


      It's a bit hackish because of the separate invocation of f at the end, but it returns the desired result.






      share|improve this answer









      $endgroup$














        Your Answer








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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Replace[l, x_List :> F[x], All]



        F[{F[{F[{a, b}], c}], d}]




        Also



        ClearAll[f]
        
f[Except[_List, x_]] := x;
        
MapAll[f, l]



        f[{f[{f[{a, b}], c}], d}]







        share|improve this answer









        $endgroup$


















          2












          $begingroup$

          Replace[l, x_List :> F[x], All]



          F[{F[{F[{a, b}], c}], d}]




          Also



          ClearAll[f]
          
f[Except[_List, x_]] := x;
          
MapAll[f, l]



          f[{f[{f[{a, b}], c}], d}]







          share|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Replace[l, x_List :> F[x], All]



            F[{F[{F[{a, b}], c}], d}]




            Also



            ClearAll[f]
            
f[Except[_List, x_]] := x;
            
MapAll[f, l]



            f[{f[{f[{a, b}], c}], d}]







            share|improve this answer









            $endgroup$



            Replace[l, x_List :> F[x], All]



            F[{F[{F[{a, b}], c}], d}]




            Also



            ClearAll[f]
            
f[Except[_List, x_]] := x;
            
MapAll[f, l]



            f[{f[{f[{a, b}], c}], d}]








            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 1 hour ago









            kglrkglr

            190k10209427




            190k10209427























                1












                $begingroup$

                F@*Reverse@Map[F@*Reverse, l, -2]



                F[{d, F[{c, F[{b, a}]}]}]




                Fold[F[{#2, #1}] &, Flatten[l]]



                F[{d, F[{c, F[{b, a}]}]}]







                share|improve this answer









                $endgroup$













                • $begingroup$
                  I think that Flatten works in this trivial case but it will not "find" the levels in a more complex situation
                  $endgroup$
                  – J42161217
                  2 hours ago
















                1












                $begingroup$

                F@*Reverse@Map[F@*Reverse, l, -2]



                F[{d, F[{c, F[{b, a}]}]}]




                Fold[F[{#2, #1}] &, Flatten[l]]



                F[{d, F[{c, F[{b, a}]}]}]







                share|improve this answer









                $endgroup$













                • $begingroup$
                  I think that Flatten works in this trivial case but it will not "find" the levels in a more complex situation
                  $endgroup$
                  – J42161217
                  2 hours ago














                1












                1








                1





                $begingroup$

                F@*Reverse@Map[F@*Reverse, l, -2]



                F[{d, F[{c, F[{b, a}]}]}]




                Fold[F[{#2, #1}] &, Flatten[l]]



                F[{d, F[{c, F[{b, a}]}]}]







                share|improve this answer









                $endgroup$



                F@*Reverse@Map[F@*Reverse, l, -2]



                F[{d, F[{c, F[{b, a}]}]}]




                Fold[F[{#2, #1}] &, Flatten[l]]



                F[{d, F[{c, F[{b, a}]}]}]








                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 2 hours ago









                Henrik SchumacherHenrik Schumacher

                61.3k585171




                61.3k585171












                • $begingroup$
                  I think that Flatten works in this trivial case but it will not "find" the levels in a more complex situation
                  $endgroup$
                  – J42161217
                  2 hours ago


















                • $begingroup$
                  I think that Flatten works in this trivial case but it will not "find" the levels in a more complex situation
                  $endgroup$
                  – J42161217
                  2 hours ago
















                $begingroup$
                I think that Flatten works in this trivial case but it will not "find" the levels in a more complex situation
                $endgroup$
                – J42161217
                2 hours ago




                $begingroup$
                I think that Flatten works in this trivial case but it will not "find" the levels in a more complex situation
                $endgroup$
                – J42161217
                2 hours ago











                1












                $begingroup$

                Another possibility, if you want just lists to acquire the F wrapper:



                l /. List -> F@*List



                F[{F[{F[{a, b}], c}], d}]







                share|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Another possibility, if you want just lists to acquire the F wrapper:



                  l /. List -> F@*List



                  F[{F[{F[{a, b}], c}], d}]







                  share|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Another possibility, if you want just lists to acquire the F wrapper:



                    l /. List -> F@*List



                    F[{F[{F[{a, b}], c}], d}]







                    share|improve this answer









                    $endgroup$



                    Another possibility, if you want just lists to acquire the F wrapper:



                    l /. List -> F@*List



                    F[{F[{F[{a, b}], c}], d}]








                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 55 mins ago









                    Carl WollCarl Woll

                    76.4k3100200




                    76.4k3100200























                        0












                        $begingroup$

                        This works:



                        {{{a, b}, c}, d} //. {{s_?ListQ, t_?(Not@*ListQ)} :> {f[s], t}} // f


                        It's a bit hackish because of the separate invocation of f at the end, but it returns the desired result.






                        share|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          This works:



                          {{{a, b}, c}, d} //. {{s_?ListQ, t_?(Not@*ListQ)} :> {f[s], t}} // f


                          It's a bit hackish because of the separate invocation of f at the end, but it returns the desired result.






                          share|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            This works:



                            {{{a, b}, c}, d} //. {{s_?ListQ, t_?(Not@*ListQ)} :> {f[s], t}} // f


                            It's a bit hackish because of the separate invocation of f at the end, but it returns the desired result.






                            share|improve this answer









                            $endgroup$



                            This works:



                            {{{a, b}, c}, d} //. {{s_?ListQ, t_?(Not@*ListQ)} :> {f[s], t}} // f


                            It's a bit hackish because of the separate invocation of f at the end, but it returns the desired result.







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 2 hours ago









                            ShredderroyShredderroy

                            1,6101117




                            1,6101117






























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