How do I transpose the 1st and -1th levels of an arbitrarily nested array? The Next CEO of Stack OverflowEmulating R data frame getters with UpValuesQuickly pruning elements in one structured array that exist in a separate unordered arrayJoining sub-lists nested two levels deep in a list`Part` like `Delete`: How to delete list of columns or arbitrarily deeper levelsHow to mesh a region using adaptive cubic elementsHow to efficiently Flatten nested lists while preserving select levels?Distribute elements of one line across arbitrary dimension of another listDeep level nested list addition`Transpose` nested `Association`Reformatting the pattern inside a list

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How do I transpose the 1st and -1th levels of an arbitrarily nested array?



The Next CEO of Stack OverflowEmulating R data frame getters with UpValuesQuickly pruning elements in one structured array that exist in a separate unordered arrayJoining sub-lists nested two levels deep in a list`Part` like `Delete`: How to delete list of columns or arbitrarily deeper levelsHow to mesh a region using adaptive cubic elementsHow to efficiently Flatten nested lists while preserving select levels?Distribute elements of one line across arbitrary dimension of another listDeep level nested list addition`Transpose` nested `Association`Reformatting the pattern inside a list










6












$begingroup$


Is there a straightforward way to convert



arr = 
 a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
;


to:



a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b


?



I need to swap the first and last dimension. Which should in principle be possible, because, although arr does not have a fixed structure, the 'bottom' is always uniform:



Level[arr, -2]



a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b



Had Transpose/Flatten/MapThread accepted a negative level specification it would have been easy. That is not the case.



One can think about that question as: How do I create arr2 so that arr[[whatever__, y_]] == arr2[[y, whatever__]]?



EDIT:



In general Level[arr, -2] should be a rectangular array, but rows do not need to be the same.



So this:



a1, b1, a2, b2, a3, b3, a4, b4, a5, b5, a6, b6, a7,b7 ;


should end up:



 a1, a2, a3, a4, a5, a6, a7 , ...;





list-manipulation







share|improve this question











$endgroup$











  • $begingroup$
    Not a solution, but Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]] gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though: SparseArray does not construct ragged structures.
    $endgroup$
    – Roman
    5 hours ago










  • $begingroup$
    Maybe something along the lines of arr /. a,b->a,a,b->b? Or perhaps more generally, arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?
    $endgroup$
    – Carl Woll
    5 hours ago











  • $begingroup$
    Does your list always contain a,b at the lowest level, or can there be anything there as long as they're all of same length?
    $endgroup$
    – Roman
    5 hours ago











  • $begingroup$
    @Roman Level[arr, -2]` should be a rectangular array but rows do not need to be the same.
    $endgroup$
    – Kuba♦
    5 hours ago










  • $begingroup$
    @Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
    $endgroup$
    – Roman
    5 hours ago
















6












$begingroup$


Is there a straightforward way to convert



arr = 
 a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
;


to:



a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b


?



I need to swap the first and last dimension. Which should in principle be possible, because, although arr does not have a fixed structure, the 'bottom' is always uniform:



Level[arr, -2]



a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b



Had Transpose/Flatten/MapThread accepted a negative level specification it would have been easy. That is not the case.



One can think about that question as: How do I create arr2 so that arr[[whatever__, y_]] == arr2[[y, whatever__]]?



EDIT:



In general Level[arr, -2] should be a rectangular array, but rows do not need to be the same.



So this:



a1, b1, a2, b2, a3, b3, a4, b4, a5, b5, a6, b6, a7,b7 ;


should end up:



 a1, a2, a3, a4, a5, a6, a7 , ...;





list-manipulation







share|improve this question











$endgroup$











  • $begingroup$
    Not a solution, but Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]] gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though: SparseArray does not construct ragged structures.
    $endgroup$
    – Roman
    5 hours ago










  • $begingroup$
    Maybe something along the lines of arr /. a,b->a,a,b->b? Or perhaps more generally, arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?
    $endgroup$
    – Carl Woll
    5 hours ago











  • $begingroup$
    Does your list always contain a,b at the lowest level, or can there be anything there as long as they're all of same length?
    $endgroup$
    – Roman
    5 hours ago











  • $begingroup$
    @Roman Level[arr, -2]` should be a rectangular array but rows do not need to be the same.
    $endgroup$
    – Kuba♦
    5 hours ago










  • $begingroup$
    @Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
    $endgroup$
    – Roman
    5 hours ago














6












6








6





$begingroup$


Is there a straightforward way to convert



arr = 
 a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
;


to:



a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b


?



I need to swap the first and last dimension. Which should in principle be possible, because, although arr does not have a fixed structure, the 'bottom' is always uniform:



Level[arr, -2]



a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b



Had Transpose/Flatten/MapThread accepted a negative level specification it would have been easy. That is not the case.



One can think about that question as: How do I create arr2 so that arr[[whatever__, y_]] == arr2[[y, whatever__]]?



EDIT:



In general Level[arr, -2] should be a rectangular array, but rows do not need to be the same.



So this:



a1, b1, a2, b2, a3, b3, a4, b4, a5, b5, a6, b6, a7,b7 ;


should end up:



 a1, a2, a3, a4, a5, a6, a7 , ...;





list-manipulation







share|improve this question











$endgroup$




Is there a straightforward way to convert



arr = 
 a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
;


to:



a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b


?



I need to swap the first and last dimension. Which should in principle be possible, because, although arr does not have a fixed structure, the 'bottom' is always uniform:



Level[arr, -2]



a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b



Had Transpose/Flatten/MapThread accepted a negative level specification it would have been easy. That is not the case.



One can think about that question as: How do I create arr2 so that arr[[whatever__, y_]] == arr2[[y, whatever__]]?



EDIT:



In general Level[arr, -2] should be a rectangular array, but rows do not need to be the same.



So this:



a1, b1, a2, b2, a3, b3, a4, b4, a5, b5, a6, b6, a7,b7 ;


should end up:



 a1, a2, a3, a4, a5, a6, a7 , ...;





list-manipulation




list-manipulation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 8 mins ago









Peter Mortensen

33627




33627










asked 5 hours ago









Kuba♦Kuba

107k12210531




107k12210531











  • $begingroup$
    Not a solution, but Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]] gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though: SparseArray does not construct ragged structures.
    $endgroup$
    – Roman
    5 hours ago










  • $begingroup$
    Maybe something along the lines of arr /. a,b->a,a,b->b? Or perhaps more generally, arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?
    $endgroup$
    – Carl Woll
    5 hours ago











  • $begingroup$
    Does your list always contain a,b at the lowest level, or can there be anything there as long as they're all of same length?
    $endgroup$
    – Roman
    5 hours ago











  • $begingroup$
    @Roman Level[arr, -2]` should be a rectangular array but rows do not need to be the same.
    $endgroup$
    – Kuba♦
    5 hours ago










  • $begingroup$
    @Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
    $endgroup$
    – Roman
    5 hours ago

















  • $begingroup$
    Not a solution, but Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]] gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though: SparseArray does not construct ragged structures.
    $endgroup$
    – Roman
    5 hours ago










  • $begingroup$
    Maybe something along the lines of arr /. a,b->a,a,b->b? Or perhaps more generally, arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?
    $endgroup$
    – Carl Woll
    5 hours ago











  • $begingroup$
    Does your list always contain a,b at the lowest level, or can there be anything there as long as they're all of same length?
    $endgroup$
    – Roman
    5 hours ago











  • $begingroup$
    @Roman Level[arr, -2]` should be a rectangular array but rows do not need to be the same.
    $endgroup$
    – Kuba♦
    5 hours ago










  • $begingroup$
    @Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
    $endgroup$
    – Roman
    5 hours ago
















$begingroup$
Not a solution, but Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]] gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though: SparseArray does not construct ragged structures.
$endgroup$
– Roman
5 hours ago




$begingroup$
Not a solution, but Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]] gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though: SparseArray does not construct ragged structures.
$endgroup$
– Roman
5 hours ago












$begingroup$
Maybe something along the lines of arr /. a,b->a,a,b->b? Or perhaps more generally, arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?
$endgroup$
– Carl Woll
5 hours ago





$begingroup$
Maybe something along the lines of arr /. a,b->a,a,b->b? Or perhaps more generally, arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?
$endgroup$
– Carl Woll
5 hours ago













$begingroup$
Does your list always contain a,b at the lowest level, or can there be anything there as long as they're all of same length?
$endgroup$
– Roman
5 hours ago





$begingroup$
Does your list always contain a,b at the lowest level, or can there be anything there as long as they're all of same length?
$endgroup$
– Roman
5 hours ago













$begingroup$
@Roman Level[arr, -2]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
5 hours ago




$begingroup$
@Roman Level[arr, -2]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
5 hours ago












$begingroup$
@Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
$endgroup$
– Roman
5 hours ago





$begingroup$
@Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
$endgroup$
– Roman
5 hours ago











3 Answers
3






active

oldest

votes


















8












$begingroup$

arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;

SetAttributes[f1, Listable]
Apply[f1, arr, 0, -3] /. f1 -> List



a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b







share|improve this answer









$endgroup$




















    3












    $begingroup$

    This is what the list at the lowest level looks like:



    el = First@Level[list, -2];


    Using this, we can solve it with a rules-based approach:



    list /. el -> # & /@ el


    or a recursive approach like this:



    walk[lists : __List, i_] := walk[#, i] & /@ lists
    walk[atoms : __, i_] := i
    walk[list, #] & /@ el





    share|improve this answer









    $endgroup$




















      1












      $begingroup$

      Terrible solution using Table but works:



      Table[Map[#[[i]] &, arr, -2], i, Last[Dimensions[Level[arr, -2]]]]





      share|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        8












        $begingroup$

        arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;

        SetAttributes[f1, Listable]
        Apply[f1, arr, 0, -3] /. f1 -> List



        a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b







        share|improve this answer









        $endgroup$

















          8












          $begingroup$

          arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;

          SetAttributes[f1, Listable]
          Apply[f1, arr, 0, -3] /. f1 -> List



          a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b







          share|improve this answer









          $endgroup$















            8












            8








            8





            $begingroup$

            arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;

            SetAttributes[f1, Listable]
            Apply[f1, arr, 0, -3] /. f1 -> List



            a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b







            share|improve this answer









            $endgroup$



            arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;

            SetAttributes[f1, Listable]
            Apply[f1, arr, 0, -3] /. f1 -> List



            a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b








            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 5 hours ago









            andre314andre314

            12.3k12352




            12.3k12352





















                3












                $begingroup$

                This is what the list at the lowest level looks like:



                el = First@Level[list, -2];


                Using this, we can solve it with a rules-based approach:



                list /. el -> # & /@ el


                or a recursive approach like this:



                walk[lists : __List, i_] := walk[#, i] & /@ lists
                walk[atoms : __, i_] := i
                walk[list, #] & /@ el





                share|improve this answer









                $endgroup$

















                  3












                  $begingroup$

                  This is what the list at the lowest level looks like:



                  el = First@Level[list, -2];


                  Using this, we can solve it with a rules-based approach:



                  list /. el -> # & /@ el


                  or a recursive approach like this:



                  walk[lists : __List, i_] := walk[#, i] & /@ lists
                  walk[atoms : __, i_] := i
                  walk[list, #] & /@ el





                  share|improve this answer









                  $endgroup$















                    3












                    3








                    3





                    $begingroup$

                    This is what the list at the lowest level looks like:



                    el = First@Level[list, -2];


                    Using this, we can solve it with a rules-based approach:



                    list /. el -> # & /@ el


                    or a recursive approach like this:



                    walk[lists : __List, i_] := walk[#, i] & /@ lists
                    walk[atoms : __, i_] := i
                    walk[list, #] & /@ el





                    share|improve this answer









                    $endgroup$



                    This is what the list at the lowest level looks like:



                    el = First@Level[list, -2];


                    Using this, we can solve it with a rules-based approach:



                    list /. el -> # & /@ el


                    or a recursive approach like this:



                    walk[lists : __List, i_] := walk[#, i] & /@ lists
                    walk[atoms : __, i_] := i
                    walk[list, #] & /@ el






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 5 hours ago









                    C. E.C. E.

                    50.9k399205




                    50.9k399205





















                        1












                        $begingroup$

                        Terrible solution using Table but works:



                        Table[Map[#[[i]] &, arr, -2], i, Last[Dimensions[Level[arr, -2]]]]





                        share|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          Terrible solution using Table but works:



                          Table[Map[#[[i]] &, arr, -2], i, Last[Dimensions[Level[arr, -2]]]]





                          share|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            Terrible solution using Table but works:



                            Table[Map[#[[i]] &, arr, -2], i, Last[Dimensions[Level[arr, -2]]]]





                            share|improve this answer









                            $endgroup$



                            Terrible solution using Table but works:



                            Table[Map[#[[i]] &, arr, -2], i, Last[Dimensions[Level[arr, -2]]]]






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 4 hours ago









                            RomanRoman

                            3,9761022




                            3,9761022



























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                                Torre de la Isleta Índice Véase también Referencias Bibliografía Enlaces externos Menú de navegación38°25′58″N 0°23′02″O / 38.43277778, -0.3838888938°25′58″N 0°23′02″O / 38.43277778, -0.38388889Torre de la Illeta de l’Horta o Torre Saleta. Base de datos de bienes inmuebles. Patrimonio Cultural. Secretaría de Estado de CulturaFicha BIC Torre de la Illeta de l’Horta. Dirección General de Patrimonio Cultural. Generalitat ValencianaLugares de interés. Ayuntamiento del CampelloTorre de la Isleta en CastillosNet.org

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                                Bienes de interés cultural de la provincia de AlicanteTorres defensivas de la provincia de AlicanteCampo de AlicanteTorres defensivas de la Comunidad Valenciana en el siglo XVICampello siglo XVICampelloprovincia de AlicanteEspañayacimiento arqueológico de La IsletabóvedamamposteríasillarejosilleríaménsulasVirrey de ValenciaBernardino de Cárdenas y Pachecoduque de Maquedapiratas berberiscosCastillo de Santa BárbaraAlicanteTorre de AguasCampelloBien de Interés Cultural Torre de la Isleta Torre de la Illeta Bien de Interés Cultural Patrimonio histórico de España Vista de la Torre Declaración 12 de febrero de 1996 Figura de protección Monumento Código RI-51 - 0009165 - 00000 [ 1 ] ​ Coordenadas 38°25′58″N 0°23′02″O  /  38.43277778, -0.38388889 Coordenadas: 38°25′58″N 0°23′02″O  /  38.43277778, -0.38388889 ...
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