How do I transpose the first and deepest levels of an arbitrarily nested array?












10












$begingroup$


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
    Mar 29 at 19:41










  • $begingroup$
    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
    Mar 29 at 19:43












  • $begingroup$
    @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
    Mar 29 at 19:52


















10












$begingroup$


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
    Mar 29 at 19:41










  • $begingroup$
    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
    Mar 29 at 19:43












  • $begingroup$
    @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
    Mar 29 at 19:52
















10












10








10


1



$begingroup$


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 Mar 30 at 5:04









J. M. is away♦

98.9k10311467




98.9k10311467










asked Mar 29 at 19:09









Kuba♦Kuba

107k12211533




107k12211533












  • $begingroup$
    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
    Mar 29 at 19:41










  • $begingroup$
    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
    Mar 29 at 19:43












  • $begingroup$
    @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
    Mar 29 at 19:52




















  • $begingroup$
    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
    Mar 29 at 19:41










  • $begingroup$
    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
    Mar 29 at 19:43












  • $begingroup$
    @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
    Mar 29 at 19:52


















$begingroup$
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
Mar 29 at 19:41




$begingroup$
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
Mar 29 at 19:41












$begingroup$
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
Mar 29 at 19:43






$begingroup$
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
Mar 29 at 19:43














$begingroup$
@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
Mar 29 at 19:52






$begingroup$
@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
Mar 29 at 19:52












4 Answers
4






active

oldest

votes


















11












$begingroup$

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









    $endgroup$





















      2












      $begingroup$

      Terrible solution using Table but works:



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





      share|improve this answer









      $endgroup$





















        1












        $begingroup$

        This is a bit roundabout but very versatile. First convert the structure arr into a list of rules, then reassemble these rules (after proper modification using RotateRight to transpose first and deepest levels) into a new structure.



        Using the toTree function from this answer (thanks to @b3m2a1 !):



        arr = {{a, b}, {a, b}, {{{a, b}, {a, b}, {a, b}}, {{a, b}, {a, b}, {a, b}}}};
        
toTree[Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, {-1}]]]



        {{a, a, {{a, a, a}, {a, a, a}}}, {b, b, {{b, b, b}, {b, b, b}}}}







        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









          11












          $begingroup$

          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









          $endgroup$


















            11












            $begingroup$

            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









            $endgroup$
















              11












              11








              11





              $begingroup$

              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 Mar 29 at 19:53









              andre314andre314

              12.4k12353




              12.4k12353























                  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 Mar 29 at 19:42









                      C. E.C. E.

                      51k3101206




                      51k3101206























                          2












                          $begingroup$

                          Terrible solution using Table but works:



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





                          share|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            Terrible solution using Table but works:



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





                            share|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              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 Mar 29 at 20:16









                              RomanRoman

                              4,3151027




                              4,3151027























                                  1












                                  $begingroup$

                                  This is a bit roundabout but very versatile. First convert the structure arr into a list of rules, then reassemble these rules (after proper modification using RotateRight to transpose first and deepest levels) into a new structure.



                                  Using the toTree function from this answer (thanks to @b3m2a1 !):



                                  arr = {{a, b}, {a, b}, {{{a, b}, {a, b}, {a, b}}, {{a, b}, {a, b}, {a, b}}}};
                                  
toTree[Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, {-1}]]]



                                  {{a, a, {{a, a, a}, {a, a, a}}}, {b, b, {{b, b, b}, {b, b, b}}}}







                                  share|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    This is a bit roundabout but very versatile. First convert the structure arr into a list of rules, then reassemble these rules (after proper modification using RotateRight to transpose first and deepest levels) into a new structure.



                                    Using the toTree function from this answer (thanks to @b3m2a1 !):



                                    arr = {{a, b}, {a, b}, {{{a, b}, {a, b}, {a, b}}, {{a, b}, {a, b}, {a, b}}}};
                                    
toTree[Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, {-1}]]]



                                    {{a, a, {{a, a, a}, {a, a, a}}}, {b, b, {{b, b, b}, {b, b, b}}}}







                                    share|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      This is a bit roundabout but very versatile. First convert the structure arr into a list of rules, then reassemble these rules (after proper modification using RotateRight to transpose first and deepest levels) into a new structure.



                                      Using the toTree function from this answer (thanks to @b3m2a1 !):



                                      arr = {{a, b}, {a, b}, {{{a, b}, {a, b}, {a, b}}, {{a, b}, {a, b}, {a, b}}}};
                                      
toTree[Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, {-1}]]]



                                      {{a, a, {{a, a, a}, {a, a, a}}}, {b, b, {{b, b, b}, {b, b, b}}}}







                                      share|improve this answer









                                      $endgroup$



                                      This is a bit roundabout but very versatile. First convert the structure arr into a list of rules, then reassemble these rules (after proper modification using RotateRight to transpose first and deepest levels) into a new structure.



                                      Using the toTree function from this answer (thanks to @b3m2a1 !):



                                      arr = {{a, b}, {a, b}, {{{a, b}, {a, b}, {a, b}}, {{a, b}, {a, b}, {a, b}}}};
                                      
toTree[Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, {-1}]]]



                                      {{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 Mar 30 at 21:29









                                      RomanRoman

                                      4,3151027




                                      4,3151027






























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                                          Plaza Victoria

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                                          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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                                          Puebla de Zaragoza

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                                          Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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                                          Musa

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                                          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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