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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                                          In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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                                          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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                                          How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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                                          1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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