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












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
    10 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
    10 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
    9 hours ago












  • $begingroup$
    @Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
    $endgroup$
    – Kuba♦
    9 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
    9 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
    10 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
    10 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
    9 hours ago












  • $begingroup$
    @Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
    $endgroup$
    – Kuba♦
    9 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
    9 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 42 mins ago









J. M. is slightly pensive♦

98.7k10311467




98.7k10311467










asked 10 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
    10 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
    10 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
    9 hours ago












  • $begingroup$
    @Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
    $endgroup$
    – Kuba♦
    9 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
    9 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
    10 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
    10 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
    9 hours ago












  • $begingroup$
    @Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
    $endgroup$
    – Kuba♦
    9 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
    9 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
10 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
10 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
10 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
10 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
9 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
9 hours ago














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




$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
9 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
9 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
9 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









    $endgroup$





















      2












      $begingroup$

      Terrible solution using Table but works:



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





      share|improve this answer









      $endgroup$














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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 9 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 10 hours ago









                    C. E.C. E.

                    50.9k399205




                    50.9k399205























                        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 9 hours ago









                            RomanRoman

                            4,0111022




                            4,0111022






























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                                Gregoriusmis

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                                Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
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