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
    9 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
    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










  • $begingroup$
    @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
    9 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
    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










  • $begingroup$
    @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 24 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
    9 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
    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










  • $begingroup$
    @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
    9 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
    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










  • $begingroup$
    @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
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
9 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
9 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
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






$begingroup$
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 9 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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                                Tabula Rosettana

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                                Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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