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 23 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









      $endgroup$














        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "387"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: false,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: null,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194206%2fhow-do-i-transpose-the-first-and-deepest-levels-of-an-arbitrarily-nested-array%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematica Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194206%2fhow-do-i-transpose-the-first-and-deepest-levels-of-an-arbitrarily-nested-array%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Fluorita

                                Image
                                Fluorita Fluorita (blau) amb pirita (daurat). Fórmula química CaF 2 Epònim fluor Localitat tipus Jáchymov Classificació Categoria Halurs Nickel-Strunz 10a ed. 3.AB.25 Nickel-Strunz 9a ed. 3.AB.25 Nickel-Strunz 8a ed. III/A.08 Dana 9.2.1.1 Heys 8.4.7 Propietats Sistema cristal·lí Cúbic Hàbit cristal·lí Sol presentar cubs; menys freqüentment dodecaedres. De vegades hexaoctaedres i tetrahexaedres. Per la combinació d'aquestes formes, les arestes dels cubs són sovint modificades. De vegades els cristalls poden presentar diferències de creixement entre les cares. Sovint s'observen cristalls compostos per sobrecreixement. Pot ser massiva, compacta, terrosa, columnar (poc freqüent), globular en agregats o botroïdal. Estructura cristal·lina a = 5.4626Å Simetria Classe (H-M): m 3 m (4/ m 3 2/ m ) - Hexoctaèdric; Grup espacial: Fm 3 m Color Blanc, groc, verd, violeta, vermell, rosa, blau o negre Mac
                                Read more

                                Hulsita

                                Image
                                Hulsita Hulsita de la localitat tipus Fórmula química Fe 2+ 2 Fe 3+ O 2 BO 3 Epònim Alfred Hulse Brooks Localitat tipus mont Brooks, península de Seward, Nome Borough, Alaska, Estats Units Classificació Categoria borats Nickel-Strunz 10a ed. 6.AB.45 Nickel-Strunz 9a ed. 6.AB.45 Nickel-Strunz 8a ed. V/G.03 Dana 24.2.3.1 Heys 9.6.4 Propietats Sistema cristal·lí monoclínic Estructura cristal·lina a = 10,68Å; b = 3,09Å; c = 5,43Å; β = 94,15° Color negre Macles en [001]; un membre de la macla es troba girat 120° en relació a l'altre Exfoliació bona en {110} Duresa 3 Lluïssor vítria, submetàl·lica Densitat 4,28 g/cm 3 (mesurada); Més informació Estatus IMA mineral heretat (G) i mineral heretat (G) Any d'aprovació 1908 Referències [1] La hulsita és un mineral de la classe dels borats, que pertany al grup de la pinakiolita. Rep el seu nom en honor del nord-americà
                                Read more

                                Península de Txukotka

                                Image
                                Península de Txukotka Tipus Península Ubicació 66° N, 172° O  /  66°N , 172°O  / 66; -172 Mar mar dels Txuktxis La península de Txukotka a l'extrem oriental de Sibèria La península de Txukotka o península dels Txuktxis (en rus Чуко́тский полуо́стров , Txukotski poluóstrov ) és una península que es troba a l'extrem oriental d'Euràsia. Contingut 1 Descripció 2 Població 3 Vegeu també 4 Bibliografia Descripció L'estret de Bering, entre la península de Txukotka i la península de Seward a Alaska Foto de satèl·lit de la zona amb la península de Txukotka a l'esquerra El seu punt més extrem és el cap Dezhnev a la vora del nucli d'Uelen, un dels poquíssims llocs habitats de la zona. Al nord limita amb el mar dels Txuktxis i amb el mar de Bering al sud i amb l'estret de Bering a l'est. A la costa sud es troben la badia de Sant Lawrence i la badia de Kresta. La penínsu
                                Read more