How to transpose the 1st and -1th levels of arbitrarily nested array? The Next CEO of Stack OverflowEmulating R data frame getters with UpValuesQuickly pruning elements in one structured array that exist in a separate unordered arrayJoining sub-lists nested two levels deep in a list`Part` like `Delete`: How to delete list of columns or arbitrarily deeper levelsHow to mesh a region using adaptive cubic elementsHow to efficiently Flatten nested lists while preserving select levels?Distribute elements of one line across arbitrary dimension of another listDeep level nested list addition`Transpose` nested `Association`Reformatting the pattern inside a list

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How to transpose the 1st and -1th levels of arbitrarily nested array?



The Next CEO of Stack OverflowEmulating R data frame getters with UpValuesQuickly pruning elements in one structured array that exist in a separate unordered arrayJoining sub-lists nested two levels deep in a list`Part` like `Delete`: How to delete list of columns or arbitrarily deeper levelsHow to mesh a region using adaptive cubic elementsHow to efficiently Flatten nested lists while preserving select levels?Distribute elements of one line across arbitrary dimension of another listDeep level nested list addition`Transpose` nested `Association`Reformatting the pattern inside a list










3












$begingroup$


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


?



So I need to swap the first and last dimension. Which should in principle be possible because altough 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 negative level specification it would have been easy. That is not the case.



One can think about that question as: how to 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.










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
    1 hour 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
    1 hour 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
    57 mins ago











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
















3












$begingroup$


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


?



So I need to swap the first and last dimension. Which should in principle be possible because altough 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 negative level specification it would have been easy. That is not the case.



One can think about that question as: how to 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.










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
    1 hour 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
    1 hour 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
    57 mins ago











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














3












3








3





$begingroup$


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


?



So I need to swap the first and last dimension. Which should in principle be possible because altough 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 negative level specification it would have been easy. That is not the case.



One can think about that question as: how to 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.










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


?



So I need to swap the first and last dimension. Which should in principle be possible because altough 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 negative level specification it would have been easy. That is not the case.



One can think about that question as: how to 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.







list-manipulation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 52 mins ago







Kuba

















asked 1 hour 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
    1 hour 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
    1 hour 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
    57 mins ago











  • $begingroup$
    @Roman Level[arr, -2]` should be a rectangular array but rows do not need to be the same.
    $endgroup$
    – Kuba♦
    53 mins 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
    51 mins 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
    1 hour 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
    1 hour 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
    57 mins ago











  • $begingroup$
    @Roman Level[arr, -2]` should be a rectangular array but rows do not need to be the same.
    $endgroup$
    – Kuba♦
    53 mins 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
    51 mins 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
57 mins 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
57 mins ago













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




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











3 Answers
3






active

oldest

votes


















3












$begingroup$

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









$endgroup$




















    0












    $begingroup$

    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









    $endgroup$




















      0












      $begingroup$

      Terrible solution using Table but works:



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





      share|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        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












          $begingroup$

          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












            3








            3





            $begingroup$

            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 49 mins ago









            andre314andre314

            12.2k12352




            12.2k12352





















                0












                $begingroup$

                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









                $endgroup$

















                  0












                  $begingroup$

                  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









                  $endgroup$















                    0












                    0








                    0





                    $begingroup$

                    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 1 hour ago









                    C. E.C. E.

                    50.8k399205




                    50.8k399205





















                        0












                        $begingroup$

                        Terrible solution using Table but works:



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





                        share|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          Terrible solution using Table but works:



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





                          share|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            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 27 mins ago









                            RomanRoman

                            3,9361022




                            3,9361022



























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                                When We Were Young (canção de Adele) Índice Antecedentes e lançamento | Faixas e formatos | Performances e covers | Desempenho nas tabelas musicais | Créditos | Histórico de lançamento | Referências Menu de navegação«Best albums of 2015»«Adele - When We Were Young (Radio Date: 22-01-2016)»«When We Were Young - Single by Adele»«Adele: Inside Her Private Life and Triumphant Return»«adele interview: world exclusive first interview in three years»«Tobias Jesso Jr: since Adele tweeted his song, he's even bigger than his dad»«Adele interviews Tobias Jesso Jr: 'I think that a couple of the ideas we had could be rap songs'»«Adele on Her Return: 'I Was So Frightened That No One Cared'»«Tobias Jesso Jr. on working with Adele: 'I was as nervous as shit'»«How Ariel Rechtshaid Pushed Adele To Her Limit»«Adele Previews 'When We Were Young' on '60 Minutes' Teaser, Tops Trending 140»«Adele Performs New '25' Ballad, "When We Were Young" Live: Watch»«Which '25' Song Should Be Adele's Next Single?»«Adele's 'When We Were Young' Confirmed As Second Single From '25'»«Adele's new single artwork for 'When We Were Young' is perfectly adorable»«Adele at the BBC review: honest, funny and spectacular – Celebrity News News»«'Saturday Night Live': Adele Sings 'Hello' and 'When We Were Young'»«'Adele: Live in New York City' NBC Special – Set List Revealed!»«Adele Closes Out the 2016 Brit Awards With 'When We Were Young'»«What Is The Adele Live Tour Set List? There's No Way She'd Leave Out These 8 Songs»«See Demi Lovato's Soaring Cover of Adele's 'When We Were Young'»«'The Voice': 5 Best Moments From Week 1 Blind Auditions»«Adele – When We Were Young (Media Control Charts)»«Adele – When We Were Young (Entertainment Monitoring Africa)»«Adele – When We Were Young (ARIA Charts)»«Adele – When We Were Young (Ö3 Austria Top 40)»«Adele – When We Were Young (Ultratop 50)»«Adele – When We Were Young (Ultratop 40)»«Adele – When We Were Young (Canadian Hot 100)»«Adele – When We Were Young (Tracklisten)»«Adele – When We Were Young (The Official Charts Company)»«Adele – Hello (IFPI Slovenská Republika)»«Adele – When We Were Young (Productores de Música de España)»«Adele – When We Were Young (Billboard Hot 100)»«Adele – When We Were Young (Pop Songs)»«Adele – When We Were Young (Adult Pop Songs)»«Adele – When We Were Young (Hot Adult Contemporary Charts)»«Adele – When We Were Young (Hot Dance Club Songs)»«Adele – When We Were Young (Rock Airplay)»«Adele – When We Were Young (IFPI Finlândia)»«Adele – When We Were Young (Syndicat National de l'Éditon Phonographique)»«Adele – When We Were Young (Magyar Hanglemezkiadók Szövetsége)»«Adele – When We Were Young (Irish Recorded Music Association)»«Adele – When We Were Young (Mexico Airplay)»«Adele – When We Were Young (VG-lista)»«Adele – When We Were Young (NZ Top 40 Singles)»«Adele – When We Were Young (MegaCharts)»«Adele – When We Were Young (Związek Producentów Audio Video)»«Adele – When We Were Young (Portugal Digital Songs)»«Adele – When We Were Young (UK Indie Singles Chart)»«Adele – When We Were Young (UK Singles Chart)»«Adele – When We Were Young (Sverigetopplistan)»«Adele – When We Were Young (Schweizer Hitparade)»«Adele – When We Were Young (Portugal Digital Songs)»«Music Canada – Gold/Platinum – When We Were Young»«NZ Top 40 Singles Chart»«Certified Awards»e