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



The Next CEO of Stack OverflowA question about transforming one List into two Lists with additional requirementsEmulating R data frame getters with UpValuesQuickly pruning elements in one structured array that exist in a separate unordered array`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`How to extract the first element in nested lists










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 18 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





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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

                            3,9961022




                            3,9961022



























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                                Умершие в 420 годуПерсоналии по алфавитуПравители Европы V векаПерсонажи артурианыПерсонажи английского фольклора англ.Старого СевераКолчестерХарлеанских генеа...
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                                Roughly how much would it cost to hire a team of dwarves to build a home in the mountainside? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How much does a house cost?How long does it take to mine rock?How much does a house cost?How much gold would the construction of a forge cost?How much does a door cost?How much would it cost to make this magic item?How much would a glue bomb cost?How much does mandrake root cost?How much does a slave cost?How much does equipment cost?How much do sheep cost?How much would firearms cost?

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