How to invert MapIndexed on a ragged structure? How to construct a tree from rules? The Next CEO of Stack OverflowFrom a list to a list of rulesHow to partition a list according to a nested table structure?Visualize a tree structure using TreeGraphExplore a nested listDrop selection of columns from a ragged arrayHow to check if two nested lists have the same structure?Rule-based branching construction of listsHow to convert a tree in a list?Replacement Rule for “flattening” list whilst adding attributesRagged Transpose

How do I go from 300 unfinished/half written blog posts, to published posts?

Limits on contract work without pre-agreed price/contract (UK)

Bold, vivid family

What can we do to stop prior company from asking us questions?

How do scammers retract money, while you can’t?

If/When UK leaves the EU, can a future goverment conduct a referendum to join the EU?

Contours of a clandestine nature

Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?

Non-deterministic sum of floats

What is the result of assigning to std::vector<T>::begin()?

Skipping indices in a product

Rotate a column

multiple labels for a single equation

How fast would a person need to move to trick the eye?

What does "Its cash flow is deeply negative" mean?

Is it ever safe to open a suspicious html file (e.g. email attachment)?

How to avoid supervisors with prejudiced views?

Example of a Mathematician/Physicist whose Other Publications during their PhD eclipsed their PhD Thesis

Make solar eclipses exceedingly rare, but still have new moons

How to Reset Passwords on Multiple Websites Easily?

How do we know the LHC results are robust?

Why do remote companies require working in the US?

Preparing Indesign booklet with .psd graphics for print

Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?



How to invert MapIndexed on a ragged structure? How to construct a tree from rules?



The Next CEO of Stack OverflowFrom a list to a list of rulesHow to partition a list according to a nested table structure?Visualize a tree structure using TreeGraphExplore a nested listDrop selection of columns from a ragged arrayHow to check if two nested lists have the same structure?Rule-based branching construction of listsHow to convert a tree in a list?Replacement Rule for “flattening” list whilst adding attributesRagged Transpose










4












$begingroup$


I have an arbitrary ragged nested list-of-lists (a tree) like



A = a, b, c, d, e, f, g, h, i, j, k, l, m, n;


Its structure is given by the rules



B = Flatten[MapIndexed[#2 -> #1 &, A, -1]]



1, 1 -> a, 1, 2 -> b, 2, 1 -> c, 2, 2 -> d, 3, 1, 1, 1 -> e, 3, 1, 1, 2 -> f, 3, 1, 1, 3 -> g, 3, 1, 1, 4 -> h, 3, 1, 1, 5 -> i, 3, 1, 2, 1 -> j, 3, 1, 2, 2 -> k, 3, 1, 2, 3 -> l, 3, 2 -> m, 4 -> n




How can I invert this operation? How can I construct A solely from the information given in B?










share|improve this question









$endgroup$
















    4












    $begingroup$


    I have an arbitrary ragged nested list-of-lists (a tree) like



    A = a, b, c, d, e, f, g, h, i, j, k, l, m, n;


    Its structure is given by the rules



    B = Flatten[MapIndexed[#2 -> #1 &, A, -1]]



    1, 1 -> a, 1, 2 -> b, 2, 1 -> c, 2, 2 -> d, 3, 1, 1, 1 -> e, 3, 1, 1, 2 -> f, 3, 1, 1, 3 -> g, 3, 1, 1, 4 -> h, 3, 1, 1, 5 -> i, 3, 1, 2, 1 -> j, 3, 1, 2, 2 -> k, 3, 1, 2, 3 -> l, 3, 2 -> m, 4 -> n




    How can I invert this operation? How can I construct A solely from the information given in B?










    share|improve this question









    $endgroup$














      4












      4








      4





      $begingroup$


      I have an arbitrary ragged nested list-of-lists (a tree) like



      A = a, b, c, d, e, f, g, h, i, j, k, l, m, n;


      Its structure is given by the rules



      B = Flatten[MapIndexed[#2 -> #1 &, A, -1]]



      1, 1 -> a, 1, 2 -> b, 2, 1 -> c, 2, 2 -> d, 3, 1, 1, 1 -> e, 3, 1, 1, 2 -> f, 3, 1, 1, 3 -> g, 3, 1, 1, 4 -> h, 3, 1, 1, 5 -> i, 3, 1, 2, 1 -> j, 3, 1, 2, 2 -> k, 3, 1, 2, 3 -> l, 3, 2 -> m, 4 -> n




      How can I invert this operation? How can I construct A solely from the information given in B?










      share|improve this question









      $endgroup$




      I have an arbitrary ragged nested list-of-lists (a tree) like



      A = a, b, c, d, e, f, g, h, i, j, k, l, m, n;


      Its structure is given by the rules



      B = Flatten[MapIndexed[#2 -> #1 &, A, -1]]



      1, 1 -> a, 1, 2 -> b, 2, 1 -> c, 2, 2 -> d, 3, 1, 1, 1 -> e, 3, 1, 1, 2 -> f, 3, 1, 1, 3 -> g, 3, 1, 1, 4 -> h, 3, 1, 1, 5 -> i, 3, 1, 2, 1 -> j, 3, 1, 2, 2 -> k, 3, 1, 2, 3 -> l, 3, 2 -> m, 4 -> n




      How can I invert this operation? How can I construct A solely from the information given in B?







      list-manipulation data-structures trees






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 3 hours ago









      RomanRoman

      3,9661022




      3,9661022




















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Here's a procedural way:



          Block[
          Nothing,
          Module[
          m = Max[Length /@ Keys[B]], arr,
          arr = ConstantArray[Nothing, Max /@ Transpose[PadRight[#, m] & /@ Keys[B]]];
          Map[Function[arr[[Sequence @@ #[[1]]]] = #[[2]]], B];
          arr
          ]
          ]

          a, b, c, d, e, f, g, h, i, j, k, l, m, n





          share|improve this answer









          $endgroup$




















            1












            $begingroup$

            Here's an inefficient but reasonably simple way:



            groupMe[rules_] :=
            If[Head[rules[[1]]] === Rule,
            Values@GroupBy[
            rules,
            (#[[1, 1]] &) ->
            (If[Length[#[[1]]] === 1, #[[2]], #[[1, 2 ;;]] -> #[[2]]] &),
            groupMe
            ],
            rules[[1]]
            ]

            groupMe[B]

            a, b, c, d, e, f, g, h, i, j, k, l, m, n





            share|improve this answer









            $endgroup$




















              1












              $begingroup$

              Here's a convoluted way using pattern replacements:



              DeleteCases[
              With[m = Max[Length /@ Keys[B]],
              Array[
              List,
              Max /@ Transpose[PadRight[#, m] & /@ Keys[B]]
              ] /.
              Map[
              Fold[
              Insert[
              #, ___,
              _,
              Append[ConstantArray[1, #2], -1]] &,
              #[[1]],
              Range[m - Length[#[[1]]]]
              ] -> #[[2]] &,
              B
              ]
              ],
              __Integer,
              Infinity
              ]

              a, b, c, d, e, f, g, h, i, j, k, l, m, n





              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%2f194217%2fhow-to-invert-mapindexed-on-a-ragged-structure-how-to-construct-a-tree-from-rul%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









                2












                $begingroup$

                Here's a procedural way:



                Block[
                Nothing,
                Module[
                m = Max[Length /@ Keys[B]], arr,
                arr = ConstantArray[Nothing, Max /@ Transpose[PadRight[#, m] & /@ Keys[B]]];
                Map[Function[arr[[Sequence @@ #[[1]]]] = #[[2]]], B];
                arr
                ]
                ]

                a, b, c, d, e, f, g, h, i, j, k, l, m, n





                share|improve this answer









                $endgroup$

















                  2












                  $begingroup$

                  Here's a procedural way:



                  Block[
                  Nothing,
                  Module[
                  m = Max[Length /@ Keys[B]], arr,
                  arr = ConstantArray[Nothing, Max /@ Transpose[PadRight[#, m] & /@ Keys[B]]];
                  Map[Function[arr[[Sequence @@ #[[1]]]] = #[[2]]], B];
                  arr
                  ]
                  ]

                  a, b, c, d, e, f, g, h, i, j, k, l, m, n





                  share|improve this answer









                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    Here's a procedural way:



                    Block[
                    Nothing,
                    Module[
                    m = Max[Length /@ Keys[B]], arr,
                    arr = ConstantArray[Nothing, Max /@ Transpose[PadRight[#, m] & /@ Keys[B]]];
                    Map[Function[arr[[Sequence @@ #[[1]]]] = #[[2]]], B];
                    arr
                    ]
                    ]

                    a, b, c, d, e, f, g, h, i, j, k, l, m, n





                    share|improve this answer









                    $endgroup$



                    Here's a procedural way:



                    Block[
                    Nothing,
                    Module[
                    m = Max[Length /@ Keys[B]], arr,
                    arr = ConstantArray[Nothing, Max /@ Transpose[PadRight[#, m] & /@ Keys[B]]];
                    Map[Function[arr[[Sequence @@ #[[1]]]] = #[[2]]], B];
                    arr
                    ]
                    ]

                    a, b, c, d, e, f, g, h, i, j, k, l, m, n






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 2 hours ago









                    b3m2a1b3m2a1

                    28.3k358163




                    28.3k358163





















                        1












                        $begingroup$

                        Here's an inefficient but reasonably simple way:



                        groupMe[rules_] :=
                        If[Head[rules[[1]]] === Rule,
                        Values@GroupBy[
                        rules,
                        (#[[1, 1]] &) ->
                        (If[Length[#[[1]]] === 1, #[[2]], #[[1, 2 ;;]] -> #[[2]]] &),
                        groupMe
                        ],
                        rules[[1]]
                        ]

                        groupMe[B]

                        a, b, c, d, e, f, g, h, i, j, k, l, m, n





                        share|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          Here's an inefficient but reasonably simple way:



                          groupMe[rules_] :=
                          If[Head[rules[[1]]] === Rule,
                          Values@GroupBy[
                          rules,
                          (#[[1, 1]] &) ->
                          (If[Length[#[[1]]] === 1, #[[2]], #[[1, 2 ;;]] -> #[[2]]] &),
                          groupMe
                          ],
                          rules[[1]]
                          ]

                          groupMe[B]

                          a, b, c, d, e, f, g, h, i, j, k, l, m, n





                          share|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            Here's an inefficient but reasonably simple way:



                            groupMe[rules_] :=
                            If[Head[rules[[1]]] === Rule,
                            Values@GroupBy[
                            rules,
                            (#[[1, 1]] &) ->
                            (If[Length[#[[1]]] === 1, #[[2]], #[[1, 2 ;;]] -> #[[2]]] &),
                            groupMe
                            ],
                            rules[[1]]
                            ]

                            groupMe[B]

                            a, b, c, d, e, f, g, h, i, j, k, l, m, n





                            share|improve this answer









                            $endgroup$



                            Here's an inefficient but reasonably simple way:



                            groupMe[rules_] :=
                            If[Head[rules[[1]]] === Rule,
                            Values@GroupBy[
                            rules,
                            (#[[1, 1]] &) ->
                            (If[Length[#[[1]]] === 1, #[[2]], #[[1, 2 ;;]] -> #[[2]]] &),
                            groupMe
                            ],
                            rules[[1]]
                            ]

                            groupMe[B]

                            a, b, c, d, e, f, g, h, i, j, k, l, m, n






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 3 hours ago









                            b3m2a1b3m2a1

                            28.3k358163




                            28.3k358163





















                                1












                                $begingroup$

                                Here's a convoluted way using pattern replacements:



                                DeleteCases[
                                With[m = Max[Length /@ Keys[B]],
                                Array[
                                List,
                                Max /@ Transpose[PadRight[#, m] & /@ Keys[B]]
                                ] /.
                                Map[
                                Fold[
                                Insert[
                                #, ___,
                                _,
                                Append[ConstantArray[1, #2], -1]] &,
                                #[[1]],
                                Range[m - Length[#[[1]]]]
                                ] -> #[[2]] &,
                                B
                                ]
                                ],
                                __Integer,
                                Infinity
                                ]

                                a, b, c, d, e, f, g, h, i, j, k, l, m, n





                                share|improve this answer









                                $endgroup$

















                                  1












                                  $begingroup$

                                  Here's a convoluted way using pattern replacements:



                                  DeleteCases[
                                  With[m = Max[Length /@ Keys[B]],
                                  Array[
                                  List,
                                  Max /@ Transpose[PadRight[#, m] & /@ Keys[B]]
                                  ] /.
                                  Map[
                                  Fold[
                                  Insert[
                                  #, ___,
                                  _,
                                  Append[ConstantArray[1, #2], -1]] &,
                                  #[[1]],
                                  Range[m - Length[#[[1]]]]
                                  ] -> #[[2]] &,
                                  B
                                  ]
                                  ],
                                  __Integer,
                                  Infinity
                                  ]

                                  a, b, c, d, e, f, g, h, i, j, k, l, m, n





                                  share|improve this answer









                                  $endgroup$















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Here's a convoluted way using pattern replacements:



                                    DeleteCases[
                                    With[m = Max[Length /@ Keys[B]],
                                    Array[
                                    List,
                                    Max /@ Transpose[PadRight[#, m] & /@ Keys[B]]
                                    ] /.
                                    Map[
                                    Fold[
                                    Insert[
                                    #, ___,
                                    _,
                                    Append[ConstantArray[1, #2], -1]] &,
                                    #[[1]],
                                    Range[m - Length[#[[1]]]]
                                    ] -> #[[2]] &,
                                    B
                                    ]
                                    ],
                                    __Integer,
                                    Infinity
                                    ]

                                    a, b, c, d, e, f, g, h, i, j, k, l, m, n





                                    share|improve this answer









                                    $endgroup$



                                    Here's a convoluted way using pattern replacements:



                                    DeleteCases[
                                    With[m = Max[Length /@ Keys[B]],
                                    Array[
                                    List,
                                    Max /@ Transpose[PadRight[#, m] & /@ Keys[B]]
                                    ] /.
                                    Map[
                                    Fold[
                                    Insert[
                                    #, ___,
                                    _,
                                    Append[ConstantArray[1, #2], -1]] &,
                                    #[[1]],
                                    Range[m - Length[#[[1]]]]
                                    ] -> #[[2]] &,
                                    B
                                    ]
                                    ],
                                    __Integer,
                                    Infinity
                                    ]

                                    a, b, c, d, e, f, g, h, i, j, k, l, m, n






                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered 2 hours ago









                                    b3m2a1b3m2a1

                                    28.3k358163




                                    28.3k358163



























                                        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%2f194217%2fhow-to-invert-mapindexed-on-a-ragged-structure-how-to-construct-a-tree-from-rul%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

                                        Can not update quote_id field of “quote_item” table magento 2Magento 2.1 - We can't remove the item. (Shopping Cart doesnt allow us to remove items before becomes empty)Add value for custom quote item attribute using REST apiREST API endpoint v1/carts/cartId/items always returns error messageCorrect way to save entries to databaseHow to remove all associated quote objects of a customer completelyMagento 2 - Save value from custom input field to quote_itemGet quote_item data using quote id and product id filter in Magento 2How to set additional data to quote_item table from controller in Magento 2?What is the purpose of additional_data column in quote_item table in magento2Set Custom Price to Quote item magento2 from controller

                                        How to solve knockout JS error in Magento 2 Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?(Magento2) knockout.js:3012 Uncaught ReferenceError: Unable to process bindingUnable to process binding Knockout.js magento 2Cannot read property `scopeLabel` of undefined on Product Detail PageCan't get Customer Data on frontend in Magento 2Magento2 Order Summary - unable to process bindingKO templates are not loading in Magento 2.1 applicationgetting knockout js error magento 2Product grid not load -— Unable to process binding Knockout.js magento 2Product form not loaded in magento2Uncaught ReferenceError: Unable to process binding “if: function()return (isShowLegend()) ” magento 2

                                        Nissan Patrol Зміст Перше покоління — 4W60 (1951-1960) | Друге покоління — 60 series (1960-1980) | Третє покоління (1980–2002) | Четверте покоління — Y60 (1987–1998) | П'яте покоління — Y61 (1997–2013) | Шосте покоління — Y62 (2010- ) | Посилання | Зноски | Навігаційне менюОфіційний український сайтТест-драйв Nissan Patrol 2010 7-го поколінняNissan PatrolКак мы тестировали Nissan Patrol 2016рвиправивши або дописавши її