How to invert MapIndexed on a ragged structure? How to construct a tree from rules? The Next...
What's the best way to handle refactoring a big file?
Preparing Indesign booklet with .psd graphics for print
Is it possible to search for a directory/file combination?
Non-deterministic sum of floats
How does the mv command work with external drives?
What does "Its cash flow is deeply negative" mean?
Elegant way to replace substring in a regex with optional groups in Python?
Why am I allowed to create multiple unique pointers from a single object?
Do I need to enable Dev Hub in my PROD Org?
Which tube will fit a -(700 x 25c) wheel?
Why don't programming languages automatically manage the synchronous/asynchronous problem?
How did the Bene Gesserit know how to make a Kwisatz Haderach?
Interfacing a button to MCU (and PC) with 50m long cable
Why do remote companies require working in the US?
If a black hole is created from light, can this black hole then move at speed of light?
Sending manuscript to multiple publishers
How do scammers retract money, while you can’t?
If the heap is initialized for security, then why is the stack uninitialized?
If/When UK leaves the EU, can a future goverment conduct a referendum to join the EU?
What is the result of assigning to std::vector<T>::begin()?
Anatomically Correct Strange Women In Ponds Distributing Swords
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?
Contours of a clandestine nature
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
$begingroup$
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?
list-manipulation data-structures trees
asked 4 hours ago
RomanRoman
3,9761022
$endgroup$
add a comment |
$begingroup$
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?
list-manipulation data-structures trees
asked 4 hours ago
RomanRoman
3,9761022
$endgroup$
add a comment |
$begingroup$
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?
list-manipulation data-structures trees
asked 4 hours ago
RomanRoman
3,9761022
$endgroup$
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
list-manipulation data-structures trees
asked 4 hours ago
RomanRoman
3,9761022
asked 4 hours ago
RomanRoman
3,9761022
asked 4 hours ago
RomanRoman
3,9761022
asked 4 hours ago
RomanRoman
3,9761022
asked 4 hours ago
RomanRoman
3,9761022
3,9761022
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
add a comment |
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
add a comment |
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
add a comment |
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
add a comment |
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
28.3k358163
add a comment |
add a comment |
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
add a comment |
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
add a comment |
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
28.3k358163
add a comment |
add a comment |
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
add a comment |
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
add a comment |
$begingroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
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}
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
answered 4 hours ago
b3m2a1b3m2a1
28.3k358163
28.3k358163
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
