How to invert MapIndexed on a ragged structure? How to construct a tree from rules?
$begingroup$
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 8 hours ago
RomanRoman
4,0111022
$endgroup$
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 8 hours ago
RomanRoman
4,0111022
$endgroup$
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 8 hours ago
RomanRoman
4,0111022
$endgroup$
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 8 hours ago
RomanRoman
4,0111022
asked 8 hours ago
RomanRoman
4,0111022
asked 8 hours ago
RomanRoman
4,0111022
asked 8 hours ago
RomanRoman
4,0111022
asked 8 hours ago
RomanRoman
4,0111022
4,0111022
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
answered 8 hours ago
b3m2a1b3m2a1
28.4k358163
answered 8 hours ago
b3m2a1b3m2a1
28.4k358163
answered 8 hours ago
b3m2a1b3m2a1
28.4k358163
28.4k358163
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
answered 8 hours ago
b3m2a1b3m2a1
28.4k358163
answered 8 hours ago
b3m2a1b3m2a1
28.4k358163
answered 8 hours ago
b3m2a1b3m2a1
28.4k358163
28.4k358163
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
$endgroup$
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 8 hours ago
b3m2a1b3m2a1
28.4k358163
answered 8 hours ago
b3m2a1b3m2a1
28.4k358163
answered 8 hours ago
b3m2a1b3m2a1
28.4k358163
answered 8 hours ago
b3m2a1b3m2a1
28.4k358163
28.4k358163
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
