How to invert MapIndexed on a ragged structure? How to construct a tree from rules? The Next...
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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
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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 |
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