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
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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 3 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 3 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 3 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 3 hours ago
RomanRoman
3,9761022
asked 3 hours ago
RomanRoman
3,9761022
asked 3 hours ago
RomanRoman
3,9761022
asked 3 hours ago
RomanRoman
3,9761022
asked 3 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 3 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 3 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 3 hours ago
b3m2a1b3m2a1
28.3k358163
$endgroup$
add a comment |
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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 3 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 3 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 3 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 3 hours ago
b3m2a1b3m2a1
28.3k358163
answered 3 hours ago
b3m2a1b3m2a1
28.3k358163
answered 3 hours ago
b3m2a1b3m2a1
28.3k358163
answered 3 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 3 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 3 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 3 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 3 hours ago
b3m2a1b3m2a1
28.3k358163
answered 3 hours ago
b3m2a1b3m2a1
28.3k358163
answered 3 hours ago
b3m2a1b3m2a1
28.3k358163
answered 3 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 3 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 3 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 3 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 3 hours ago
b3m2a1b3m2a1
28.3k358163
answered 3 hours ago
b3m2a1b3m2a1
28.3k358163
answered 3 hours ago
b3m2a1b3m2a1
28.3k358163
answered 3 hours ago
b3m2a1b3m2a1
28.3k358163
28.3k358163
add a comment |
add a comment |
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