How do I transpose the first and deepest levels of an arbitrarily nested array?How to invert MapIndexed on a ragged structure? How to construct a tree from rules?A question about transforming one List into two Lists with additional requirementsEmulating R data frame getters with UpValuesQuickly pruning elements in one structured array that exist in a separate unordered array`Part` like `Delete`: How to delete list of columns or arbitrarily deeper levelsHow to mesh a region using adaptive cubic elementsHow to efficiently Flatten nested lists while preserving select levels?Distribute elements of one line across arbitrary dimension of another listDeep level nested list addition`Transpose` nested `Association`How to extract the first element in nested lists
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How do I transpose the first and deepest levels of an arbitrarily nested array?
How to invert MapIndexed on a ragged structure? How to construct a tree from rules?A question about transforming one List into two Lists with additional requirementsEmulating R data frame getters with UpValuesQuickly pruning elements in one structured array that exist in a separate unordered array`Part` like `Delete`: How to delete list of columns or arbitrarily deeper levelsHow to mesh a region using adaptive cubic elementsHow to efficiently Flatten nested lists while preserving select levels?Distribute elements of one line across arbitrary dimension of another listDeep level nested list addition`Transpose` nested `Association`How to extract the first element in nested lists
$begingroup$
Is there a straightforward way to convert
arr =
a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
;
to:
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
?
I need to swap the first and last dimension. Which should in principle be possible, because, although arr does not have a fixed structure, the 'bottom' is always uniform:
Level[arr, -2]
a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
Had Transpose/Flatten/MapThread accepted a negative level specification, it would have been easy. That is not the case.
One can think about that question as: How do I create arr2 so that arr[[whatever__, y_]] == arr2[[y, whatever__]]?
EDIT:
In general Level[arr, -2] should be a rectangular array, but rows do not need to be the same.
So this:
a1, b1, a2, b2, a3, b3, a4, b4, a5, b5, a6, b6, a7,b7 ;
should end up:
a1, a2, a3, a4, a5, a6, a7 , ...;
list-manipulation
edited Mar 30 at 5:04
J. M. is away♦
98.9k10311467
asked Mar 29 at 19:09
Kuba♦Kuba
107k12211534
$endgroup$
add a comment |
$begingroup$
Is there a straightforward way to convert
arr =
a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
;
to:
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
?
I need to swap the first and last dimension. Which should in principle be possible, because, although arr does not have a fixed structure, the 'bottom' is always uniform:
Level[arr, -2]
a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
Had Transpose/Flatten/MapThread accepted a negative level specification, it would have been easy. That is not the case.
One can think about that question as: How do I create arr2 so that arr[[whatever__, y_]] == arr2[[y, whatever__]]?
EDIT:
In general Level[arr, -2] should be a rectangular array, but rows do not need to be the same.
So this:
a1, b1, a2, b2, a3, b3, a4, b4, a5, b5, a6, b6, a7,b7 ;
should end up:
a1, a2, a3, a4, a5, a6, a7 , ...;
list-manipulation
edited Mar 30 at 5:04
J. M. is away♦
98.9k10311467
asked Mar 29 at 19:09
Kuba♦Kuba
107k12211534
$endgroup$
$begingroup$
Not a solution, butFlatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]]gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though:SparseArraydoes not construct ragged structures.
$endgroup$
– Roman
Mar 29 at 19:41
$begingroup$
Maybe something along the lines ofarr /. a,b->a,a,b->b? Or perhaps more generally,arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?
$endgroup$
– Carl Woll
Mar 29 at 19:43
$begingroup$
@Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
$endgroup$
– Roman
Mar 29 at 19:52
add a comment |
$begingroup$
Is there a straightforward way to convert
arr =
a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
;
to:
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
?
I need to swap the first and last dimension. Which should in principle be possible, because, although arr does not have a fixed structure, the 'bottom' is always uniform:
Level[arr, -2]
a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
Had Transpose/Flatten/MapThread accepted a negative level specification, it would have been easy. That is not the case.
One can think about that question as: How do I create arr2 so that arr[[whatever__, y_]] == arr2[[y, whatever__]]?
EDIT:
In general Level[arr, -2] should be a rectangular array, but rows do not need to be the same.
So this:
a1, b1, a2, b2, a3, b3, a4, b4, a5, b5, a6, b6, a7,b7 ;
should end up:
a1, a2, a3, a4, a5, a6, a7 , ...;
list-manipulation
edited Mar 30 at 5:04
J. M. is away♦
98.9k10311467
asked Mar 29 at 19:09
Kuba♦Kuba
107k12211534
$endgroup$
Is there a straightforward way to convert
arr =
a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
;
to:
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
?
I need to swap the first and last dimension. Which should in principle be possible, because, although arr does not have a fixed structure, the 'bottom' is always uniform:
Level[arr, -2]
a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b
Had Transpose/Flatten/MapThread accepted a negative level specification, it would have been easy. That is not the case.
One can think about that question as: How do I create arr2 so that arr[[whatever__, y_]] == arr2[[y, whatever__]]?
EDIT:
In general Level[arr, -2] should be a rectangular array, but rows do not need to be the same.
So this:
a1, b1, a2, b2, a3, b3, a4, b4, a5, b5, a6, b6, a7,b7 ;
should end up:
a1, a2, a3, a4, a5, a6, a7 , ...;
list-manipulation
list-manipulation
edited Mar 30 at 5:04
J. M. is away♦
98.9k10311467
asked Mar 29 at 19:09
Kuba♦Kuba
107k12211534
edited Mar 30 at 5:04
J. M. is away♦
98.9k10311467
asked Mar 29 at 19:09
Kuba♦Kuba
107k12211534
edited Mar 30 at 5:04
J. M. is away♦
98.9k10311467
edited Mar 30 at 5:04
J. M. is away♦
98.9k10311467
edited Mar 30 at 5:04
J. M. is away♦
98.9k10311467
98.9k10311467
asked Mar 29 at 19:09
Kuba♦Kuba
107k12211534
asked Mar 29 at 19:09
Kuba♦Kuba
107k12211534
asked Mar 29 at 19:09
Kuba♦Kuba
107k12211534
107k12211534
$begingroup$
Not a solution, butFlatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]]gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though:SparseArraydoes not construct ragged structures.
$endgroup$
– Roman
Mar 29 at 19:41
$begingroup$
Maybe something along the lines ofarr /. a,b->a,a,b->b? Or perhaps more generally,arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?
$endgroup$
– Carl Woll
Mar 29 at 19:43
$begingroup$
@Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
$endgroup$
– Roman
Mar 29 at 19:52
add a comment |
$begingroup$
Not a solution, butFlatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]]gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though:SparseArraydoes not construct ragged structures.
$endgroup$
– Roman
Mar 29 at 19:41
$begingroup$
Maybe something along the lines ofarr /. a,b->a,a,b->b? Or perhaps more generally,arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?
$endgroup$
– Carl Woll
Mar 29 at 19:43
$begingroup$
@Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
$endgroup$
– Roman
Mar 29 at 19:52
$begingroup$
Not a solution, but
Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]] gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though: SparseArray does not construct ragged structures.$endgroup$
– Roman
Mar 29 at 19:41
$begingroup$
Not a solution, but
Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]] gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though: SparseArray does not construct ragged structures.$endgroup$
– Roman
Mar 29 at 19:41
$begingroup$
Maybe something along the lines of
arr /. a,b->a,a,b->b? Or perhaps more generally, arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?$endgroup$
– Carl Woll
Mar 29 at 19:43
$begingroup$
Maybe something along the lines of
arr /. a,b->a,a,b->b? Or perhaps more generally, arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?$endgroup$
– Carl Woll
Mar 29 at 19:43
$begingroup$
@Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
$endgroup$
– Roman
Mar 29 at 19:52
$begingroup$
@Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
$endgroup$
– Roman
Mar 29 at 19:52
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;
SetAttributes[f1, Listable]
Apply[f1, arr, 0, -3] /. f1 -> List
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
answered Mar 29 at 19:53
andre314andre314
12.4k12353
$endgroup$
add a comment |
$begingroup$
This is what the list at the lowest level looks like:
el = First@Level[list, -2];
Using this, we can solve it with a rules-based approach:
list /. el -> # & /@ el
or a recursive approach like this:
walk[lists : __List, i_] := walk[#, i] & /@ lists
walk[atoms : __, i_] := i
walk[list, #] & /@ el
answered Mar 29 at 19:42
C. E.C. E.
51k3101206
$endgroup$
add a comment |
$begingroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, -2], i, Last[Dimensions[Level[arr, -2]]]]
answered Mar 29 at 20:16
RomanRoman
4,61511128
$endgroup$
add a comment |
$begingroup$
This is a bit roundabout but very versatile. First convert the structure arr into a list of rules, then reassemble these rules (after proper modification using RotateRight to transpose first and deepest levels) into a new structure.
Using the toTree function from this answer (thanks to @b3m2a1 !):
arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;
toTree[Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]]]
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
answered Mar 30 at 21:29
RomanRoman
4,61511128
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;
SetAttributes[f1, Listable]
Apply[f1, arr, 0, -3] /. f1 -> List
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
answered Mar 29 at 19:53
andre314andre314
12.4k12353
$endgroup$
add a comment |
$begingroup$
arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;
SetAttributes[f1, Listable]
Apply[f1, arr, 0, -3] /. f1 -> List
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
answered Mar 29 at 19:53
andre314andre314
12.4k12353
$endgroup$
add a comment |
$begingroup$
arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;
SetAttributes[f1, Listable]
Apply[f1, arr, 0, -3] /. f1 -> List
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
answered Mar 29 at 19:53
andre314andre314
12.4k12353
$endgroup$
arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;
SetAttributes[f1, Listable]
Apply[f1, arr, 0, -3] /. f1 -> List
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
answered Mar 29 at 19:53
andre314andre314
12.4k12353
answered Mar 29 at 19:53
andre314andre314
12.4k12353
answered Mar 29 at 19:53
andre314andre314
12.4k12353
answered Mar 29 at 19:53
andre314andre314
12.4k12353
12.4k12353
add a comment |
add a comment |
$begingroup$
This is what the list at the lowest level looks like:
el = First@Level[list, -2];
Using this, we can solve it with a rules-based approach:
list /. el -> # & /@ el
or a recursive approach like this:
walk[lists : __List, i_] := walk[#, i] & /@ lists
walk[atoms : __, i_] := i
walk[list, #] & /@ el
answered Mar 29 at 19:42
C. E.C. E.
51k3101206
$endgroup$
add a comment |
$begingroup$
This is what the list at the lowest level looks like:
el = First@Level[list, -2];
Using this, we can solve it with a rules-based approach:
list /. el -> # & /@ el
or a recursive approach like this:
walk[lists : __List, i_] := walk[#, i] & /@ lists
walk[atoms : __, i_] := i
walk[list, #] & /@ el
answered Mar 29 at 19:42
C. E.C. E.
51k3101206
$endgroup$
add a comment |
$begingroup$
This is what the list at the lowest level looks like:
el = First@Level[list, -2];
Using this, we can solve it with a rules-based approach:
list /. el -> # & /@ el
or a recursive approach like this:
walk[lists : __List, i_] := walk[#, i] & /@ lists
walk[atoms : __, i_] := i
walk[list, #] & /@ el
answered Mar 29 at 19:42
C. E.C. E.
51k3101206
$endgroup$
This is what the list at the lowest level looks like:
el = First@Level[list, -2];
Using this, we can solve it with a rules-based approach:
list /. el -> # & /@ el
or a recursive approach like this:
walk[lists : __List, i_] := walk[#, i] & /@ lists
walk[atoms : __, i_] := i
walk[list, #] & /@ el
answered Mar 29 at 19:42
C. E.C. E.
51k3101206
answered Mar 29 at 19:42
C. E.C. E.
51k3101206
answered Mar 29 at 19:42
C. E.C. E.
51k3101206
answered Mar 29 at 19:42
C. E.C. E.
51k3101206
51k3101206
add a comment |
add a comment |
$begingroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, -2], i, Last[Dimensions[Level[arr, -2]]]]
answered Mar 29 at 20:16
RomanRoman
4,61511128
$endgroup$
add a comment |
$begingroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, -2], i, Last[Dimensions[Level[arr, -2]]]]
answered Mar 29 at 20:16
RomanRoman
4,61511128
$endgroup$
add a comment |
$begingroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, -2], i, Last[Dimensions[Level[arr, -2]]]]
answered Mar 29 at 20:16
RomanRoman
4,61511128
$endgroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, -2], i, Last[Dimensions[Level[arr, -2]]]]
answered Mar 29 at 20:16
RomanRoman
4,61511128
answered Mar 29 at 20:16
RomanRoman
4,61511128
answered Mar 29 at 20:16
RomanRoman
4,61511128
answered Mar 29 at 20:16
RomanRoman
4,61511128
4,61511128
add a comment |
add a comment |
$begingroup$
This is a bit roundabout but very versatile. First convert the structure arr into a list of rules, then reassemble these rules (after proper modification using RotateRight to transpose first and deepest levels) into a new structure.
Using the toTree function from this answer (thanks to @b3m2a1 !):
arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;
toTree[Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]]]
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
answered Mar 30 at 21:29
RomanRoman
4,61511128
$endgroup$
add a comment |
$begingroup$
This is a bit roundabout but very versatile. First convert the structure arr into a list of rules, then reassemble these rules (after proper modification using RotateRight to transpose first and deepest levels) into a new structure.
Using the toTree function from this answer (thanks to @b3m2a1 !):
arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;
toTree[Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]]]
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
answered Mar 30 at 21:29
RomanRoman
4,61511128
$endgroup$
add a comment |
$begingroup$
This is a bit roundabout but very versatile. First convert the structure arr into a list of rules, then reassemble these rules (after proper modification using RotateRight to transpose first and deepest levels) into a new structure.
Using the toTree function from this answer (thanks to @b3m2a1 !):
arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;
toTree[Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]]]
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
answered Mar 30 at 21:29
RomanRoman
4,61511128
$endgroup$
This is a bit roundabout but very versatile. First convert the structure arr into a list of rules, then reassemble these rules (after proper modification using RotateRight to transpose first and deepest levels) into a new structure.
Using the toTree function from this answer (thanks to @b3m2a1 !):
arr = a, b, a, b, a, b, a, b, a, b, a, b, a, b, a, b;
toTree[Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]]]
a, a, a, a, a, a, a, a, b, b, b, b, b, b, b, b
answered Mar 30 at 21:29
RomanRoman
4,61511128
answered Mar 30 at 21:29
RomanRoman
4,61511128
answered Mar 30 at 21:29
RomanRoman
4,61511128
answered Mar 30 at 21:29
RomanRoman
4,61511128
4,61511128
add a comment |
add a comment |
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$begingroup$
Not a solution, but
Flatten[MapIndexed[RotateRight[#2] -> #1 &, arr, -1]]gives you a list of rules of what needs to be constructed. I don't know of a way to construct it though:SparseArraydoes not construct ragged structures.$endgroup$
– Roman
Mar 29 at 19:41
$begingroup$
Maybe something along the lines of
arr /. a,b->a,a,b->b? Or perhaps more generally,arr /. a_?VectorQ :> First@a, a_?VectorQ :> Last@a?$endgroup$
– Carl Woll
Mar 29 at 19:43
$begingroup$
@Kuba maybe you can come up with a recursion that constructs the result from the list of rules I gave 4 lines up? That would be a handy tool to have in any case.
$endgroup$
– Roman
Mar 29 at 19:52