How do I transpose the 1st and -1th levels of an arbitrarily nested array? The Next CEO of...
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How do I transpose the 1st and -1th levels of an arbitrarily nested array?
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How do I transpose the 1st and -1th levels of an arbitrarily nested array?
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$begingroup$
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 14 mins ago
Peter Mortensen
33627
asked 5 hours ago
Kuba♦Kuba
107k12210531
$endgroup$
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 14 mins ago
Peter Mortensen
33627
asked 5 hours ago
Kuba♦Kuba
107k12210531
$endgroup$
$begingroup$
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
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
Does your list always contain{a,b}at the lowest level, or can there be anything there as long as they're all of same length?
$endgroup$
– Roman
5 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
5 hours ago
$begingroup$
@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
5 hours ago
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 14 mins ago
Peter Mortensen
33627
asked 5 hours ago
Kuba♦Kuba
107k12210531
$endgroup$
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 14 mins ago
Peter Mortensen
33627
asked 5 hours ago
Kuba♦Kuba
107k12210531
edited 14 mins ago
Peter Mortensen
33627
asked 5 hours ago
Kuba♦Kuba
107k12210531
edited 14 mins ago
Peter Mortensen
33627
edited 14 mins ago
Peter Mortensen
33627
edited 14 mins ago
Peter Mortensen
33627
33627
asked 5 hours ago
Kuba♦Kuba
107k12210531
asked 5 hours ago
Kuba♦Kuba
107k12210531
asked 5 hours ago
Kuba♦Kuba
107k12210531
107k12210531
$begingroup$
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
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
Does your list always contain{a,b}at the lowest level, or can there be anything there as long as they're all of same length?
$endgroup$
– Roman
5 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
5 hours ago
$begingroup$
@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
5 hours ago
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
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
Does your list always contain{a,b}at the lowest level, or can there be anything there as long as they're all of same length?
$endgroup$
– Roman
5 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
5 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
Does your list always contain
{a,b} at the lowest level, or can there be anything there as long as they're all of same length?$endgroup$
– Roman
5 hours ago
$begingroup$
Does your list always contain
{a,b} at the lowest level, or can there be anything there as long as they're all of same length?$endgroup$
– Roman
5 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
5 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
5 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
@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
5 hours ago
add a comment |
3 Answers
3
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 5 hours ago
andre314andre314
12.3k12352
$endgroup$
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 5 hours ago
C. E.C. E.
50.9k399205
$endgroup$
add a comment |
$begingroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, {-2}], {i, Last[Dimensions[Level[arr, {-2}]]]}]
answered 4 hours ago
RomanRoman
3,9761022
$endgroup$
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
$begingroup$
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 5 hours ago
andre314andre314
12.3k12352
$endgroup$
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 5 hours ago
andre314andre314
12.3k12352
$endgroup$
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 5 hours ago
andre314andre314
12.3k12352
$endgroup$
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 5 hours ago
andre314andre314
12.3k12352
answered 5 hours ago
andre314andre314
12.3k12352
answered 5 hours ago
andre314andre314
12.3k12352
answered 5 hours ago
andre314andre314
12.3k12352
12.3k12352
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 5 hours ago
C. E.C. E.
50.9k399205
$endgroup$
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 5 hours ago
C. E.C. E.
50.9k399205
$endgroup$
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 5 hours ago
C. E.C. E.
50.9k399205
$endgroup$
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 5 hours ago
C. E.C. E.
50.9k399205
answered 5 hours ago
C. E.C. E.
50.9k399205
answered 5 hours ago
C. E.C. E.
50.9k399205
answered 5 hours ago
C. E.C. E.
50.9k399205
50.9k399205
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 4 hours ago
RomanRoman
3,9761022
$endgroup$
add a comment |
$begingroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, {-2}], {i, Last[Dimensions[Level[arr, {-2}]]]}]
answered 4 hours ago
RomanRoman
3,9761022
$endgroup$
add a comment |
$begingroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, {-2}], {i, Last[Dimensions[Level[arr, {-2}]]]}]
answered 4 hours ago
RomanRoman
3,9761022
$endgroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, {-2}], {i, Last[Dimensions[Level[arr, {-2}]]]}]
answered 4 hours ago
RomanRoman
3,9761022
answered 4 hours ago
RomanRoman
3,9761022
answered 4 hours ago
RomanRoman
3,9761022
answered 4 hours ago
RomanRoman
3,9761022
3,9761022
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
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
Does your list always contain
{a,b}at the lowest level, or can there be anything there as long as they're all of same length?$endgroup$
– Roman
5 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
5 hours ago
$begingroup$
@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
5 hours ago