How do I transpose the first and deepest levels of an arbitrarily nested array? The Next CEO...
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How do I transpose the first and deepest levels of an arbitrarily nested array?
The Next CEO of Stack OverflowA 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 50 mins ago
J. M. is slightly pensive♦
98.7k10311467
asked 10 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 50 mins ago
J. M. is slightly pensive♦
98.7k10311467
asked 10 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
10 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
10 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
10 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
10 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
10 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 50 mins ago
J. M. is slightly pensive♦
98.7k10311467
asked 10 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 50 mins ago
J. M. is slightly pensive♦
98.7k10311467
asked 10 hours ago
Kuba♦Kuba
107k12210531
edited 50 mins ago
J. M. is slightly pensive♦
98.7k10311467
asked 10 hours ago
Kuba♦Kuba
107k12210531
edited 50 mins ago
J. M. is slightly pensive♦
98.7k10311467
edited 50 mins ago
J. M. is slightly pensive♦
98.7k10311467
edited 50 mins ago
J. M. is slightly pensive♦
98.7k10311467
98.7k10311467
asked 10 hours ago
Kuba♦Kuba
107k12210531
asked 10 hours ago
Kuba♦Kuba
107k12210531
asked 10 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
10 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
10 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
10 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
10 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
10 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
10 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
10 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
10 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
10 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
10 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
10 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
10 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
10 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
10 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
10 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
10 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
10 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
10 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
10 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
10 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 10 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 10 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 9 hours ago
RomanRoman
4,0111022
$endgroup$
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
$begingroup$
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 10 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 10 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 10 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 10 hours ago
andre314andre314
12.3k12352
answered 10 hours ago
andre314andre314
12.3k12352
answered 10 hours ago
andre314andre314
12.3k12352
answered 10 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 10 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 10 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 10 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 10 hours ago
C. E.C. E.
50.9k399205
answered 10 hours ago
C. E.C. E.
50.9k399205
answered 10 hours ago
C. E.C. E.
50.9k399205
answered 10 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 9 hours ago
RomanRoman
4,0111022
$endgroup$
add a comment |
$begingroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, {-2}], {i, Last[Dimensions[Level[arr, {-2}]]]}]
answered 9 hours ago
RomanRoman
4,0111022
$endgroup$
add a comment |
$begingroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, {-2}], {i, Last[Dimensions[Level[arr, {-2}]]]}]
answered 9 hours ago
RomanRoman
4,0111022
$endgroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, {-2}], {i, Last[Dimensions[Level[arr, {-2}]]]}]
answered 9 hours ago
RomanRoman
4,0111022
answered 9 hours ago
RomanRoman
4,0111022
answered 9 hours ago
RomanRoman
4,0111022
answered 9 hours ago
RomanRoman
4,0111022
4,0111022
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
10 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
10 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
10 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
10 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
10 hours ago