How do I transpose the 1st and -1th levels of an arbitrarily nested array? The Next CEO of...
Won the lottery - how do I keep the money?
Complex fractions
How do I make a variable always equal to the result of some calculations?
Can we say or write : "No, it'sn't"?
Unreliable Magic - Is it worth it?
How did the Bene Gesserit know how to make a Kwisatz Haderach?
Would a completely good Muggle be able to use a wand?
Are there any unintended negative consequences to allowing PCs to gain multiple levels at once in a short milestone-XP game?
Why am I allowed to create multiple unique pointers from a single object?
If the heap is initialized for security, then why is the stack uninitialized?
Why do remote companies require working in the US?
Why does the UK parliament need a vote on the political declaration?
Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?
sp_blitzCache results Memory grants
What happened in Rome, when the western empire "fell"?
Bold, vivid family
Interfacing a button to MCU (and PC) with 50m long cable
How to solve a differential equation with a term to a power?
What connection does MS Office have to Netscape Navigator?
What exact does MIB represent in SNMP? How is it different from OID?
Is there a difference between "Fahrstuhl" and "Aufzug"
Why don't programming languages automatically manage the synchronous/asynchronous problem?
Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?
How do scammers retract money, while you can’t?
How do I transpose the 1st and -1th levels of an arbitrarily nested array?
The Next CEO of Stack OverflowEmulating R data frame getters with UpValuesQuickly pruning elements in one structured array that exist in a separate unordered arrayJoining sub-lists nested two levels deep in a list`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`Reformatting the pattern inside a list
$begingroup$
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 32 mins ago
Peter Mortensen
33627
asked 6 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 32 mins ago
Peter Mortensen
33627
asked 6 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 32 mins ago
Peter Mortensen
33627
asked 6 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 32 mins ago
Peter Mortensen
33627
asked 6 hours ago
Kuba♦Kuba
107k12210531
edited 32 mins ago
Peter Mortensen
33627
asked 6 hours ago
Kuba♦Kuba
107k12210531
edited 32 mins ago
Peter Mortensen
33627
edited 32 mins ago
Peter Mortensen
33627
edited 32 mins ago
Peter Mortensen
33627
33627
asked 6 hours ago
Kuba♦Kuba
107k12210531
asked 6 hours ago
Kuba♦Kuba
107k12210531
asked 6 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 5 hours ago
RomanRoman
3,9761022
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194206%2fhow-do-i-transpose-the-1st-and-1th-levels-of-an-arbitrarily-nested-array%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 5 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 5 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 5 hours ago
RomanRoman
3,9761022
$endgroup$
Terrible solution using Table but works:
Table[Map[#[[i]] &, arr, {-2}], {i, Last[Dimensions[Level[arr, {-2}]]]}]
answered 5 hours ago
RomanRoman
3,9761022
answered 5 hours ago
RomanRoman
3,9761022
answered 5 hours ago
RomanRoman
3,9761022
answered 5 hours ago
RomanRoman
3,9761022
3,9761022
add a comment |
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194206%2fhow-do-i-transpose-the-1st-and-1th-levels-of-an-arbitrarily-nested-array%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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