How do I transpose the first and deepest levels of an arbitrarily nested array? The Next CEO...
Unreliable Magic - Is it worth it?
Does it take more energy to get to Venus or to Mars?
Would a completely good Muggle be able to use a wand?
Do I need to enable Dev Hub in my PROD Org?
Anatomically Correct Strange Women In Ponds Distributing Swords
Why has the US not been more assertive in confronting Russia in recent years?
Several mode to write the symbol of a vector
How to safely derail a train during transit?
Is "for causing autism in X" grammatical?
How does the Z80 determine which peripheral sent an interrupt?
Won the lottery - how do I keep the money?
How to count occurrences of text in a file?
How do I go from 300 unfinished/half written blog posts, to published posts?
Complex fractions
Why do we use the plural of movies in this phrase "We went to the movies last night."?
What was the first Unix version to run on a microcomputer?
What is "(CFMCC)" on an ILS approach chart?
Why do airplanes bank sharply to the right after air-to-air refueling?
What does convergence in distribution "in the Gromov–Hausdorff" sense mean?
Are there any unintended negative consequences to allowing PCs to gain multiple levels at once in a short milestone-XP game?
Why didn't Khan get resurrected in the Genesis Explosion?
Rotate a column
What flight has the highest ratio of time difference to flight time?
Bold, vivid family
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 23 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 23 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
9 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
9 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
9 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
9 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
9 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 23 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 23 mins ago
J. M. is slightly pensive♦
98.7k10311467
asked 10 hours ago
Kuba♦Kuba
107k12210531
edited 23 mins ago
J. M. is slightly pensive♦
98.7k10311467
asked 10 hours ago
Kuba♦Kuba
107k12210531
edited 23 mins ago
J. M. is slightly pensive♦
98.7k10311467
edited 23 mins ago
J. M. is slightly pensive♦
98.7k10311467
edited 23 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
9 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
9 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
9 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
9 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
9 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
9 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
9 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
9 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
9 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
9 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
9 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
9 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
9 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
9 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
9 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
9 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
9 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
9 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
9 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
9 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 9 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 9 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
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-first-and-deepest-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 9 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 9 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 9 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 9 hours ago
andre314andre314
12.3k12352
answered 9 hours ago
andre314andre314
12.3k12352
answered 9 hours ago
andre314andre314
12.3k12352
answered 9 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 9 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 9 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 9 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 9 hours ago
C. E.C. E.
50.9k399205
answered 9 hours ago
C. E.C. E.
50.9k399205
answered 9 hours ago
C. E.C. E.
50.9k399205
answered 9 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 |
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-first-and-deepest-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
9 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
9 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
9 hours ago
$begingroup$
@Roman Level[arr, {-2}]` should be a rectangular array but rows do not need to be the same.
$endgroup$
– Kuba♦
9 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
9 hours ago