Group consecutive and non-consecutive values in a vector

5 vues (au cours des 30 derniers jours)
Daniel van Huyssteen
Daniel van Huyssteen le 7 Mar 2022
Commenté : Raymond MacNeil le 3 Fév 2024
Consider a vector V2 containing certain entries of some other larger vector V1 comprising nV1 entries.
How would one split up the vector V2 into segments comprising consecutive entries and output these segments as the cells of a cell array C?
Additionally, it could be the case that V2 'wraps', that is, if the first segment of C contains the value 1, and the last segment of C contains the value nV1, these two segments should be merged.
I would like to avoid dynamic re-sizing of vectors/arrays if possible.
Possible test case #1:
Input:
nV1 = 8;
V2 = [1];
Output:
C = {[1]};
Possible test case #2:
Input:
nV1 = 8;
V2 = [3,4,5];
Output:
C = {[3,4,5]};
Possible test case #3:
Input:
nV1 = 8;
V2 = [1,2,4,6,7];
Output:
C = {{[1,2]},{[4]},{[6,7]}};
Possible test case #4:
Input:
nV1 = 8;
V2 = [1,2,4,7,8];
Output:
C = {{[1,2,7,8]},{[4]}};
or
C = {{[4]},{[1,2,7,8]}};
  1 commentaire
Jan
Jan le 7 Mar 2022
I assume you do not need a nested cell
C = {{[4]},{[1,2,7,8]}}
but a cell is sufficient:
C = {[4], [1,2,7,8]}

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Jan
Jan le 7 Mar 2022
Modifié(e) : Jan le 17 Nov 2022
nV1 = 8;
V2 = [1,2,4,6,7];
grp = cumsum([true, diff(V2)~=1]);
C = splitapply(@(x) {x}, V2, grp);
if V2(end) == nV1
C{1} = [C{1}, C{end}];
C(end) = [];
end
C
C = 1×3 cell array
{[1 2]} {[4]} {[6 7]}
  2 commentaires
Hyeongseok
Hyeongseok le 17 Nov 2022
Nice solution! Thanks!
Raymond MacNeil
Raymond MacNeil le 3 Fév 2024
Brilliant, thanks.

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