Group consecutive and non-consecutive values in a vector

조회 수: 2 (최근 30일)
Daniel van Huyssteen
Daniel van Huyssteen 2022년 3월 7일
댓글: Raymond MacNeil 2024년 2월 3일
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]}};
  댓글 수: 1
Jan
Jan 2022년 3월 7일
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 2022년 3월 7일
편집: Jan 2022년 11월 17일
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]}

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