I have a one-dimensional cell array where each element of the array contains a timetable with identical format. Is there a direct way of creating a vector that contains the last element of one specific variable in each timetable (without looping)?
For example, something like:
CellArray{:}.Variable(end)

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Dave O
Dave O 2022년 6월 30일
편집: Dave O 2022년 6월 30일
Thanks for your reply, Rik and Adam!
Rik
Rik 2022년 6월 30일
You're welcome. If either answer solved your problem, feel free to mark it as accepted answer. If the other was helpful as well, consider giving it an upvote.

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Rik
Rik 2022년 6월 30일
편집: Rik 2022년 6월 30일

1 개 추천

No. You could use cellfun, but that will only hide the loop. Using a loop directly tends to have better efficiency.
But don't worry: loops are not as bad as you might think (especially if you pre-allocate the output). They are only bad when there isn't a builtin equivalent that operates on the entire array.

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Adam Danz
Adam Danz 2022년 6월 30일
+1 Rik, @Dave O if a loop is more readable to you than cellfun, go with the loop.

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Adam Danz
Adam Danz 2022년 6월 30일

2 개 추천

In this example, the cell array is named myCellArray and the variable is named myVar.
y = cellfun(@(A)A.myVar(end), myCellArray)
If the values you are extracting are non-numeric or non-scalar,
y = cellfun(@(A){A.myVar(end)}, myCellArray)

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2022년 6월 30일

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