How to create a new column that depends on the stabilization of another

조회 수: 1 (최근 30일)
The title is pretty vague so here's my example:
I have a column on a table whose values are true and false (1 for when it's on and 0 for off)
Table.On = [0;0;0;1;1;1;1;1;0;0;1;1;1;1;0;1;1] (and so on)
However, I need a column that will provide stable values, since the first or second or third consecutive 1's, it's still "heating up", so it's only stable on the forth 1.
So I would like to create a new column like this (if it's 0 again it resets the same procedure):
Table.Stable = [0;0;0;0;0;0;1;1;0;0;0;0;0;1;0;0;0] ---- only consecutive 1's after 3 consecutive 1's.
This could be easily done in a for loop or while but I have 90000 rows and thats would take a long time. Is there a way to do this without a for/while loop?
Thank you!!

채택된 답변

Dyuman Joshi
Dyuman Joshi 2023년 3월 2일
Using find() will be slower than using logical indexing, specially given the number of rows you have
in = [0;0;0;1;1;1;1;1;0;0;1;1;1;1;0;1;1];
out = movsum(in,[3 0])==4
out = 17×1 logical array
0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0

추가 답변 (1개)

Cameron
Cameron 2023년 3월 2일
Table.On = [0;0;0;1;1;1;1;1;0;0;1;1;1;1;0;1;1];
indx = find(movsum(Table.On,4) == 4) + 1;
Table.Stable = zeros(length(Table.On),1);
Table.Stable(indx) = 1;
  댓글 수: 2
Margarida
Margarida 2023년 3월 2일
ahhhh! so cool! thank you so much :)
Cameron
Cameron 2023년 3월 2일
@Dyuman Joshi is correct. Using find() is less efficient so use this instead.
Table.On = [0;0;0;1;1;1;1;1;0;0;1;1;1;1;0;1;1];
Table.Stable = movsum(Table.On,[3 0]) == 4;

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