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

1 次查看（过去 30 天）
Margarida
Margarida 2023-3-2
评论： Margarida 2023-3-2
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!!

请先登录，再回答此问题。

采纳的回答

Dyuman Joshi
Dyuman Joshi 2023-3-2
Ran in:
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

更多回答（1 个）

Cameron
Cameron 2023-3-2
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;

类别

在 Help Center 和 File Exchange 中查找有关 Characters and Strings 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by