Find a position of last changing value in array

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Nik Rocky
Nik Rocky 2020-6-7
评论： Nik Rocky 2020-6-7
Hello,
i'm looking for function that detect a position of detection of last changing value (value stay constant or have just little changes).
In this case tube_dimension = 37;
I'm realize this in for-loop, but maybe thereare better way?
Sensivity_old = Sensivity;
TP ...
FN ....
Sensivity = (TP ./ (TP + FN));
if (Sensivity > Sensivity_old + 1.0e-03)
Sencivity_end = tube_cnt;
else
end
disp(Sencivity_end)
Thank you!

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Umair Hassan
Umair Hassan 2020-6-7
编辑：Umair Hassan 2020-6-7
Hi Nikita,
You can use MATLAB's built in function differentrial (diff) to find out the minimum variability points of your data. The first index of the those points would be where your data would have the plateau. If you can attach some data, I would give you a code as well, otherwise psuedo code is:
plateau_points=diff(sensitivity);
plateau_points=find(plateau_points ==0) % finding where the points have 0 change
plateau_tube_dimension=tube_dimension(plateau_points(1));
Hope this helps,
Thanks!
  1 个评论
Nik Rocky
Nik Rocky 2020-6-7
Hi Hassan,
thanks for your response!
I trieded to do your Answer:
My input is:
Sensivity =
Columns 1 through 18
0.8749 0.8806 0.8860 0.9100 0.9114 0.9130 0.9361 0.9374 0.9480 0.9493 0.9539 0.9559 0.9576 0.9586 0.9596 0.9606 0.9609 0.9612
Columns 19 through 36
0.9619 0.9622 0.9626 0.9632 0.9639 0.9649 0.9659 0.9665 0.9672 0.9675 0.9675 0.9679 0.9682 0.9685 0.9685 0.9712 0.9728 0.9745
Columns 37 through 50
0.9768 0.9771 0.9771 0.9771 0.9771 0.9771 0.9771 0.9771 0.9771 0.9771 0.9771 0.9771 0.9771 0.9771
after
plateau_points=diff(Sensivity)
I get:
plateau_points =
Columns 1 through 18
0.0057 0.0054 0.0241 0.0013 0.0017 0.0231 0.0012 0.0106 0.0013 0.0046 0.0020 0.0017 0.0010 0.0010 0.0010 0.0003 0.0003 0.0007
Columns 19 through 36
0.0003 0.0003 0.0007 0.0007 0.0010 0.0010 0.0007 0.0007 0.0003 0 0.0003 0.0003 0.0003 0 0.0027 0.0017 0.0017 0.0023
Columns 37 through 49
0.0003 0 0 0 0 0 0 0 0 0 0 0 0
after:
plateau_points=diff(Sensivity);
plateau_points=find(plateau_points > 0); % finding where the points have 0 change
plateau_tube_dimension = nonzeros(plateau_points(end));
I get 37! Is a right one. Thank you very much!

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