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Removing invalid results knowing the trend

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marco esteves
marco esteves on 22 Aug 2019
Commented: Adam Danz on 23 Aug 2019
Hello,
I have a matrix, which is an output from angle measurement from a motor. So, over time the angle increases over time until it reaches a maximum (360 i.e) and then stops or descreases.
. For instance:
[0 10 -1 40 50 0 60 10 90 0 ]
How can I remove the invalid values, that in that example are the -1 , second and third '0's and the second 10 ?
Thank you

  2 Comments

the cyclist
the cyclist on 22 Aug 2019
What is the general rule for "invalid"?
Adam Danz
Adam Danz on 22 Aug 2019
Your rule should produce a logical vector where True indicates data that should be removed and False indicates data that should be retained. Then you have 2 options.
data(idx) = NaN; % replace the unwanted values with NaN
% or
data(idx) = []; %remove the unwanted values.
The benefit of the first method is that the index of each data point within the vector is retained. So data(4) will always be 40 before and after you replace the unwanted data.

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Accepted Answer

Ted Shultz
Ted Shultz on 22 Aug 2019
Edited: Ted Shultz on 22 Aug 2019
From what you say, it sounds like your valid rule is that if the reported angle is invalid if it is less than the previous value.
A simple slow way to test for this is:
ang = [0 10 -1 40 50 0 60 10 90 0 ];
% find ang(n) < ang(n-1) and remove
for ii = numel(ang):-1:2
if ang(ii) < ang(ii-1)
ang(ii) = [];
end
end
disp(ang)
output: 0 10 40 50 60 90

  3 Comments

Adam Danz
Adam Danz on 22 Aug 2019
Or without a loop,
ang = [0 10 -1 40 50 0 60 10 90 0 ];
ang([false,diff(ang)<=0]) = [];
ang =
0 10 40 50 60 90
marco esteves
marco esteves on 23 Aug 2019
I think this the most elegant choice. thank you!
@adam is it possible to save the indexes of the nonvalid values ?
Adam Danz
Adam Danz on 23 Aug 2019
Sure. The indices are computed with in the square brackets above. It assumes the first value is never a nonvalid value. Any following value is nonvalid if it's less than the previous value as Ted explained.
[false,diff(ang)<=0]

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