Need help with a while loop to take numbers off.

Alright so I have counting algorithm but i need it to extract elements from an original source and after extracting them and storing I need matlab to discard them so they dont get count again. How do i do this? My code: if true
DATA = [ 60 0 40 -20 40 -40 50 -30 30 -60 0 -30 40 -30 60];
L = length(DATA);
E = [1:L];
N = 1;
while N < L
N = N + 1;
if N < 3
continue
end
X(N) = [abs(DATA(N)-DATA(N-1))];
Y(N) = [abs(DATA(N-1)-DATA(N-2))];
if X < Y
continue
else
RANGE(N) = Y(N);
XMEAN(N) = [.5*(DATA(N-1)+DATA(N-2))]
DATA (N -2) = []
DATA (N-1) = []
N = N - 2
end
for I = N:L
L = L - 2
E(I) = E(I+1);
end
end
end

1 Comment

Sorry to not finish expressing my question. So what I'm using right now is if true
DATA(N-2) = []
&
DATA(N-1) = []
end
to delete the values used to calculate RANGE because once range is calculated I dont need to calculate it again for those same points. Is there a better matlab function/code to accomplish this need?

Answers (1)

Yes, that will delete elements but you're indexing elements and that will change, not to mention the length of your DATA array keeps getting shorter as you delete elements to eventually as N approaches L you'll get an index out of range error.
What are you really trying to do? Can't you just extract arrays and subtract them, and use diff? Or use hist if you want to count values?
data_n = DATA(3:end)
data_nm1 = DATA(2:end-1)
data_nm2 = DATA(1:end-2)
X = abs(data_n - data_nm1)
Y = abs(data_n - data_nm2)
RANGE = y
xMean = 0.5 * (data_nm1 - data_nm2)
% I have no idea what E is supposed to represent.

3 Comments

To be honest I do not know what E is representing. I am performing something called Rainflow Counting. So what I'm trying to do is to count one max value and min value (RANGE), then discard those points in the array and do it again for the secondmax second min do the same, thirdmax thirdmin and so on. everytime discarding the value already calculated
Yes I've seen it but is written in C and is hard to manipulate for me. I dont have much programming knowledge. Besides the program I need is pretty simple compared to that

This question is closed.

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Asked:

on 21 Jul 2012

Closed:

on 20 Aug 2021

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