How can I remove repeating values from the vector?

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Mariusz Sledz
Mariusz Sledz 2020년 3월 21일
댓글: Mariusz Sledz 2020년 3월 29일
Hi,
I've got a loop that simulates Simulink model, collects the data and plots everything on the graphs, here is simplified code:
for v = 4:3:26
air_velocity = v;
while abs(old_u - new_u) >= errortolerance % siumlate untill model converges
simOut = sim('TurbineWithVariablePitch_new_.slx',time);
umax = round(max(simOut.Velocity), 2);
old_u = new_u;
new_u = umax;
disp(['Current iteration: ',num2str(iteration), ' for time: ', num2str(time, '%10.5e') ])
if abs(old_u - new_u) <= errortolerance
cprintf('*blue',['Solution fully converged with the current error of ', num2str(abs(old_u - new_u))]);
cprintf('\n');
else
disp(['Solution did not converged with the current error of ', num2str(abs(old_u - new_u))])
end
if iteration >= maxiteration
cprintf('*red','Solution did not converged\n');
break
end
iteration = iteration +1;
time = time + timestep;
end
simulationtime = [0:0.01:2000];
a = round(simOut.AoA.signals.values);
n=max (numel(simulationtime)); %make the vectors the same size
u(end+1:n)=nan;
a(end+1:n)=nan;
AoAout = [AoAout , a]; %add to the results
SimulationTime = [SimulationTime , simulationtime'];
end
f4 = figure('Name','AoA vs Time');
plot(SimulationTime, AoAout, 'linewidth', 2);
ylabel('AoA (deg)');
xlabel('Time (s)');
The problem is, I'm getting repitetive values for AoA:
They are not always the same so unique command doesn't work... I've been stuck on this problem for a while. I'm fairly new to MATLAB and tring to achieve something similar to this:
  댓글 수: 6
Mariusz Sledz
Mariusz Sledz 2020년 3월 24일
Still have no clue how to do it. I'm alredy using round() function to get the values. I've tride to define position as a single vector from 0 to 360, does'n work.
How can I select single cycle? Or how can I average them?
Mariusz Sledz
Mariusz Sledz 2020년 3월 26일
Hi, I'm actually really desperate for the solution and still haven't figured it out.

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Mariusz Sledz
Mariusz Sledz 2020년 3월 27일
So... I found the solution. It may not be the most elegant but it works. The code goes troug the vector untill it finds the value that's smaller than 10. Then it starts saving the values untill it reaches 350. After that it fills the rest with NaN.
position=[];
CountOn = 0;
row=1;
End = 0;
for x = 1:1:1000
b=p(row);
d=NaN;
if b <= 10
CountOn = 1;
end
if CountOn == 1
d=b;
end
if b>=345 && CountOn==1
CountOn = 0;
End=1;
end
if End == 1
d=NaN;
end
position=[position, d];
row=row+1;
end
position = position';
Posting if anyone will encouter the same issue.
  댓글 수: 1
Mariusz Sledz
Mariusz Sledz 2020년 3월 29일
I've optimised the code a bit:
position=[];
CountOn = 0;
row=1;
End = 0;
newb = 10;
for x = 1:1:1000
oldb=newb;
newb=p(row);
d=NaN;
if newb > oldb && oldb < 10
CountOn = 1;
end
if CountOn == 1
d=oldb;
end
if newb<oldb && CountOn==1
CountOn = 0;
End=1;
end
if End == 1
d=NaN;
end
if row > 1
position=[position, d];
end
row=row+1;
end
position = position';

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