Delete every point that is nearer than a variable

Hello everyone,

I'm trying to delete every point that is nearer than the variable vel_over, in this example 1.4mm, to the extreme values (these are around 339mm and 439mm). This program worked fine until I tried to change the duration of the period. The error message says that the index exceeds the number of array elements and I tried to just delete the last element, but it still doesn't work (the last element suddendly appears again). Can someone help me and find the mistake in my code? I'm thankful for all your ideas.

clear;

vel_over = 1.4;

A = 50;     %Amplitude

t = 0.05;       %Frequenz

off = 389.387;      %Offset von x-Achse

vel_document = readtable("C:\Users\Bruce Rogers\Documents\RoboDK\1pi\RoboDK\Velocity_1pi.txt");

vel = vel_document{:,1};

factor = 0.5;

x = 0:t:(1/factor)*10*pi;

y = -A*sin(factor * x) + off; 

T = (1:length(x))*t;

figure(1)
plot(T,y,'b.')
xlabel('time')
ylabel('distance')

hold on
%%
%y = y(:,1:end-1);
newpoints = zeros(1,(length(y)));

newpoints(1,1) = y(1,1);
%newpoints(1,2) = y(1,2);

z = zeros(1,(length(y)));

%z(numel(y)) = 0;

for i = 2:length(y)
    
    z(i) = (y(i) - y(i-1));

end

z = abs(z);

for i = 1:length(z)
    
    z(i) = z(i) < vel_over;

end

z(numel(y)) = 0;

z(1,1) = 0; 

%y(isnan(y))=0;


%z(numel(y)) = 0;

%%

for i = 1:length(y)
    
    %newpoints(i) = y(i);
   
    if y(i) > 339.30 && y(i) < 339.45
        
        j = i;
        
        while z(j) == 1                 %%The error message appears here
             
            y(j+2) = NaN;
            
            j = j + 2;
        
        end
        
        j = i;
        
        while z(j) == 1
             
            y(j-2) = NaN;
            
            j = j - 2;
        
        end

    end
    
    if y(i) > 439.33 && y(i) < 439.39
        
        j = i;
        
        while z(j) == 1
             
            y(j+2) = NaN;
            
            j = j + 2;
        
        end
        
        j = i;
        
        while z(j) == 1
             
            y(j-2) = NaN;
            
            j = j - 2;
        
        end
        
    end
    
    
end    

newpoints(1,1) = y(1,1);

for i = 2:length(y)
    
    newpoints(i) = y(i);
    
    while isnan(y(i))
        
        vel(i) = ((vel(i)+vel(i-1))/2);
        
        %vel(i-1) = 0;
      
         if isnan(newpoints(i))
         
             newpoints(i) = [];
        
         end
       
        i = i + 1;
        
    end
    
end



figure(1)
plot(T,newpoints,'r.')
xlabel('time')
ylabel('distance')

newpoints = newpoints';

%xlswrite('neue_punkte.xlsx',newpoints)
%xlswrite('neue_geschwindigkeit.xlsx',vel)

Answer 1

[Walter Roberson]

    while isnan(y(i))
        
        vel(i) = ((vel(i)+vel(i-1))/2);
        
        %vel(i-1) = 0;
      
         if isnan(newpoints(i))
         
             newpoints(i) = [];
        
         end
       
        i = i + 1;

After you delete a point, newpoints is now shorter, but you continue to treat it as if it were the original length.

Remember that deleting an element in a matrix is like Tetris: something "falls down" to occupy the slot. If you have an array [1 2 3 4] and you delete 3 then the array becomes [1 2 4] with 4 moving into the space where 3 was, and the vector no longer being 4 long. And you increrement your focus to position 4, but notice you never examined it! When you delete a point and you are moving forward, then you need to retain your focus at the same location, in order to examine the data that fell down into the current location.

There are two ways to proceed:

• create a logical vector mask that indicates whether an element is to be kept (true) or not (false). Then at the end, newpoints = newpoints(mask); OR
• delete backwards . End towards the beginning. If you do that, then anything that "falls down" is data that has already been examined and does not need to be examined again, and it does not matter that the array gets shorter since you still have the same number of elements to examine heading towards the beginning.

[Bruce Rogers]

Hello Walter,

you're absolutly right. I like the comparison with tetris. But even if I delete the last while loop, the error remains. Any ideas why my variable y keeps getting its old length, although I deleted the last element?

[Walter Roberson]

You never delete anything from y. You assign NaN to some of the elements of y.

Your code is odd. Why do you do

z = abs(z);
for i = 1:length(z)
    
    z(i) = z(i) < vel_over;
end

when you could just do

z = abs(z) < vel_over;

?

Or better yet, change variable names as you switch between z as some kind of distance, and z as some kind of mask.

[Bruce Rogers]

No, I mean if I try something like

y = y(:,1:end-1);

The length of the vector will stay the same. I justs replace the numbers I don't need with NaN, because I can work better with it later in Excel.

If I use

z = abs(z) < vel_over;

it's the same result, I get 1s and 0s, only that my attempt is no logical vector. The array element length error is the same