Issues updating vector inside a for loop

clc, clear, close all;

x3 = 0.5*200; % Half the domain width

x = load("t_10s.txt");

L = x(:,1); % Length/coordinates

[l, m] = size(x);

r1 = zeros(m-1,1);

for i = 1:m-1

n = x(:,i+1); % value of n at the corresponding (L) location

% Find values of L for which n = 1

ind1 = find(n>=0.999);

x1 = L(ind1,1);

r = max(x1) - min(x1); % 2*radius?

r1(i,1) = r

end

% r = max(x1) - min(x1)

% r1(i) = r

Hi all, I am having an issue with updating my vector output from this code. I want to store all the calculated r values from each iteration into a column vector r1 of size (m-1,1), however, I keep getting a size incompatibility errors at line r1(i,1)=r. Could someone help out, please. Thank you.

By the way, I load x (attached) in as a matrix the first column of which I assign to vector L, and columns 2 through m are the different n-vectors for each iteration. Thank you!

Answer 1

[Chris]

When find doesn't find any values > 0.999, it returns an empty vector, with a size of 0x1. Furthermore, max returns an empty vector even if the other value is a number. 0x1 empty vectors don't fit in a 1x1 slot like r(i,1).

The error text is often helpful: Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 0-by-1.

r1(i,1)     =     r

left side (1x1) right side (0x1)

You could test for the empty value and, for instance, convert it to a NaN:

if isempty(r)
    r(i,1) = nan;
else
    r(i,1) = r;
end

Incidentally, for your future reference, r(i) is equivalent to r(i,1) in this case (and L(i) to L(i,1)). Since the second dimension of r (and L) has size 1, the value only has one column in which it is possible to be placed.

[Buhari Ibrahim]

Thanks Chris, for your response. I have tried that and I am still getting r1 output only zero entries like the preallocation. Perhaps, is there a better way to find the indexes of elements in n that are >=0.999, and use those indexes to find their corresponding entries in L? Thank you.

[Chris]

Sorry...I left out the "1".

I'm not sure whether this code accomplishes what you want, though.

x3 = 0.5*200; % Half the domain width in the neck growth direction
x = load("t_10s.txt");
L = x(:,1); % Length/coordinates along of line through neck
[l, m] = size(x);
r1 = zeros(m-1,1);
for i = 1:m-1
    n = x(:,i+1); % value of order parameter at the corresponding (L) location
 
    % Find values of L for which n = 1
    ind1 = find(n>=0.999);
    x1 = L(ind1,1);
    r = max(x1) - min(x1);  % Neck radius?

    if isempty(r)
        r1(i,1) = nan;
    else
        r1(i,1) = r;
    end
end

figure
plot(r1)
xlim([0 inf])

[Buhari Ibrahim]

Hi Chris, I really appreciate the effort, unfortunately, it is still not accomplishing the goal. With the attached text file, r1 is uspposed to be a 21-by-1 vector with a few first zero entries and the remaining are positive real numbers. However, when I run your modified code, I get the entries of r1 to be almost all NaNs. Thanks once again.

[Chris]

Hi Buhari,

Max values >.999 don't occur until i==13, so you only get nine out of 21 non-zero points, based on your code.

[Buhari Ibrahim]

Hi Chris,

Thank you for your input, I really appreciate the help.