Error in the loop

Hi all, I am trying to plot the y co-ordinates for different files with respect to one x value but when i execute the program it is showing the following error:

 Attempted to access x(133); index out of bounds because numel(x)=132.Error in ==> new2 at 48 if x(j) == Z1

It is pointing the following lines of code:

for i=1:4 prompt1 = {'enter next input file name including extension'}; fname1 = inputdlg(prompt1); a = importdata(char(fname1), DELIMITER, HEADERLINES); new2()

   for j = 1:200
    if x(j) == Z1
        Z2 = y(j)
       break;
    end
   end
   figure(1);
   plot(Z1,Z2,'r*');
   hold on; 
end
Can anyone pls help me out,
regards

[bym]

please post the code that defines y,Z1 and x before the loop.

essentially x(133) (and higher) don't exist

[santosh]

function [Z1,Z2,Z3,Z4]= new2(x,y)

% read .cor file

DELIMITER = '\t';

HEADERLINES = 74;

prompt1 = {'enter the input file name including extension'};

fname1 = inputdlg(prompt1);

% Import the file change the fileName

a = importdata(char(fname1), DELIMITER, HEADERLINES);

function [] = new2(~,~)

x = a.data(:,1);

y = a.data(:,2);

% clf;

figure(2);

plot(x,y);

hold on;

%%%%%%% joining the end points

figure(1);

plot(x([1 end]),y([1 end]));

end

new2()

Z1= median(x)

for i = 1:100

if x(i) == Z1

Z2 = y(i)

break;

end

end

figure(1);

plot(Z1,Z2,'r*');

hold on;

%%%%%%%%%% loop to take different files

for i=1:4

prompt1 = {'enter next input file name including extension'};

fname1 = inputdlg(prompt1);

a = importdata(char(fname1), DELIMITER, HEADERLINES);

new2()

for j = 1:200

if x(j) == Z1

Z2 = y(j)

break;

end

end

figure(1);

plot(Z1,Z2,'r*');

hold on;

end

end

this is the complete code and input files contains x whose index ends at 132

Answer 1

[Matt Tearle]

Well, the obvious answer is basically what the error message is saying: x only has 132 elements, but you're looping j up to 200. So when j gets to 133, the line if x(j) ... is going to fail.

What may be causing the problem is the ==. If x and Z1 are doubles, you may not get exact equality. If you were planning on that, that could be causing your indexing error. Try

if abs(x(j)-Z1) < tol

where tol is some small value.

Trying to interpret your code, it seems like you're trying to define Z2 to be the value of y at whichever index x is equal to Z1. If so, why not use logical indexing instead:

Z2 = y(x==Z1);

or, as discussed above,

Z2 = y(abs(x-Z1)<tol);

(This line replaces the whole for-loop, BTW)

[santosh]

Hi Matt, thanks alot for ur answer but still i m getting the same error if i use the line Z2 = y(x(i)==Z1);

inside the for loop

[Matt Tearle]

Again, this shouldn't be in a loop.

But, regardless, now you've posted the full code I can see the problem. I'll add another answer.

Answer 2

[Walter Roberson]

Your function is named new2, and inside that function you define a subfunction that is also named new2. At least one of us is confused.

[santosh]

I changed it but stil i m getting the same error

[Jan]

Please update the posted code by editing your original question - and please use code formatting to improve the readability.

Answer 3

[Matt Tearle]

The problem is that Z1 is the median of x, and x has an even number of elements. In that case, the median is the average of the two middle values. If they're not the same, the median will not be any of the values in x. The real question is what y you want back in that case. The average of the two x values in the middle?

Z1 = median(x);
tmp = abs(x-Z1);
Z2 = mean(y(tmp == min(tmp)));

This will work for any number of elements of x.

[Matt Tearle]

Just to clarify, these three lines replace

Z1= median(x)

for i = 1:100

if x(i) == Z1

Z2 = y(i)

break;

end

end