Error in the loop

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santosh
santosh 2011-4-6
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
  2 个评论
bym
bym 2011-4-6
please post the code that defines y,Z1 and x before the loop.
essentially x(133) (and higher) don't exist
santosh
santosh 2011-4-6
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

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回答（3 个）

Matt Tearle
Matt Tearle 2011-4-6
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)
  2 个评论
santosh
santosh 2011-4-6
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
Matt Tearle 2011-4-6
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.

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Walter Roberson
Walter Roberson 2011-4-6
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.
  2 个评论
santosh
santosh 2011-4-6
I changed it but stil i m getting the same error
Jan
Jan 2011-4-6
Please update the posted code by editing your original question - and please use code formatting to improve the readability.

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Matt Tearle
Matt Tearle 2011-4-6
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.
  1 个评论
Matt Tearle
Matt Tearle 2011-4-6
Just to clarify, these three lines replace
Z1= median(x)
for i = 1:100
if x(i) == Z1
Z2 = y(i)
break;
end
end

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