"Matrix dimensions must agree"

Am trying to do some calculation between two matrix. unfortunately i keep getting this error. Other methods that i have tried such as bsxfun(@times run my computer into space. Any help given will be much appreciated. Here is the line of code

val = load ('edata.txt'); 
for data=2:1:length(val)-1
    if(val(data) > val(data-1) & val(data) > val(data+1) & val(data) > 4)
        val1(data) = 1;
    else
        val1(data) = 0;
    end
    end
figure
plot((val-min(val))/(max(val)-min(val)), '-g'); title('\bf Prominent Detection Peaks');
hold on
stem(val1'.*((val-min(val))/(max(val)-min(val)))', ':k');
hold off

The of dimension of val1 = 1x7167 and val contain 7168x1 Thanks in advance

[Stephen23]

"The of dimension of val1 = 1x7167 and val contain 7168x1"

What do you expect element 7168 of val to get multiplied with?

[Evans Gyan]

Thanks for the comment Stephen. What i wanted to do is multiply val1 with ((val-min(val))/(max(val)-min(val))) and plot the results.

[Guillaume]

• Which line is giving the error, the stem line? Please give us the full error message.
• Which version of matlab?
• Why all the tranpose in stem(val1' .* expr') ?
• Is the intent of the above to obtain a 7167x7168 matrix, or a vector of 7167 or 7168 length?
• If the latter, then you haven't answered Stephen's question.

[Evans Gyan]

Thanks Guillaume for your comment.

The version of Matlab am using is R2015a. This is the error which the system return to me.

Error using  .* 
Matrix dimensions must agree.
Error in coded (line 30)
stem(val1'.*((val-min(val))/(max(val)-min(val)))', ':k');

When i use stem(val1'*((val-min(val))/(max(val)-min(val)))', ':k'); , my computer run into space or low memory. Actuallly the intent is to multiply val1 with ((val-min(val))/(max(val)-min(val))) but it looks like val1 and val have different length and i cant fix it

[Guillaume]

"The intent is [repeat of code that is in the question]"

There's absolutely no point in restating (for the 2nd time) the code that is in the question. We understand the code just fine. It can't work since the matrix dimensions don't agree.

We'd be a lot further if you answered the question we asked. So I repeat:

Is the intent to obtain a 7167x7168 matrix (which is what you'd use bsxfun for) or a vector of some length, possibly 7167 or 7168 (for which you'd never use bsxfun)?

If it's a vector, what do you expect element 7168 of val to get multiplied with?

[Evans Gyan]

The intent is to obtain a vector of length. Element 7168 should multiply with the last element of val (if its possible). The length of val1 fell short of 1 and i dont know why it occurred after going over the loop.

Answer 1

[Guillaume]

Because val1 is not preallocated in the code you've showned, it isn't obvious that is created by the loop.

val1 is shorter by 1 element than val because you explicitly construct it so. Your loop is

for loopindex 2 : numel(val)-1  %data as a name for a loop index is extremely misleading
   val1(loopindex) = ...  
end %last index is numel(val)-1. Hence val1 has length numel(val)-1

If you want val1 to be the same length as val then you can assign 0 to the last element, the same way 0 is automatically assigned to val1(0) at the first step of the loop when it does val(2) = .... So after the loop you could have:

val1(end + 1) = 0;

But the loop is completely unneeded and is a very slow way to construct val1 (particularly without the aforementioned preallocation). This would achieve the same, including creating val1 the correct length:

valdiff = diff(val);
val1 = [false, valdiff(1:end-1) > 0 & valdiff(2:end) < 0 & val(2:end-1) > 4, false];

With that fix your stem line will work.

No idea why you tried to use bsxfun or why you mentioned it. There is nowhere in your code where it could be used.

[Evans Gyan]

Thank you Guillaume for taking pains to look at my witty codes. Although am still hovering around it, you have saved me a great deal of head-knocking. When i tried this valdiff = diff(val); val1 = [false, valdiff(1:end-1) > 0 & valdiff(2:end) < 0 & val(2:end-1) > 4, false]; This is what Matlab returns :Error using horzcat Dimensions of matrices being concatenated are not consistent. I appreciate your effort, Man

[Guillaume]

I assumed val was a row vector. If val is a column vector then replace the , by ; for vertical concatenation:

val1 = [false; valdiff(1:end-1) > 0 & valdiff(2:end) < 0 & val(2:end-1) > 4; false];

[Evans Gyan]

It worked Guillaume. Thank you so much. You saved me the hassle. Perfect touch