Effacer les filtres
Effacer les filtres

First time matlab user problem Matrix dimensions must agree?

1 vue (au cours des 30 derniers jours)
Azfar
Azfar le 23 Avr 2013
Hi,
I'm a first time matlab user and I've run into a bit of a problem I keep getting an error. Here is my sequence of codes:
derivefilter(FREQSPEC,600);
handle =
1
y =
173.0011
ans =
173.0011
filtering=applyfilter(pic1,ans);
Error using .*
Matrix dimensions must agree.
Error in applyfilter (line 3)
filteredspec=imgspec.*filter;
Here is the script "applyfilter":
function y=applyfilter(img,filter)
imgspec=fftshift(fft2(img));
filteredspec=imgspec.*filter;
y=real(ifft2(fftshift(filteredspec)));
Any help would be appreciated thanks.

Réponse acceptée

Image Analyst
Image Analyst le 23 Avr 2013
What does derivefilter() return? You're not accepting any matrix for the filter that you can then pass in to applyfilter() instead of ans. You may need to have something like
myFilter = derivefilter(FREQSPEC,600);
filtering=applyfilter(pic1, myFilter);
but we can't tell until we know what's inside derivefilter().
  6 commentaires
Walter Roberson
Walter Roberson le 24 Avr 2013
pic1 = rgb2gray(pic1); %convert to 600x600
but you also need to solve the other issues.
Azfar
Azfar le 24 Avr 2013
Thanks for your help, re-sizing the pic1 variable resolved the issue.

Connectez-vous pour commenter.

Plus de réponses (2)

Leah
Leah le 23 Avr 2013
imgspec and filter must have the same dimension to do a element-wise multiplications
size(imspec)
size(filter)

Jan
Jan le 23 Avr 2013
Modifié(e) : Jan le 23 Avr 2013
Your code would be easier to read, if you spend the time for reading the manuals for this forum and format the code properly.
It looks like:
imgspec .* filter
cannot work, because the array sizes do not match. This is surprising, because the value of filter seems to be a scalar.
Using ans explicitly is a bad idea, because this value is very volatile and debugging can change the value. filter is not a good choice also, because this is a name of an important built-in function.
  2 commentaires
Azfar
Azfar le 23 Avr 2013
Thanks for the response, how exactly would I go about resolving this situation?
The scripts I use are laid out by my supervisor for my project. So I have to adhere to his methodology.
Jan
Jan le 24 Avr 2013
@Azfar: Then it is the job of your supervisor to find the solution. We cannot guess how this should be fixed, because all we see is the failing code. To find a solution, one has to know the intention of the code, but wrong code cannot carry this kind of information.

Connectez-vous pour commenter.

Catégories

En savoir plus sur Logical dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by