Azzera filtri
Azzera filtri

Output matrix dimension mismatch in for loop

1 visualizzazione (ultimi 30 giorni)
Hi guys, I am having problem with storing the peaks of my data in a for loop. It said subscripted assignment dimension mismatch. I am guessing may be the number of peaks is not the same for all the values in the loop. Here is a simplified version of my codes:
q = 0;
g= 0;
for A = 0:pi/2:4*pi;
for B = 1:1:10;
y(q+1) = sin(A*B);
H(q+1) = B;
q = q+1;
end
q = 0;
C(g+1,:) = y;
[pks,locs] = findpeaks(C(g+1,:));
D(g+1,:) = pks;
E(g+1,:) = B(locs);
g = g+1;
end
plot(H,y)
g = 0;
How can I store my peaks and the corresponding x values?
Thanks guys!

Risposta accettata

Jan
Jan il 27 Feb 2017
Modificato: Jan il 27 Feb 2017
Please post the complete error message, such that we do not have to guess the line, which causes the problem. I guess boldly:
C(g+1,:) = y;
Here y is a [1, q+1] row vector. Assigning it to a row of C requires, the C has the same number of columns. But this cannot be true over multiple iterations.
You did not include comments in the code. Without a description of what you want to achieve, suggestion a solution is an educated guess only.
[EDITED] The number of peaks differ between the iterations (I guess), then store them in a cell array:
Av = 0:pi/2:4*pi;
D = cell(1, numel(Av));
for A = Av
...
[pks,locs] = findpeaks(C(g+1,:));
D{g+1} = pks;
...
  3 Commenti
Jan
Jan il 27 Feb 2017
See [EDITED] in my answer.
Catherine
Catherine il 27 Feb 2017
Hi Jan, thanks so much for that. That worked perfectly! Just one last thing, is it possible to convert the cell array back to matrix with each entry occupying a new row? I have tried cell2mat but that just change it to a row vector with all the results which I can't identify which results belong to which iteration.The following is my results:
[] [1,1] [3.67394039744206e-16,6.12323399573677e-16,8.57252759403147e-16,1.10218211923262e-15] [1,1] [] [1,1] [1.10218211923262e-15,5.38968387752153e-15,-9.80955400591059e-16,6.85926003649836e-15] [1,1] []
I know the different length of the results will be a problem. Can it fill the empty cell with zeros to make up the same length? Something like the following matrix:
0 0 0 0
1 1 0 0
3.67394039744206e-16 6.12323399573677e-16 8.57252759403147e-16 1.10218211923262e-15
1 1 0 0
0 0 0 0
....... so on

Accedi per commentare.

Più risposte (0)

Categorie

Scopri di più su Shifting and Sorting Matrices in Help Center e File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by