RMS of matrices in a cell or array

3 Ansichten (letzte 30 Tage)
Mike
Mike am 15 Mär. 2016
Kommentiert: Mike am 17 Mär. 2016
Hi all,
If I have an n x n matrix, P, I find the rms by RMS_P = sqrt(mean(Prv(:).^2));.
Please, how can I find the rms of the ith elements in an array, P(;,:,i), along the ith dimension? Thanks.
This is what I have that outputs the rms of a matrix:
ang = 0:90:270;
low_lim = ang - 1;
up_lim = ang + 1;
%The input, inp_file is a 1001 x 1001 matrix, and it's attached.
for i = 1:length(ang)
p_rot = zeros([1001 1001 4]);
ang_var = low_lim + (up_lim - low_lim).*rand(1,length(ang));
p_rot(:,:,i) = imrotate(inp_file,ang_var(i),'crop');
mean_p_rot = mean(p_rot,3);
P = (p_rot(:,:,1)) - mean_p_rot;
P(isnan(P)==1)=0;
RMS_P = sqrt(mean(P(:).^2));
end
I'm trying to repeat this loop multiple times. What I tried is:
for ii = 1: 5
for i = 1:length(ang)
p_rot = zeros([1001 1001 4]);
ang_var = low_lim + (up_lim - low_lim).*rand(1,length(ang));
p_rot(:,:,i) = imrotate(inp_file,ang_var(i),'crop');
mean_p_rot = mean(p_rot,3);
P = (p_rot(:,:,1)) - mean_p_rot;
P(isnan(P)==1)=0;
RMS_P = sqrt(mean(P(:).^2));
end
P_new(:,:,ii) = P;
RMS_P_new(ii) = sqrt(mean(P_new(:).^2));
end
I'm having trouble putting the first loop in another loop, and finding the rms of each ii in P_new(:,:,ii).
Please help. Thanks!
  4 Kommentare
dpb
dpb am 16 Mär. 2016
Hmmm....oh, I'd forgotten rms is in Signal Processing toolbox. Try
sqrt(mean(x.*x))
instead or if x can be imaginary
sqrt(mean(x .* conj(x)))
Think should be noticeably faster than .^2.
But, that aside, does rms(P(:)) solve the issue however you compute it?
Mike
Mike am 17 Mär. 2016
Thanks! I stored each rms result from the inner loop, as an array element in the second loop.
end
P_new(:,:,ii) = P;
RMS_P_new(:,:,ii) = RMS_P;
end
I want click "accept your answer" but there's no button for that. I think it's because you replied as a comment, not as an answer. Thanks dpb.

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Antworten (1)

John BG
John BG am 17 Mär. 2016
Mike
have you noticed that rms has an option to aim on the dimension you want to sweep along?
P=randi(10,3,3,3)
P(:,:,1) =
2.00 8.00 1.00
5.00 10.00 9.00
10.00 7.00 10.00
P(:,:,2) =
7.00 4.00 8.00
8.00 7.00 1.00
8.00 2.00 3.00
P(:,:,3) =
1.00 7.00 1.00
1.00 4.00 5.00
9.00 10.00 4.00
rms(P,1)
ans(:,:,1) =
6.56 8.43 7.79
ans(:,:,2) =
7.68 4.80 4.97
ans(:,:,3) =
5.26 7.42 3.74
rms(P,2)
ans(:,:,1) =
4.80
8.29
9.11
ans(:,:,2) =
6.56
6.16
5.07
ans(:,:,3) =
4.12
3.74
8.10
rms(P,3)
ans =
4.24 6.56 4.69
5.48 7.42 5.97
9.04 7.14 6.45
If you find this answer of any help solving your question, please click on the thumbs-up vote link,
thanks in advance
John
  1 Kommentar
Mike
Mike am 17 Mär. 2016
Thanks John. That'd have worked if I could use the built-in rms function. I get an error message when I try that. I stored each rms result from the inner loop, as an array element in the second loop.
I clicked the thumbs up...thanks!
end
P_new(:,:,ii) = P;
RMS_P_new(:,:,ii) = RMS_P;
end

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