can someone help me loop this??
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n=4 t=1:21
std_t1 = (((pc(1,:)- MA(1,:)).^2)/(n-1)+...
((pc(2,:)- MA(1,:)).^2)/(n-1)+...
((pc(3,:)- MA(1,:)).^2)/(n-1)+...
((pc(4,:)- MA(1,:)).^2)/(n-1)).^(1/2)
std_t2 = (((pc(2,:)- MA(2,:)).^2)/(n-1)+...
((pc(3,:)- MA(2,:)).^2)/(n-1)+...
((pc(4,:)- MA(2,:)).^2)/(n-1)+...
((pc(5,:)- MA(2,:)).^2)/(n-1)).^(1/2)
std_t3 = (((pc(3,:)- MA(3,:)).^2)/(n-1)+...
((pc(4,:)- MA(3,:)).^2)/(n-1)+...
((pc(5,:)- MA(3,:)).^2)/(n-1)+...
((pc(6,:)- MA(3,:)).^2)/(n-1)).^(1/2)
etc .. say I would like to continue this pattern for a total of 21 time periods
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Réponse acceptée
Roger Stafford
le 30 Sep 2013
Rather than sweating out a vectorization, it's a lot easier to do it with a single for-loop. Also it may be just as fast or faster.
n = 4;
m = 21;
std_t = zeros(m,size(pc,2));
for k = 1:m
std_t(k,:) = std(pc(k:k+n-1,:)-repmat(MA(k,:),n,1),1);
end
The rows of 'std_t' are your vectors std_t1, vtd_t2, etc. Make sure 'pc' is defined up to m+n-1 rows.
Plus de réponses (2)
Walter Roberson
le 29 Sep 2013
T1 = reshape(pc, 4, size(pc,1)/4, []);
T2 = permute( reshape( repmat(MA,[4 1 1]), size(MA,1), 4, size(MA,2) ), [2 1 3] );
T3 = (T1 - T2).^2;
T4 = squeeze( sum(T3,1) );
T5 = sqrt(T4 ./ (n-1));
Now one of the dimensions of T5 corresponds to the stdx* index (that is, corresponds to the subscript of MA that you used), and the other dimension of T5 corresponds to the second dimension of pc and MA. Provided, that is, that I worked out all the manipulations properly in my head.
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