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PO
primarykey
Id
9695742
data
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0
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0
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CreationDate
2012-03-14T04:01:25.010
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2012-03-14T15:22:44.617
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2012-03-14T15:22:44.617
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232610
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1214731
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9694129
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2
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1
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0
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text
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I'll give this a shot based on the information you've provided. I don't have timing information though. <ol> <li>Make <code>px</code> a 3D matrix <code>DxVxC</code> (instead of <code>1xC</code> cell array of <code>DxV</code> 2D matrices)</li> <li><code>px_new=px(:,x,:)</code> reorganizes the matrix, putting your values of interest into the main diagonal of the first two dimensions</li> <li>Create a logical index with ones on the diagonal (<code>mask=eye(D,D);</code>); repmat so it is the size of C in the third dimension.</li> <li>Index into <code>px_new</code> and <code>reshape</code>, so that there are <code>C</code> columns.</li> <li><code>prod</code> the columns, leaving a <code>1xC</code> vector</li> <li>Multiply this (element-wise) with <code>py</code> to get your output <pre><code>px = nan(3,4,5); %# create test 3D matrix px(:, :, 1) = [1 2 3 4; 4 5 6 4; 7 8 9 4]; px(:,:,2)=px(:,:,1)*1.5; px(:,:,3)=px(:,:,2)*1.5; px(:,:,4)=px(:,:,3)*1.5; px(:,:,5)=px(:,:,4)*1.5; x = [4 2 3]; %# 1xD vector discrete on 1-V px_new=px(:,x,:); %# reorganize into DxDxC idx=logical(repmat(eye(size(pd_new,1))),[1,1,size(pd_new,3)])); %# logical index P = prod(reshape(pd_new(idx),[],size(pd_new,3))); %#P is now 1xC vector </code></pre></li> </ol> This code was tested at <a href="http://www.online-utility.org/math/math_calculator.jsp" rel="nofollow">http://www.online-utility.org/math/math_calculator.jsp</a> Edit: I was taking a couple of unnecessary steps originally. I've updated it to be more concise.
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1. POVectorizing a short loop such that Octave matches Matlab speed
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PostTypePostTypeId
1. PTAnswer
UserLastEditorUserId
1. USJonas
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1. UStmpearce
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1. VO
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