index out of bounds because numel(averg)=1
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This is the code that I'm currently using:
A = interpolate;
averg = mean([A(1:end-2),A(3:end)],2);
streaking = [];
for idx = 2:size(A,2)-1
streaking(:,idx) = (abs(A(:,idx)-averg(idx))./averg(idx))*100;
end
The full error message:
>> Attempted to access averg(2); index out of bounds because numel(averg)=1.
>> Error in rad_cal2 (line 817)
streaking(:,idx) = (abs(A(:,idx)-averg(idx))./averg(idx))*100;
interpolate is a matrix
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採用された回答
Laura Proctor
2012 年 8 月 8 日
編集済み: Laura Proctor
2012 年 8 月 8 日
You are using linear indexing to index into A, and so the output is a row vector. For example: A(1:end-2) will take all terms except for the last two in the matrix A, and create a row vector. Concatenating A(1:end-2) with A(3:end) creates a really long row vector. The mean of a row vector along the row, takes the average of all the elements which produces a scalar value for averg.
If you want to take the ROWS 1:end-2, then you can index into A using row column indexing:
A(1:end-2,:)
For what values are you trying to find the means? How are you trying to arrange the values in A to find the means? Be careful how you find that matrix ([A(1:end-2),A(3:end)]) - you may want to add an intermediate variable to calculate it, and then find the mean.
A = interpolate;
B = (A(1:end-2,:) + A(3:end,:))/2;
averg = mean(B,2)
modified accordingly.
7 件のコメント
Sean de Wolski
2012 年 8 月 8 日
No, just a different approach. More adaptable if say you wanted to use a longer window length. E.g. to change mine to include 5 elements one each side:
averg2 = conv2(A,ones(11,1)./11,'valid');
To adapt it to your above:
averg2 = conv2(A,[0.5;0;0.5],'valid')
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