Multidimensional array indexing question

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Jason Nicholson
Jason Nicholson el 8 de Ag. de 2016
Editada: Jason Nicholson el 9 de Ag. de 2016
I have a matrix x that is of size [61 2 45].
linearIndex = find(x(:,1,:) < x(:,2,:));
xAverage = (x(:,1,:) + x(:,2,:))/2;
Now I want to assign the average to anywhere x(:,1,:) < x(:,2,:). I come up with the following but it seems a bit verbose and un-elegant. Thoughts on how to do this better?
[subScriptIndex1, subScriptIndex2, subScriptIndex3] = ind2sub(size(linearIndex), linearIndex);
x(subScriptIndex1, 1, subScriptIndex3) = xAverage(subScriptIndex1, 1, subScriptIndex3);
x(subScriptIndex1, 2, subScriptIndex3) = xAverage(subScriptIndex1, 1, subScriptIndex3);
  1 comentario
Jason Nicholson
Jason Nicholson el 8 de Ag. de 2016
The new code looks like this:
linearIndex = linearIndex(:,[1,1],:);
x(linearIndex) = xAverage(linearIndex);
This is cleaner.

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Respuesta aceptada

Stephen23
Stephen23 el 8 de Ag. de 2016
Editada: Stephen23 el 8 de Ag. de 2016
Your understanding is correct: if a logical index is shorter than the array it is being used on, then the index is not expanded in any way. The solution is to make the index the exact size that you require:
x = reshape((1:18)',[3 2 3])
xx = x;
idx = x(:,1,:) < x(:,2,:);
idx = idx(:,[1,1],:) % or repmat
x(idx) = nan
[xx(:) x(:)]
  3 comentarios
Stephen23
Stephen23 el 8 de Ag. de 2016
Note that this behavior is closely related:
>> X = [1,2,3,4];
>> X([false,true]) % shorter than X
ans =
2
>> X([false,true,false(1,200)]) % longer than X, but only false..
ans =
2
Jason Nicholson
Jason Nicholson el 9 de Ag. de 2016
Interesting.

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Más respuestas (1)

Fangjun Jiang
Fangjun Jiang el 8 de Ag. de 2016
x=rand(6,2,4);
MeanX=mean(x,2);
idx=x(:,1,:) < x(:,2,:);
x(idx)=MeanX(idx);
  3 comentarios
Fangjun Jiang
Fangjun Jiang el 8 de Ag. de 2016
How about this? I think it works but what if the second dimension is larger than 2? There must be a a better way.
x=rand(6,2,4);
MeanX=mean(x,2);
MeanX(:,2,:)=MeanX(:,1,:);
idx=x(:,1,:) < x(:,2,:);
idx(:,2,:)=idx(:,1,:);
x(idx)=MeanX(idx);
Jason Nicholson
Jason Nicholson el 9 de Ag. de 2016
Editada: Jason Nicholson el 9 de Ag. de 2016
Your second suggestion does work.
When the dimension is greater than 2, I think repmat may be the solution.

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