Multidimensional array indexing question

1 vue (au cours des 30 derniers jours)
Jason Nicholson
Jason Nicholson le 8 Août 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 commentaire
Jason Nicholson
Jason Nicholson le 8 Août 2016
The new code looks like this:
linearIndex = linearIndex(:,[1,1],:);
x(linearIndex) = xAverage(linearIndex);
This is cleaner.

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Réponse acceptée

Stephen23
Stephen23 le 8 Août 2016
Modifié(e) : Stephen23 le 8 Août 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 commentaires
Stephen23
Stephen23 le 8 Août 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 le 9 Août 2016
Interesting.

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Plus de réponses (1)

Fangjun Jiang
Fangjun Jiang le 8 Août 2016
x=rand(6,2,4);
MeanX=mean(x,2);
idx=x(:,1,:) < x(:,2,:);
x(idx)=MeanX(idx);
  3 commentaires
Fangjun Jiang
Fangjun Jiang le 8 Août 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 le 9 Août 2016
Modifié(e) : Jason Nicholson le 9 Août 2016
Your second suggestion does work.
When the dimension is greater than 2, I think repmat may be the solution.

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