Multidimensional array indexing question

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Jason Nicholson
Jason Nicholson 2016-8-8
编辑： Jason Nicholson 2016-8-9
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);
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Jason Nicholson
Jason Nicholson 2016-8-8
The new code looks like this:
linearIndex = linearIndex(:,[1,1],:);
x(linearIndex) = xAverage(linearIndex);
This is cleaner.

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采纳的回答

Stephen23
Stephen23 2016-8-8
编辑：Stephen23 2016-8-8
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(:)]
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Stephen23
Stephen23 2016-8-8
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

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更多回答（1 个）

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

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