Multidimensional array indexing question

2016 8 8
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更新时间：2016 8 9
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0 个投票

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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Stephen23
Stephen23 2016-8-8
编辑：Stephen23 2016-8-8

0 个投票

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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Fangjun Jiang
Fangjun Jiang 2016-8-8

1 个投票

x=rand(6,2,4);
MeanX=mean(x,2);
idx=x(:,1,:) < x(:,2,:);
x(idx)=MeanX(idx);

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Jason Nicholson
Jason Nicholson 2016-8-8
编辑：Jason Nicholson 2016-8-8
The meanX line is better than what I did for sure.
The line starting with x(idx) doesn't work though. You don't get an error but it is wrong. The problem is the size of idx is [6 1 4] while x is of size [6 2 4]. I think what is happening is logical indexing relies on the linear index of an array. Therefore size of idx has 24 elements while x has 48 elements. Only the first 24 elements of x are modified while the 2nd 24 elements are not modified. Try the following to see what I am saying. Note that the 2nd half according to the linear index of x is not modified. The [xx(:) x(:)] when output to the command window, should make this clear.
x = reshape((1:18)',[3 2 3])
xx = x;
idx = x(:,1,:) < x(:,2,:);
x(idx) = nan
[xx(:) x(:)]
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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