Vectorizing a loop calculation
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Hi,
Is it possible to convert the loop below into a vector calculation? Thank you.
dataset=table();
dataset.value=rand(30,1);
dataset.time=sort(rand(30,1));
timefilter=dataset.time-0.2;
result1=repmat(NaN,height(dataset),1);
result2=repmat(NaN,height(dataset),1);
result3=repmat(NaN,height(dataset),1);
%Loop:
for row=5:height(dataset)
startrow=min(find(dataset.time >=timefilter(row)));
result1(row,1)=nansum(dataset.value(startrow:row));
result2(row,1)=length(dataset.value(startrow:row));
result3(row,1)=nanstd(dataset.value(startrow:row)) ;
end
3 个评论
Roger Stafford
2018-2-24
编辑：Roger Stafford
2018-2-24
|I would say that the chance of finding a reasonable vectorization of your for-loop is very slim. The trouble is that 'startrow' is a quite unpredictable quantity dependent on the vagaries of the random ascending values in 'dataset.time' and is not easily calculated without some kind of for-loop. It is the index of the first 'dataset.time' value whose difference from its value at index 'row' is less than or equal to .2 as 'row' ascends and that is difficult to vectorize. I believe it would be best done as an iterative process where one step depends on the results of the previous step, and that is not really the nature of most vectorization schemes. I think there exists a possibly better iterative method than is involved in your for-loops with their inefficient 'find' operation, however it would require a bit of time working it out.
I would point out that 'nansum' and 'nanstd' are surely unnecessary operations as opposed to 'sum' and 'std' since 'dataset.value' contains no NaN values. |
Azura Hashim
2018-2-26
Meade
2018-2-26
The sum, length, and std calculations are so trivially fast that it might be fastest to calculate them for the entire array, then just sort. If this doesn't seem unreasonable for your use, I would use bsxfun or arrayfun on your data, then sort. That would be fastest I think.
That would get rid of the loop entirely.
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2018-2-24
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