Replace missing elements in an array with NaN
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Hello! for i=1:n
scanindex = round((i-1)*pointsperscan+1):round(i*pointsperscan);
filterindex == (i-1)*pointsperscan+find(vlos(scanindex)<=blade_flicker_threshold&...
quality(scanindex)>=quality_range(1)&...
quality(scanindex)<=quality_range(2)&...
power(scanindex)>=power_range(1)&...
power(scanindex)<=power_range(2))';
So i have this code. Scan index reads the entire second values with 312 entries and the filterindex filters based on the scan parameters which results in an array less than 312 entries. I want to keep the size of the filterindex same as the scanindex as I need to rewrite the code for more general purposes and so I was thinking if there was a way to replace the indeces which do not match the filter parameters with NaN values. So the output should read that the size of scanindex and the filterindex are the same!
Thanks
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采纳的回答
Thorsten
2016-10-4
You can add the needed number of NaNs to the filterindex;
filterindex(end+1:numel(scanindex) = NaN;
2 个评论
Thorsten
2016-10-4
编辑:Thorsten
2016-10-4
I see. I think that you can use logical indexing to solve your problem:
filterindex = nan(size(scanindex));
fi = (i-1)*pointsperscan+find(vlos(scanindex)<=blade_flicker_threshold&...
quality(scanindex)>=quality_range(1)&...
quality(scanindex)<=quality_range(2)&...
power(scanindex)>=power_range(1)&...
power(scanindex)<=power_range(2))';
filterindex(fi) = true;
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