Does NaN used in confusion matrix affects results?
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I have used confusion matrix and for making two matrices of equal dimensions I had used NaN but is it wise to use that? for example: A= [2 3 7 ]; B= [3 4 NaN]; C= A+B; C= 5 7 NaN I am getting where result should have been [5 7 7] according to me.Can anyone help? Thanx in advance.
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Accepted Answer
Chad Greene
on 16 Jun 2016
Here's one way:
C = nansum([A; B])
C =
5 7 7
2 Comments
Chad Greene
on 17 Jun 2016
Using NaN as a place filler is often the best way to do it. The reason 1+NaN=NaN is because Matlab doesn't what the answer is if you add a number to something that is not a number. But the nansum function and its sisters (nanmean, nanmedian, etc) are available to simply ignore NaNs the way you want.
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