# How to deal with NaN statistical analysis?

조회 수: 6 (최근 30일)
Tony Castillo 2020년 3월 2일
댓글: Tony Castillo 2020년 3월 2일
First block
[cmin, indice_min]=min(IRR(:,4));
min_dia=IRR(indice_min, 1:3);
disp(min_dia)
Second block
[cmean, indice_mean]=mean(IRR(:,4), ('omitnan')); %
mean_dia=IRR(indice_mean, 2:3);
disp(mean_dia)
Dear all,
I have been analysing a dataset, whereas I have noticed that for some statistical operations such as "min", "max", and also "mode", there are not problems if datastes contains blank spaces ("NaN"), nonetheless, for "median" as well as "mean" the statistical MATLAB functions present some issues, even if I include in the code 'omitnan'. The structure portrayed at "First Block" works correctly, nevertheless, "Second block" does not work properly presenting some issues. Would you helping me with this?
I hope you can help me to cope with this problem.
Sincerely,
##### 댓글 수: 2없음 표시없음 숨기기
You didn't tell us what the issues are and we don't have the data to run your code.
Tony Castillo 2020년 3월 2일
You are rigth

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### 채택된 답변

Jonas Allgeier 2020년 3월 2일
Mean only gives you a single output argument, the mean value; so requesting a second output argument will not work.
##### 댓글 수: 6이전 댓글 4개 표시이전 댓글 4개 숨기기
Tony Castillo 2020년 3월 2일
You have the reason, sorry for it.
Tony Castillo 2020년 3월 2일
However, as I need a day for my analysis with mean value. I prepared this script in order to find that day. In this way I get one day with the exactly characteristics from a mean value.
cmean=mean(IRR(:,4), ('omitnan'));
[row,col]=find(IRR>=(cmean-1)&IRR<=(cmean+1), 1);
mean_dia=IRR(row, 1:3);
disp(mean_dia)

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