How to deal with NaN statistical analysis?

    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,

[Adam]

You didn't tell us what the issues are and we don't have the data to run your code.

[Tony Castillo]

You are rigth

Answer 1

[Jonas Allgeier]

Mean only gives you a single output argument, the mean value; so requesting a second output argument will not work.

[Tony Castillo]

"min", "mode", etc give a single value as well, is not it?

[Jonas Allgeier]

no, min() has a second output argument, which gives you the index of the minimum value. Check help min and help mean.

[Tony Castillo]

The problem is this if I do not include de index I get a value, but as I need the index it returns to me a error:

mean(IRR(:,4),'omitnan')

ans =

3333.20

>> [cmen, indi_mean]=mean(IRR(:,4),'omitnan')

Error using mean

Too many output arguments.

[Adam]

How can you 'need the index' for a mean? It has no index, it's just an average.

[Tony Castillo]

You have the reason, sorry for it.

[Tony Castillo]

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)