# Average values with nan value

7 vues (au cours des 30 derniers jours)
Commenté : Adam Danz le 22 Mar 2023
Hi
I have a matrix with 3 dimentions; i,j,t(time).
I try to calculate mean value for each time step but the matter is we have some nan-values. I have four basins and decide to calculate avrage for each basin seperately. It doesnot work properly
............................................
for t= 1:length(time)
if j>85
U1(t)=nanmean(Uwind,'all')
else if j>65 & j<85
U2(t)=nanmean(Uwind,'all')
else if j>25 & j<65
U3(t)=nanmean(Uwind,'all')
else
U4(t)=nanmean(Uwind,'all')
end
end
end
end
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens

Connectez-vous pour commenter.

### Réponse acceptée

Antoni Garcia-Herreros le 16 Mar 2023
Not exacty sure what you mean by U1(t). I you want to calculate the mean (excluding NaNs) for each timestep for all i and a specified range of j this should work.
U4=nanmean(Uwind(:,1:25,:),[1,2]);
U3=nanmean(Uwind(:,26:64,:),[1,2]);
U2=nanmean(Uwind(:,65:84,:),[1,2]);
U1=nanmean(Uwind(:,65:end,:),[1,2]);
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
Adam Danz le 22 Mar 2023
Instead of nanmean, use mean with the omitnan flag.
For example,
U4 = mean(Uwind(:,1:25,:),[1,2],'omitnan');
mean has more capabilities such as supporting tall arrays, GPU arrays, and code generation, works better with tables, etc. nanmean is discouraged starting in MATLAB R2020b.

Connectez-vous pour commenter.

### Plus de réponses (1)

Cameron le 16 Mar 2023
myArray = nan(8,1); %array of nan values
indx = 1:2:7; %index to populate that array
myArray(indx,1) = rand(length(indx),1) %have some nan and some numbers
myArray = 8×1
0.1698 NaN 0.0233 NaN 0.8454 NaN 0.6111 NaN
mean(myArray(~isnan(myArray))) %take the average of only the numbers
ans = 0.4124
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens

Connectez-vous pour commenter.

### Catégories

En savoir plus sur ROI-Based Processing dans Help Center et File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by