finding average of a percentile of rows in a matrix?
3 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Hi All,
In the below given matrix, I wish to calculate the mean/average of first 20% rows and subtract it from the mean of the last 20% rows.
x =
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
this is a sample matrix, so there can be 100 rows and 100 columns but I need to average the values in the first 20% and subtract it from the average of last 20%. No sorting is needed in my matrix. therefore, in the above matrix, I would need to average the first two rows and subtract the resultant from the average of the last two rows.
any help will be appreciated.
Regards,
AMD.
0 commentaires
Réponse acceptée
Azzi Abdelmalek
le 26 Juil 2012
Modifié(e) : Azzi Abdelmalek
le 26 Juil 2012
n=size(x,1);n1=round(n/5);row_mean=mean(x(n-n1+1:end,:))-mean(x(1:n1,:))
result_mean=mean(row_mean)
% you can use the result: row_mean wich is a line % or the result: result_mean which is the mean of the line row_mean
2 commentaires
Plus de réponses (2)
Wayne King
le 26 Juil 2012
Modifié(e) : Wayne King
le 26 Juil 2012
Assume A is your matrix.
numrows = round(0.2*size(A,1));
lastrowstart = size(A,1)-numrows+1;
lowermean = mean(A(1:numrows,:),2);
uppermean = mean(A(lastrowstart:end,:),2);
uppermean-lowermean
Or did you mean take the mean over all elements in the bottom 20% of rows and upper 20% of rows, so you end up with a single number?
If that is the case:
numrows = round(0.2*size(A,1));
lastrowstart = size(A,1)-numrows+1;
lowermean = mean(mean(A(1:numrows,:),2));
uppermean = mean(mean(A(lastrowstart:end,:),2));
uppermean-lowermean
5 commentaires
Azzi Abdelmalek
le 30 Juil 2012
Modifié(e) : Azzi Abdelmalek
le 30 Juil 2012
%replacing nan by zero
x(find(isnan(x)))=0 % add this to replace nan by zero
% the previous code
n=size(x,1);n1=round(n/5);row_mean=mean(x(n-n1+1:end,:))-mean(x(1:n1,:))
result_mean=mean(row_mean)
%in case you want calculate the mean, ignoring nan , that means: if i have [1 3 nan 4]; the mean will not (1+3+0+4)/4, but (1+3+4)/3. in this case add this code
ind=arrayfun(@(y) ~isnan(y),x)
x(find(isnan(x)))=0;
n=size(x,1);n1=round(n/5);
row_mean1=sum(x(n-n1+1:end,:))./sum(ind(n-n1+1:end,:))-sum(x(1:n1,:))./sum(ind(1:n1,:))
1 commentaire
Voir également
Catégories
En savoir plus sur Time Series dans Help Center et File Exchange
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!