finding average of a percentile of rows in a matrix?
Asked by Ahmad on 26 Jul 2012
Latest activity Commented on by Ahmad on 31 Jul 2012
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.
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3 Answers
Answer by Azzi Abdelmalek on 26 Jul 2012
Edited by Azzi Abdelmalek on 26 Jul 2012
Accepted answer
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 Comments
Ahmad on 30 Jul 2012
Hi Azzi,
Sorry for being late. Couldn't reply due to illness.
My data has NaNs and the answer results in a NaN. How can I avoid NaNs in this calculation.
Regards,
AMD.
Answer by Wayne King on 26 Jul 2012
Edited by Wayne King on 26 Jul 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
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Ahmad on 30 Jul 2012
Hi Wayne,
Sorry for being late. Couldn't reply due to illness. Yes there are NaNs in my data. Please tell how to skip NaNs in this mean calculation.
Regards,
AMD.
Answer by Azzi Abdelmalek on 30 Jul 2012
Edited by Azzi Abdelmalek on 30 Jul 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,:))
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