finding average of a percentile of rows in a matrix?

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.

Answer 1

[Azzi Abdelmalek]

   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

[ARS]

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.

[Azzi Abdelmalek]

can we replace them with 0?

Answer 2

[Wayne King]

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

[ARS]

Hi Wayne,

Thanks for ur reply. It produces NaN as result. lowermeans is producing output but uppermeans is all NaNs.

any solution to this. Also, I need a single number at the end as difference of averages.

[Wayne King]

are there any NaNs in your matrix?

[ARS]

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.

[Oleg Komarov]

Use nanmean instead of mean.

[ARS]

Thanks.

Answer 3

[Azzi Abdelmalek]

%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,:))

[Azzi Abdelmalek]

if you dn't want replace nan by 0, what do you want to do?