Finding average of every nth row?

I have a column vector that is 9985x1. I want to create a new vector that lists the average of every 6 rows in the column. How can I do this? Is there a way to do it with a loop?

Thank you!

[Jan]

"of every 6 rows" or "of every 6th row"?

[farkab95]

The average of rows 1 to 6 and then rows 7 to 12 and then 13 to 18 etc. Thanks!

Answer 1

[Jan]

This is the mean over 6 subsequent rows:

x  = rand(9985, 1);
S  = numel(x);
xx = reshape(x(1:S - mod(S, 6)), 6, []);
y  = sum(xx, 1).' / 6;

The trailing rows are ignored.

[EDITED] General method for matrices:

x  = rand(9985, 14);
p  = 6;
n  = size(x, 1);          % Length of first dimension
nc = n - mod(n, p);       % Multiple of p
np = nc / p;              % Length of result
xx = reshape(x(1:nC, :), p, np, []); % [p x np x size(x,2)]
y  = sum(xx, 1) / p;      % Mean over 1st dim
y  = reshape(y, np, []);  % Remove leading dim of length 1

See also FEX: BlockMean

[Gervasio Batista]

Hi, how would you do this for several columns in the same matrix using a for loop? Could you provide an example?

Thanks!

[Jan]

Do you mean that x is a matrix and you want to reduce the size in the 1st dimension only? Then:

x  = rand(9985, 16);
S  = size(x, 1);
xx = reshape(x(1:S - mod(S, 6), :), 6, S/6, []);
y  = squeeze(sum(xx, 1) / 6);

[Parth Swaroop]

I have an (n,m) matrix. I want to reduce this matrix size by averaging upto every pth element. How can i modify your code for this?

[Jan]

@Parth Swaroop: I've posted the code to sum up every 6 elements. You want every p elements. Then you have to replace each 6 by p:

n  = 9985;
m  = 16;
x  = rand(n, m);
xx = reshape(x(1:n - mod(n, p), :), p, n/p, []);
y  = squeeze(sum(xx, 1) / p);

[Jan]

@vicsim: See [EDITED]

[Ade Narayana]

Super!! Thank you Jan!!

Answer 2

[Andrei Bobrov]

Let A - your array (9985 x 1).

out = splitapply(@mean,A,ceil((1:numel(A))'/6));

or

out = accumarray(ceil((1:numel(A))'/6),A(:),[],@mean);

[Luis Camacho]

Thanks Andrei!

Answer 3

[Birdman]

mean(A(1:6:9985),:)

[Jan]

This creates a vector of every 6th element of A at first. Then mean(AA, :) is most likely a typo and mean(A(1:6:end, :)) is meant.

Answer 4

[Joe S]

These answers work when dim=1 but appears to fail for ndims =2 (multiple columns). To average nRows together (variable "gps") for ndims=2, see below. You have the option of including the average of the "modulus rows" with: includeMods=1.

%create simple matrix, easier to check vs. random numbers
nRows = 11;
nCols = 3;
includeMods=1;
x = reshape(1:(nRows*nCols),[nRows,nCols]);
%provide row number to average over
grps = 3;

%determine remainder rows from Modulo operation 
exRows = mod(nRows,grps);

% work with the top portion that group "cleanly"
topM = x(1:end-exRows,:);

%gather dims for reshaping
nRowsTop = size(topM,1);
numTop = numel(topM);

%Reshape so nrows = grps (row number to average over)
flatTop = reshape(topM,[grps numTop./grps]);

%take averages
meanTop = mean(flatTop,1);

% reshape back to original number of columns
final = reshape(meanTop,[nRowsTop./grps nCols]);

%optional, average the modulus rows, add back into final answer
if includeMods
    if ~(exRows==0)
        exMat= x(end-exRows+1:end,:);
        meanEx = mean(exMat,1);
        final = vertcat(final,meanEx);
    end
end