# How do I take the average of every n values in a vector?

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Brooks on 27 Jun 2013
Commented: Imran Yasin on 31 Jul 2020
I have some data on Pulse Rate and the sample was taken at 1000 Hz (One sample every millisecond), way too big for what I want to see. My vector is 399277x1 and I want to be able to average every 1000 values and get that number in a new vector of somewhere around 400x1. Is there any way to do this?
Thanks

Matthew Eicholtz on 27 Jun 2013
Try this...
n = 1000; % average every n values
a = reshape(cumsum(ones(n,10),2),[],1); % arbitrary data
b = arrayfun(@(i) mean(a(i:i+n-1)),1:n:length(a)-n+1)'; % the averaged vector
COOL! thank you

Jan on 28 Jun 2013
Edited: Jan on 17 May 2019
x = rand(399277, 1); % Some example data
n = 1000; % Number of elements to create the mean over
s1 = size(x, 1); % Find the next smaller multiple of n
m = s1 - mod(s1, n);
y = reshape(x(1:m), n, []); % Reshape x to a [n, m/n] matrix
Avg = transpose(sum(y, 1) / n); % Calculate the mean over the 1st dim
Stelios Fanourakis on 19 May 2019
@Jan
x = rand(399277, 1); % Some example data
n = 1000; % Number of elements to create the mean over
s1 = size(x, 1); % Find the next smaller multiple of n
m = s1 - mod(s1, n);
y = reshape(x(1:m), n, []); % Reshape x to a [n, m/n] matrix
Avg = transpose(sum(y, 1) / n);
Although, it is consist of 6 simple code lines I wasn't familiar with the though behind it.
For instance, in this case, as your example data you used a vector of 399277 rows in a single column. You chose to find the mean per 1000 elements (n). I didn't quite understand why you are trying to find the next smaller multiple of n. I suppose you want integers to perfectly divide the x and not decimals. Am I correct?
You then reshape the matrix for faster computation, probably?
And what this transpose stands for? You calculate the mean over the first dimension of y, since the second one is the n. Right?

Image Analyst on 27 Jun 2013
Edited: Image Analyst on 5 Jan 2019
Here is how I'd do it (an alternate way), using blockproc to average in 100 element long blocks in "jumps":
% Create sample data
PulseRateF = rand(399277, 1);
% Define the block parameter. Average in a 100 row by 1 column wide window.
blockSize = [1000, 1];
% Block process the image to replace every element in the
% 100 element wide block by the mean of the pixels in the block.
% First, define the averaging function for use by blockproc().
meanFilterFunction = @(theBlockStructure) mean2(theBlockStructure.data(:));
% Now do the actual averaging (block average down to smaller size array).
blockAveragedDownSignal = blockproc(PulseRateF, blockSize, meanFilterFunction);
% Let's check the output size.
[rows, columns] = size(blockAveragedDownSignal)
Jan on 16 May 2019
Look at the code: These numbers are the size of the variable blockAveragedDownSignal.

Andrei Bobrov on 28 Jun 2013
Edited: Andrei Bobrov on 28 Jul 2017
x = randi(1000,399277,1);
n = 1000;
m = numel(x);
out = nanmean(reshape( [x(:);nan(mod(-m,n),1)],n,[]));
or
out = accumarray(ceil((1:numel(x))/1000)',x(:),[],@mean);
Imran Yasin on 31 Jul 2020
Very nice job!

Aubai on 20 Jan 2017
Alexey on 1 Nov 2018
Great solution! Thanks!