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Asked by Ede gerlderlands
on 30 Jun 2013

I have a matrix of y=[9, 5346]

I want to find the the mean of every six values of each row. I tried this function but can't really figure out how to do it

for ii=1:9 oo(ii)=accumarray(1:6:5346,y(ii,:),,[],@mean); end

it's saying there is error in this code . I don't really know how this accumarray works.

Any help is appreciated

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Answer by the cyclist
on 30 Jun 2013

Edited by the cyclist
on 30 Jun 2013

Accepted answer

accumarray() can be a bit tricky to learn. I think this solves your problem. I commented it to help you see what is happening.

A = rand(9,5346); % Make up some fake data that is the same size as yours At = A'; % Take the transpose, because accumarray works down columns.

numberPerGroup = 6; numberGroups = 5346/6; % 931, but wanted you to see where this comes from

% The subs array indicates which rows (of the transposed array) are going to get grouped. You have 931 groups total, each gather 6 rows. subs = repmat(1:numberGroups,[numberPerGroup,1]); subs = subs(:);

% Preallocate the output array output = zeros(numberGroups,9);

% Accumulate each column in turn for nr = 1:9 output(:,nr) = accumarray(subs,At(:,nr),[],@mean); end

% Transpose the output back output = output';

Answer by Matt J
on 30 Jun 2013

Although you can do this with accumarray, it will probably be faster to do

oo=downsampn(y,[1,6]);

using the function below.

function M=downsampn(M,bindims) %DOWNSAMPN - simple tool for downsampling n-dimensional nonsparse arrays % % M=downsampn(M,bindims) % %in: % % M: an array % bindims: a vector of integer binning dimensions % %out: % % M: the downsized array

nn=length(bindims);

[sz{1:nn}]=size(M); %M is the original array sz=[sz{:}];

newdims=sz./bindims;

args=num2cell([bindims;newdims]);

M=reshape(M,args{:});

for ii=1:nn

M=mean(M,2*ii-1);

end

M=reshape(M,newdims);

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