How to use accumarray?
4 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
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
0 comentarios
Respuesta aceptada
the cyclist
el 30 de Jun. de 2013
Editada: the cyclist
el 30 de Jun. de 2013
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';
Más respuestas (1)
Matt J
el 30 de Jun. de 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);
Ver también
Categorías
Más información sobre Logical en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!