I have 2 large 3D matrices, and I've noticed that passing a DIM argument for the ttest2 function decreases the computational time on my computer substantially as opposed to looping through each index in the first 2 dimensions (near 60-fold for the data I'm working with):
[~,~,~,stats]=ttest2(A,B,[],[],[],3);
vs.
stats=cell(rmax,cmax);
for r=1:rmax
for c=1:cmax
[~,~,~,stats{r,c}]=ttest2(A(r,c,:),B(r,c,:));
end
end
Now I would like to do the same thing, but adding another matrix C and using anova1 instead of ttest2. The documentation for anova1 suggests that it won't accept a similar DIM argument, but I'm wondering if there's a similar way to cut down on time and avoid looping through each element.
Prashant Kumar answered . 2024-04-26 15:27:55
function [F,DFR,DFE]=NDfstat(X,GDIM,MDIM)
if exist('GDIM','var') && exist('MDIM','var')
D=size(X);
X=permute(X,[GDIM MDIM setdiff(1:numel(D),[GDIM MDIM])]);
end
D=size(X);
DFE = D(1)-1;
DFR = D(1)*(D(2)-1);
% X = X-repmat(mean(X,2),[1 D(2) ones(1,numel(D)-2)]);
% anova1 does this, supposedly to improve accuracy, but it
% seems to destroy calculation. Not sure what the deal is.
xm = squeeze(mean(X,2));
gm = squeeze(mean(xm));
SSE = squeeze(D(2)*sum((xm-repmat(shiftdim(gm,-1),[D(1) ones(1,numel(D)-1)])).^2));
SST = squeeze(sum(sum((X-repmat(shiftdim(gm,-2),[D(1:2) ones(1,numel(D)-1)])).^2)));
SSR = SST-SSE;
F = (SSE/DFE)./(SSR/DFR);
end
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