how to take matrix element (i.e row) one by one and save each new matrices in a new variable example:
original matrix: a=[2 6 7 8;4 5 3 7;9 7 6 9;5 3 1 9]
become: a1=[4 5 3 7;9 7 6 9;5 3 1 9] a2=[2 6 7 8;9 7 6 9;5 3 1 9] a3=[2 6 7 8;4 5 3 7;5 3 1 9] a4=[2 6 7 8;4 5 3 7;9 7 6 9]

 采纳的回答

a=[2 6 7 8;4 5 3 7;9 7 6 9;5 3 1 9]
id = ~eye(size(a));
h = arrayfun(@(i1)a(id(:,i1),:),1:size(id,2),'un',0);
[out,idxout] = min(cellfun(@(x)det(x*x.'),h))
OR
a=[2 6 7 8;4 5 3 7;9 7 6 9;5 3 1 9]
id = ~eye(size(a));
s = size(a,2);
h = zeros(s*[1 1 1] - [1 0 0]);
kt = zeros(s,1);
for j1 = 1:s
k = a(id(:,j1),:);
kt(j1) = det(k*k.');
h(:,:,j1) = k;
end
[out,outidx] = min(kt)

5 个评论

error:
??? Undefined command/function 'arrayfun'.
release of MATLAB ?
Based on the error message style, Eri must be using MATLAB 5, I think.
@andrei bobrov
your second code seem to work fine, could you explain how it works?
@andrei bobrov and Walter Roberson
i am using matlab 7
@andrei bobrov
can you explain your code, since i need to actually work with larger matrix and more complicated function
i don't understand from line 4 onwards

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更多回答(2 个)

Sean de Wolski
Sean de Wolski 2012-2-1

1 个投票

Don't do it. That is all.
If you explain your end goal to us we can help you find a better way.

4 个评论

my goal is to find the product of the result matrices * their transpose, find its determinant, find which one has the smallest determinant, then repeat the process all over again until there is only one row left
@eri: And this is more efficient and more flexible, if you use dynamic indices instead of a copy to a separate variable.
@Jan Simon
and how is that?
Use cell arrays, as described in that FAQ entry.

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eri
eri 2012-2-2

0 个投票

can someone help me?

1 个评论

What does:
which -all arrayfun
return? And what is the output from:
ver
at the command line?

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