matrix, using variables like a1,a2,a3,...an, minimum

2 ビュー (過去 30 日間)
eri
eri 2012 年 2 月 17 日
編集済み: Michael 2013 年 10 月 15 日
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]
then i want to find determinant from each of them * their transpose (ie: det(a1*a1')) and then find which one of them has minimum value
i have asked before and i receive this answer
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)
the problem is i don't know how this code works, and since the matrix i am going to use is different than that i use as example, i can't adjust that code

回答 (3 件)

Walter Roberson
Walter Roberson 2012 年 2 月 17 日
  1 件のコメント
eri
eri 2012 年 2 月 17 日
yes i already read it, but i still don't get it

サインインしてコメントする。


N/A
N/A 2012 年 2 月 17 日
eval( sprintf( 'a%d=a(%d, :), ', [1;1]*(1:size(a,1))) )

carmen
carmen 2012 年 2 月 17 日
ok heres a brute force version without fancy commands.
a=[2 6 7 8;4 5 3 7;9 7 6 9;5 3 1 9]
rows=size(a,1);
columns=size(a,2);
for i=1:rows
b=1:rows;
b(i)=[];
tmp=a([b],:)
eval(sprintf('a%d=tmp;',i));
determinante(i)=det(tmp*tmp');
end
min=find(determinante==min(determinante)); %gives you the index of
% determinante where the minimum values is found
i hope this works for arbitrary n*m matrices
  1 件のコメント
carmen
carmen 2012 年 2 月 17 日
add a
determinante(min)
to see what the minimum value is

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeLogical についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by