Prev: bluring an image with 1D FIR filter
Next: Execute an m file from the command line
From: Bruno Luong on 28 Apr 2010 16:08
OK, as everyone come up to the same solution; here is a different method without using sub2ind

M=rand(8,20,20)

% Engine
[M3 idx3] = min(M, [], 3);
[M2 idx2] = min(M3, [], 2);
[M1 idx1] = min(M2, [], 1);
idx1
idx2=idx2(idx1)
idx3=idx3(idx1,idx2)

% Check
M(idx1,idx2,idx3)
min(M(:))

% Bruno
From: Bruno Luong on 28 Apr 2010 17:43
In recursively way just for fun:

function varargout = ndmin(M)
nd = ndims(M);
if nd==2 && size(M,2)==1
[~, varargout{1}] = min(M, [], 1);
else
[M I] = min(M, [], nd);
[varargout{1:nd-1}] = ndmin(M);
varargout{nd} = I(varargout{1:nd-1});
end
end % ndmin



%% Command line
M=rand(8,20,20);
[n m t]=ndmin(M)

% Bruno
From: Matt J on 29 Apr 2010 02:19
"Erik Andersson" <annerk02(a)student.umu.se> wrote in message <hra24c$di4$1(a)fred.mathworks.com>...
> Hello!
> I have a n*m*t matrix wher a=8, b=20, c=20. In this matrix a have calculated error in different values. Now I want to find the indexvalues, i.e. the values of n, m and t where the minimumerrorvalue of the matrix is?
=======

Are you sure it's not enough just to find the linear lindex of the minimizer?

[trash, LinearIndex]=min(TheMatrix(:));

If so, it could save you computation.
First  |  Prev  | 
Pages: 1 2
Prev: bluring an image with 1D FIR filter
Next: Execute an m file from the command line