Set conditions between two matrices

First question :* I have two matrices (360x1) and I want to set conditions to generate a third matrix (360x1). I want to compare the first row of the first matrice [A] to the first row of the second matrice [B], up until the 360th row of the first matrix to the 360th rows of the second matrix.

This is what I tried to write:

    E = ones(360,1);
 for r=1:360
   for c=1:1
    for k=1:360
    if A(k,1) == B(k,1)
         E(r,c)= 2;
    elseif A(k,1) > B(k,1)
        E(r,c)= -1;
    else A(k,1) < B(k,1)
        E(r,c)= 0;
    end
    end
   end

...and it doesn't work.

Second question : I would like to display a different text instead of the value '2', '-1' and '0'.

Thank you in advance !

[Jan]

"It doesn't work" is too lean to describe the problem you have. Do you get an error message or does the result differ from your expectations? Currently the comparison does not depend on r.

Answer 1

[Guillaume]

If you're happy with having 1 when A(k)>B(k), 0 when A(k)==B(k) and -1 when A(k)<B(k), then the answer is simply:

E = sign(A-B)

No loop needed. If you do want -1, 2, 0 in the above case, respectively, then:

vals = [0 2 -1];
E = vals(sign(A-B) + 2)

"I would like to display a different text instead of the value '2', '-1' and '0'"

First, you're mixing up text and numeric. Your code, and the answers above, generate numerical arrays. There is no text. You could have text instead, e.g.:

vals = {'smaller', 'equal', 'greater'};  %for text, has to be a cell array
E = vals(sign(A-B) + 2)

but be aware that working with text is a bit harder than working with numerical matrices.

Finally, if you're working with vectors use linear indexing (i.e. A(k)) rather than subscript indexing (i.e A(k, 1)). The former will work regardless of the direction of the vector, the latter will not.

[Stephen23]

+1 very clear explanation

[Jan]

+1. Maybe Camille Fong want to compare the vectors elementwise like:

E = sign(A - B.')   % >= Matlab R2016b, auto-expanding

to get E as a matrix. For older Matlab versions:

E = sign(bsxfun(@minus, A, B.'))

[Guillaume]

Yes, I wondered about that but Camille did state she wanted a 360x1 result. The r loop is indeed puzzling in the original code.

Answer 2

[Pawel Jastrzebski]

clear all;
clc;

% DATA
A = randi(50,1,360,'int16');
B = randi(50,1,360,'int16');

% LOGICAL VECTORS

AB_Equal  = A == B
A_Greater = A > B
B_Greater = ~(AB_Equal | A_Greater)

% NEW MATRIX
C         = zeros(1,length(A));
C(A == B) = 2;
C(A > B)  = -1;

% IF you want to swap the -1,0,2 for text classification, i.e.:
% -1 = 'a'
%  0 = 'b'
%  2 = 'c'

cVal            = '0';
cTxt            = repmat(cVal,1,length(A));
cTxt(AB_Equal)  = 'c';
cTxt(A_Greater) = 'a';
cTxt(B_Greater) = 'b';