MATLAB: Return the number of the row that contains the largest value in column vectors

matrix

Hi there,

Is there a way in matlab to return the number of a row that contains the largest value?

For example, for the following matrix

0.14  0  0.99  1  0
0.73  0.93  0.02  0  0
0  0  0  0  0.99
0  0  0.01  0  0
0.01  0.05  0.03  0.01  0
0  0.02  0  0  0.02
0.01  0  0.01  0  0

How could you make it return 2 2 1 1 3?

Many thanks John

Best Answer

The second output argument of the max() function will do this:

 [m index] = max(x)

where x is your matrix. Note that by default, max() works down the columns of a matrix. Read "doc max" for more details.

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Hi, max(A) returns maximum element in case A is a vector. If A is a matrix then it returns a row vector of the maximum value of each column.

Here, you can do max(max(A)) to get the maximum element.Similarly min(min(A)) to get minimum element.

MATLAB: Largest Candidate Rule

Hi Jason,

Ok. I am assuming some things. I assume that you are a very basic and new user of MATLAB. So The code which I am giving is an extremely basic code just to give you a feel of the steps. There can be more optimized and complicated and slick codes but I have made this basic code for you for your understanding.

a(:,1) = randi(1000,[1,100]); % Suppose it denotes the time taken 
a(:,2) = randi(5000,[1,100]); % Suppose it denotes constraint A
a(:,3)= randi(20000,[1,100]); % Suppose it denotes constraint B
[p,q] = sort(a(:,1),'descend'); % Please look the syntax up in the help file
a(:,[1:end]) = a(q,[1:end]); % What I have done is that I have just arranged the observations in the order of their decreasing time
work1 = []; % Suppose I have 3 workstations and I have declared three vectors to store the elements to work on them
work2=[];
work3=[];
for i = 1: size(a,1) 
    if a(i,2)+a(i,3)>2000; % IF the sum of constraints A and B is more than 2000 then work at workstation 1
        work1= horzcat(work1,a(i,1));
    else
        if a(i,2)+a(i,3)<2000 && a(i,2)+a(i,3)>1000 % Similarly the two conditions
            work2 = horzcat(work2,a(i,1));
        else
            work3=horzcat(work3,a(i,1));
        end
    end
end

Of Course the number of if statements involved would be large depending upon the constraints and their inter-relationships. In Pattern Recognition also we work on large number of constraints and in many cases it is impossible to escape the if-else statements. Dont be afraid of them. In some cases you may reduce the number of if-else statements but it takes luck and conditions. I hope this code would give you an idea of the problem. I hope I have captured ur problem and given u an apt reply. If you need more help, mail back

Happy to Help

UJJWAL

Best Answer

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