Indexing in for loop in order to remove non-zero and nan entries from vectors

Hello Friends,

I have the following code:

    P = [1, 0, 3];
    Q = [0, 4, 5];

% These AB, CD, EF are just for illustration purpose. They could be any to make AB, CD , EF equal size vectors.

    AB = [P + Q];  % A vector
    CD = [P - Q];  % A vector
    EF = [P ./ Q]; % A vector

% Now I want to pass above compute vectors AB, CD, EF in for loop. I try as follows:

    M = {'AB', 'CD', 'EF'};

    for i = 1:length(M)

        P1 = P(M{i}~=0);          %It removes those elements of P which correspond to 0 entries in M.
        t = M{i};                 %Create a temporary variable.
        M = t(M{i}~=0);           %It removes 0 entries from M.

        P = P1(~isnan(M(i)));     %It removes those elements of P which correspond to NaN entries in M.
        M = t(~isnan(M(i)));      %It removes NaN entries from M.

        if strcmp(M, 'AB')
            f = f(P,M);
        elseif strcmp(M, 'CD')
            f = g(Q,M);
        elseif strcmp(M, 'EF')
            f = h(R,M); 
        end
  end

This code is not computing P1, t, M, P values properly. For example the following line of code takes M{i} = AB for i = 1, but gives totally wrong answer.

I have tried to change {} to () and [], etc., but nothing works. I will appreciate any advice.

P1 = P(M{i}~=0);

Answer 1

[Todd Leonhardt]

The problem appears to be that your loop changes M into something with a single element on the first pass through, but then tries to access the 2nd element on the 2nd pass through.

On the first pass through (i=1), this following line sets M equal to 'AB'

    M = t(M{i}~=0)

Then this line mutates it further to just 'A':

    M = t(~isnan(M(i)));

So on the 2nd pass through your loop (i=2), this line tries to access the 2nd element in the M cell array, but there is only 1 element:

    P1 = P(M{i}~=0);

[hello_world]

I know, and I am trying to fix it, but still no success. In

P1 = P(M{i}~=0);

M{i} is AB, a cell, while P is double. If I change AB to cell to double, it does not work. It should actually take values stored in AB, and not just characters AB.