Subscript indices must either be real positive integers or logicals

Hei All,

I have been getting the error message "Subscript indices must either be real positive integers or logicals. " referring to the calculation of z(i, j) whilst working on the code below. I have spent hours trying to fix the issue, but couldn't figure out what I was missing.

a = 0.2302;
N=4;
theta = linspace(0, 2 * pi * pi, N);
r = linspace(a, 2 * a, N);

sigma_TT = r.^4 .* cos(2 .* theta);

x = zeros(N, N);
y = zeros(N, N);
z = zeros(N, N);

for i = 1:N
    for j = 1:N
        x(i, j) = (cos(theta(i)))' * r(j);
        y(i, j) = (sin(theta(i)))' * r(j);
        z(i, j) = real(sigma_TT([theta(i), r(j)]));
    end
end

The for loop part is actually what I trying to convert from python to MATLAB. In python it is expressed as:

x = r * np.cos(theta[:,None])
y = r * np.sin(theta[:,None])
z = sigma_TT(SHmax, Shmin, Pp, Pm, a, r, theta[:,None])

Any help on this issue would be much appreciated.

[Ammar Dodin]

The reason why you see that error is because theta(i) is evaluating to zero. Then when you do sigmaTT([theta(i),r(j)]), that expression throws the error that you're seeing.

[RG]

Thanks Ammar. I have in fact tried with non-zero values of theta and still get the same error because this time r(i) is not an integer. Is there a way to write the code for z without the loop? The extraction from python that I mentioned above is taken from a code that is functional with same inputs I am trying to use.

I am able to compute x and y without the loop, but get stuck with the syntax for z. The syntax for x and y that give me the results I expect are:

x = (cos(theta))' * r;
y = (sin(theta))' * r;

[Ammar Dodin]

I am not too familiar with numpy, but I think I understand what the first two lines of your python code are doing. It looks like you're creating new vectors that contain the cosine/sin of each of the values of the vector theta after multiplying them with some constant. I don't understand what z is doing? I am assuming sigma_TT is some function you wrote.

Could you possibly provide the Python code for that function?

[RG]

You are right, sigma_TT is a separate function. I managed to fix the issue, apparently I forgot to write sigma_TT function. I will post the solution below just for a reference.

Answer 1

[RG]

I had to first create a function called sigma_TT:

function sigma_TT = sigma_TT(r, theta)

sigma_TT =  r.^4 .* cos(2 * theta);

The code worked well when sigma_TT is called within compute_sTT function:

function compute_sTT

a = 3;
N = 4;

theta = linspace(0,  3 * pi * pi, N);
r = linspace(a, 3 * a, N);

x = zeros(N, N);
y = zeros(N, N);
z = zeros(N, N);

for i = 1:N
    for j = 1:N
        x(i, j) = (cos(theta(i)))' * r(j);
        y(i, j) = (sin(theta(i)))' * r(j);
        z(i, j) = sigma_TT(r(j), theta(i));
    end
end

Answer 2

[Image Analyst]

It's pretty well explained in the FAQ: http://matlab.wikia.com/wiki/FAQ#How_do_I_fix_the_error_.22Subscript_indices_must_either_be_real_positive_integers_or_logicals..22.3F

[Image Analyst]

Did you read the FAQ? Or even see my answer? OK, here's the reason

[theta(i), r(j)]

takes on the values 0 and 0.2302. It's a 1 by 2 row vector.

You have defines sigma_TT as a 1 by 4 array with these values:

sigma_TT =
        0.0028081563033616        0.0073593799306762       0.00812938329944458       -0.0093006967826725

To access those values you can pass in an index of 1, 2, 3, or 4. For example, sigma_TT(2) would get you 0.0073593799306762.

However neither 0 nor 0.2302 is of a value 1,2,3, or 4. You can't have the zeroeth index of an array. You can't have the 0.2302th index of an array. You can have the first index or second index, but not something in between. Indexes has to be positive integers like 1,2,3,4, etc. or else logical/binary arrays of true and false elements.

[RG]

The link you provided was already explained above by Ammar Dodin.

[Image Analyst]

I thought my comment would explain it more thoroughly than his original 2-line comment. Anyway, hopefully you looked over the FAQ and maybe found some other good stuff in there.

[RG]

Yes, it was useful indeed. Many thanks for it.