how to overcome error: Index in position 1 is invalid. Array indices must be positive integers or logical values.

Hello everyone

Seems I can't fix my problem

I have code like this:

alpha=0.1:0.1:1;
a=[1 2 3 4 5 6];
u=[1 3 3 3 2 1];
op=[3 3 3 3 3 3];
act=numel(a);
d=cumsum(u);
n=length(alpha)*d(act);
o=3; 
B=zeros(n,(3+(o*max(op)))); % I want to store all my result in this matrix

then I have calculation like:

for z=alpha
    for i=1:act
        for j=1:u(i)
            options=zeros(1,o*(op(i))); % I want to store the result of my calculation for each iteration in this matrix
            for k=1:op(i)
                r=a(i)-(u(i)-j+1)*op(i)+k;
                t=1;
                c=2;
                q=3; % t,c,q is my calculation. actually it was so long, so I just simplify it here
                options(1,((o*(k-1)+1):(o*k)))=[t c q];
            end
            e=z*10
            e=e*d(act)-d(act)+d(i)-u(i)+j;
            B(e,1)=z;
            B(e,2)=i;
            B(e,3)=j;
            B(e,4:end)=options;
        end
    end
end
B

I want to get matrix B.

but for this code, I got error for

Index in position 1 is invalid. Array indices must be positive integers or logical values.

Error in (line 27)

B(e,1)=z;

it stops when the alpha is 0.3

Please help.

Thank you

Answer 1

[Pier Giorgio Petrolini]

Hello, the problem is basically that the variable "e" after calculations has some decimals (zeros), and you can't use it as it is for indexing because index should be integers.

Try to rewrite the code by adding the command "int64" as follow:

....

e=z*10

e=int64(e*d(act)-d(act)+d(i)-u(i)+j);

B(e,1)=z;

....

I hope it helps, let me know!

Kind regards,

PGP

[Putri Basenda Tarigan]

It works now. Thankyou so much for the help, Sir!

[Pier Giorgio Petrolini]

You're welcome! Thank you for your positive feedback!

Answer 2

[Walter Roberson]

alpha=0.1:0.1:1;

When you multiply by 10 you do not always get integers. 0.1 does not have an exact representation in finite binary, and the error adds up.

You can round(e) but there are better ways.

[Putri Basenda Tarigan]

Thankyou so much for the answer Sir.

But, how can I round e Sir?

I multiply it by 10 Because I want to get the row number for each iteration Sir.

[Walter Roberson]

for zidx = 1 : numel(alpha)
    z = alpha(zidx);
    for i=1:act
        for j=1:u(i)
            options=zeros(1,o*(op(i))); % I want to store the result of my calculation for each iteration in this matrix
            for k=1:op(i)
                r=a(i)-(u(i)-j+1)*op(i)+k;
                t=1;
                c=2;
                q=3; % t,c,q is my calculation. actually it was so long, so I just simplify it here
                options(1,((o*(k-1)+1):(o*k)))=[t c q];
            end
            e = zidx;
            e=e*d(act)-d(act)+d(i)-u(i)+j;
            B(e,1)=z;
            B(e,2)=i;
            B(e,3)=j;
            B(e,4:end)=options;
        end
    end
end
B