Why do I get an index error?

Here in this part of my code

r3 = randi([1 N],[G 1]);
for i = 1:G
ch = ceil(r3(i,1)/3840);
[r, c] = ind2sub([4 960],r3-((ch-1)*3840));
if Ma{ch,1}(r,c) == 0
    Ma{ch,1}(r,c) = 1;
else
    Ma{ch,1}(r,c) = 0;
end
end

'r' and 'c' can't be zero, but I get:

??? Subscript indices must either be real positive integers or
logicals

Would you please help me with this?

[KSSV]

[r, c] = ind2sub([4 960],r3-((ch-1)*3840));

The values inside ind2sub should be positive integers. In your case they might be taking zeros and/or negative values. Check that.

[Sherwin]

But I think the smallest amount that this term can get:

r3-((ch-1)*3840

is 1! not negative not 0...

Answer 1

[Steven Lord]

On which line does this error occur?

If it's on the line where you define r3, you've probably created a variable named randi that is shadowing the built-in randi function.

For the line where you define ch, you've probably shadowed ceil.

In either of these cases, rename or remove the variable.

[Sherwin]

Thank you!

Answer 2

[Guillaume]

It's not clear what you're trying to do with your code, but clearly it's not going to work

r3 is a vector with values between 1 and an unknown N. So, let's assume that r3(1) is N and r3(2) is 1. Because of the way you create ch it is an integer at least 1, and more if N > 3840. For the error you see, N must be at least 3841, so let's assume that,

So, for i = 1, r3(i) = 3840, and ch = ceil(3841/3840) = 2. You then have

r3 - ((ch-1)*3840) = [3841 1] - (1 * 3840) = [3841 1] - 3840 = [1 -3839]

As you can see you've got a very negative index here. It gets even worse if N > 2*3840.

Note that there is no point in the ind2sub call in your code. You could just as well use the linear index to index Ma{ch} with absolutely no change of behaviour.

[Sherwin]

Thank you for your answer, but r3 is a random column vector with the length of G and each time one element is picked to enter this term:

r3-((ch-1)*384

and the random element is between 1 and N so the smallest amount r3 can take is 1, and if this happens,

ch = ceil(r3(i,1)/3840)= ceil(1/3849) = 1
[r, c] = ind2sub([4 960],1-((1-1)*3840)) = 
[r,c] = ind2sub([4 960], 1) = 1

this should happen.There shouldn't be negative results.

[Guillaume]

I thought my demonstration was clear. Create the following variables:

 N = 3841;  %minimum value for the error to occur
 r3 = [N; 1]; %a possible output of your randi. Don't need any more elements to break your code. G is irrelevant.
 i = 1;   %first pass of the loop

Now, look at the values of

ch = ceil(r3(i) / 3840)  %is ceil(3841/3840) = 2
r3 - ((ch - 1) * 3840

You'll see that the output of the last expression is

[1; -3839]

Anyway, it would be much better if you told us what you're trying to achieve as I'm fairly certain you're going at it completely wrong.

[Sherwin]

Thank you so much.. I got it :)

Answer 3

[David Goodmanson]

Hi Sherwin, May be there shouldn't be negative indices, but there actually are negative indices, as Guillaume has pointed out. Here is an example, running your code:

G = 2;
r3 = [1; 3841];
for i = 1:G
  ch = ceil(r3(i,1)/3840);
  [r, c] = ind2sub([4 960],r3-((ch-1)*3840))
end

which for i=2 comes up with

c = -959
       1

Perhaps you want to have r3(i,1) on the fourth line of your original code so as to make a single index instead of a vector's worth.

[Sherwin]

Thank you so much!