MATLAB array indexing error?

Question

I have been doing array indexing as follows

>> n=[1 2 3]
n =
     1     2     3

>> idx=1
idx =
     1

>> n([idx+1 idx])
ans =
     2     1

without any problem. However, today I encounter this error in my following snippet. Please forgive me. I understand I should provide MWE, but I simply cannot reproduce the error!

>> %%% interpret tree
bitmap = zeros(pix_no_per_side, pix_no_per_side);
x_pixels = 1:1:pix_no_per_side;
y_pixels = 1:1:pix_no_per_side;
rule_set = {}; % cell to accommodate rows of diff. sizes
% parse branches
trav_iter = tr.depthfirstiterator;
while tr.get(trav_iter(end)) <= 0 % still have branch unparsed
    branch = [1]; % contains the root
    node_idx = 1;
    while tr.get(branch(end)) > 0 % not bottom yet
        node_idx = node_idx+1;
        branch = [branch trav_iter(node_idx)];
    end
    rule_set{end+1} = branch; % one rule per cell
    disp(node_idx) % DISPLAYING FOR DEBUGGING
    if numel(rule_set) == 1 % only one branch found so far
        trav_iter(node_idx) = []; % remove this leaf
    else % more than one branches found so far
        trav_iter([node_idx-1 node_idx]) = []; % remove this leaf and parent node
    end
end
     3

     3

     2

Subscript indices must either be real positive integers or logicals.
>>

macduff · Accepted Answer · 2014-06-17 16:14:40Z

2

The index node_idx-1 is likely either 0 or -1. The subscript error almost always indicates this problem. However, any empty, NaN or other value can cause this as well.

Perhaps this is because the assignment node_idx = 1; sets the value and in your last loop, it is never incremented yielding a 0 index.

answered Jun 17, 2014 at 16:14

macduff

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1 Comment

Sibbs Gambling Over a year ago

Thanks, but the 3 3 2 is what I get when I do disp(node_idx).

nkjt · Accepted Answer · 2014-06-17 17:18:15Z

0

The problem with your debugging line is that it only displays the output of node_idx at the end of the while loop. If it errors after setting node_idx to 1 and before hitting disp then you will not see the value of the index that caused the error.

Here's a simple way of understanding it:

x = 1:3

for n = 3:-1:0
    y = x(n);
    disp(n)
end

As you can probably guess, the above code will error when n gets to 0, However, when it runs, I get:

     3

     2

     1

Attempted to access x(0); index must be a positive integer or logical.

i.e. disp never shows 0.

But from the command line:

>> n

n =

     0

If this is within a function, use dbstop if error and then check node_idx when the code stops.

answered Jun 17, 2014 at 17:18

nkjt

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Sibbs Gambling · Accepted Answer · 2014-06-17 18:40:03Z

0

It turns out that error message is triggered by

trav_iter(end)

where trav_iter is possibly to be empty.

Reproducing the error then becomes.

>> a = []

a =

     []

>> a(end)
Subscript indices must either be real positive integers or logicals.

Hence, to fix that, I add one logical and to make sure that the trav_iter is not empty. i.e.,

while numel(trav_iter) ~= 0 && tr.get(trav_iter(end)) <= 0

answered Jun 17, 2014 at 18:40

Sibbs Gambling

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