Looping through index number + 1

Hello, say I have a cell array called 'group' and I want to loop to do a specific calculation which is the following:

In this case, i goes from 1 to 15 and I am calculating the norm between each cell subarray, 1st row.

So when i = 15, where my code has i+1 this would be 16 and it would exceed the index number of array elements.

What would be the best way to avoid this issue? Using an if statement when i == 15?

Thanks.

Answer 1

[Steven Lord]

This code probably isn't doing what you think it is. I've commented it out below so I could run some code later in this answer.

% for i = (1:size(group))'

The size function called with 1 input returns a vector containing the size of the input in each dimension. When you call the colon operator : with one of the inputs being a vector (rather than a scalar) the operator ignores all elements after the first.

In addition, the colon operator creates a row vector. By transposing it you create a column vector. How does the for keyword handle iterating over a column vector? It uses this syntax from that documentation page: "valArray — Create a column vector, index, from subsequent columns of array valArray on each iteration. " So this loop is only ever going to run 1 iteration (unless group has 0 rows, in which case it won't run at all.)

If you want to iterate over all elements of an array I'd use the numel function instead. Compare v1 and v2 below.

c = {1, 2:3, 4:6, 7:10}

c = 1×4 cell array
    {[1]}    {[2 3]}    {[4 5 6]}    {[7 8 9 10]}

size(c)

ans = 1×2

1 4

v1 = 1:size(c)

v1 = 1

numel(c)

ans = 4

v2 = 1:numel(c)

v2 = 1×4

1 2 3 4

Of course you can perform arithmetic on numel in your loop.

v3 = 1:numel(c)-1

v3 = 1×3

1 2 3

[JIAN-HONG YE ZHU]

Thank you very much for your insight, it was a great explanation.

Answer 2

[Antoni Garcia-Herreros]

Not exactly sure what you are trying to do but, why do you want to loop through all 15 cells? isn't it enough to loop through 14?

for i=1:size(group)-1 % This only loops 14 times