How to loop through 1D array elements multiple times?

Question

Assuming size_of_array < n, to iterate through array elements multiple times, normally I would use something like (in C++):

for (size_t i = 0; i < n; ++i)
{
    elem = arr[ i % size_of_arr ];
}

If arr = {1, 2, 3};, for elem I would get:

1 2 3 1 2 3 ...

However, in MATLAB / Octave indexes start from 1 and:

for i = 1 : n

  elem = arr( mod( i, length(arr) + 1) );
end

so when n == length(arr) I get an error:

error: arr(0): subscripts must be either integers 1 to (2^31)-1 or logicals

How is this done when indexes start from 1?

The idiomatic form for that is mod(i-1,n)+1 (I call it 1-based modulus). So, perhaps elem = arr(mod(i-1, length(arr))+1); works for you (I'm not sure what you are trying to achieve) — Luis Mendo
– Luis Mendo, Commented Sep 5, 2017 at 15:19
never use length, it returns the length of the longest dimension. Instead, use numel (number of elements). — carandraug
– carandraug, Commented Sep 5, 2017 at 15:47
error: arr(0): The zero is what Matlab is trying to yell at you about. — Yvon
– Yvon, Commented Sep 5, 2017 at 19:26
@LuisMendo Thanks for the helpful comment, it worked perfectly. If you are interested for the use, check : codereview.stackexchange.com/questions/174876/… — Ziezi
– Ziezi, Commented Sep 5, 2017 at 19:58

Dev-iL · Accepted Answer · 2018-03-01 11:34:28Z

5

In c++ you would do

arr[ i % size_of_arr]

where the inner index i % size_of_arr is in range [0 size_of_arr-1]

In MATLAB, i is in range [1 size_of_arr], thus just change to

mod( i-1, numel(arr) ) + 1

the first bit (mod( i-1, numel(arr))) makes sure that the index is inside [0 size_of_arr-1], and the you just need to add 1 to that.

Dev-iL

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answered Sep 5, 2017 at 15:20

Ander Biguri

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1 Comment

Thank you! So, what we do is offsetting the index to zero-origin and adjusting the result of modulo operation to start from 1 and on, right?