Repeating a task for each row without loop

I'd like to do this expression without a loop,

x = [1, 1; 2, 2; 3, 3; 4, 7; 8, 8; 9, 9; 10, 15; 16, 16; 17, 17];

for i = 1:size(x,1)

y(i,1) = {[x(i,1):x(i,2)]}

end

Thanks in advance,

[Bob Thompson]

So, you just want to set each row of y to be a cell containing the corresponding row of x? How are you using y later in your code? Do you need to even specify y as a separate variable, or can you just index x later on?

[Jonathan Pelletier-Marcotte]

Y is a report of the numbers missing between the first column of X and the second column of X, if there is no missing numbers, only one number will be there, if not all the numbers between the first column of X and the second column of X will be there.

[Bob Thompson]

Ok. I personally don't know of a way to conduct a specific operation for each row of a matrix without defining a loop to run through the rows. It might be possible with some fancy indexing, but I don't know how.

[Jonathan Pelletier-Marcotte]

No problem,

Thank you Bob

Answer 1

[Jan]

Instead of avoiding the loop, start with a clean version with pre-allocation:

x = [1, 1; 2, 2; 3, 3; 4, 7; 8, 8; 9, 9; 10, 15; 16, 16; 17, 17];
n = size(x, 1);
y = cell(n, 1);
for k = 1:n
  y{k} = x(k,1):x(k,2);
end

This has several improvements:

• With a pre-allocation before the loop, the array does not grow iteratively. Remember that creating an array iteratively, Matlab has to create a new array and copy the old contents in each iteration. For a 1xN array this needs to reserve memory for sum(1:N) elements and this is growing extremely fast.
• y(i,1) = {...} creates a cell on the right side only to copy its elements to the left side. The creation of the temporary cell can be saved by using the curly braces on the left side: y{i,1} = ...
• a:b is a vector already. Including it in square brackets concatenates it with nothing. Therefore [a:b] is a waste of time. See Avoid square brackets .
• The link about brackets might be useful: Are you sure that storing the index vectors explicitly is useful? Remember that:

y = x(a:b);

is faster than:

v = a:b;
y = x(v);

So maybe your Nx2 matrix is the best way to store the indices already. Then searching for a vectorized way to convert it might be a bad idea in general.

[Jonathan Pelletier-Marcotte]

Very clear answer, thank you!