Convert matlab index vector to 2D array based on gaps

Hi all,

I am asking to get feedback on my code and see if there are other ways to programme this basic algorithm more efficiently or elegantly.

out = segments([1 end]); % Start point of first interval

for idx = find(diff(segments) > 1) % Iterate over gaps

out(end, 2) = segments(idx); % End point of current interval

out(end + 1, 1) = segments(idx + 1); % Start point of next interval

end

out(end, 2) = segments(end); % End point of last interval

out(:, diff(out, [], 2) == 0) = []; % Remove point intervals

E.g., for segments = [12 13 14 16 17 18], the output out = [12 14; 16 18] noting the gap between 14 and 16.

Answer 1

[Dyuman Joshi]

You can vectorize the code -

seg = [12 13 14 16 17 18 20 21 22 23 24];
idx = diff(seg)~=1;
out = [seg(1) seg([false idx]); seg([idx false]) seg(end)].'

out = 3×2

12 14 16 18 20 24

[ErikJ GiesenLoo]

This is genius!

It took me a few minutes to understand but i think it effectively conveys the first row having all the "start" indices and the second row having all the end indices.

Is there a reason why the transpose is written as .' instead of just '?

[Dyuman Joshi]

Yes, that's it!

As for the tranpose, it's a habit. Since we are dealing with real numbers, .' is equal to ' . So you can use either of them here.

[Dyuman Joshi]

Updated code -

seg = [12 13 14 16 17 18 20 21 22 23 24];
idx = diff(seg)~=1;
out = [seg([true idx]); seg([idx true])].'

out = 3×2

12 14 16 18 20 24

[Stephen23]

"Is there a reason why the transpose is written as .' instead of just '?"

Because .' is transpose (even if beginners often use the complex conjugate transpose instead):

https://www.mathworks.com/help/matlab/ref/transpose.html

Using the correct operator has several benefits:

• it makes the intent clear
• it makes your code more generalized (assuming the rest of your algorithm can also handle complex numbers)
• it increases trust in your code
• it is the correct operator