Octave/Matlab: Adding new elements to a vector

Question

Having a vector x and I have to add an element (newElem) .

Is there any difference between -

x(end+1) = newElem; 

and

x = [x newElem];

?

Solution 1

x(end+1) = newElem is a bit more robust.

x = [x newElem] will only work if x is a row-vector, if it is a column vector x = [x; newElem] should be used. x(end+1) = newElem, however, works for both row- and column-vectors.

In general though, growing vectors should be avoided. If you do this a lot, it might bring your code down to a crawl. Think about it: growing an array involves allocating new space, copying everything over, adding the new element, and cleaning up the old mess...Quite a waste of time if you knew the correct size beforehand :)

Solution 2

Just to add to @ThijsW's answer, there is a significant speed advantage to the first method over the concatenation method:

big = 1e5;
tic;
x = rand(big,1);
toc

x = zeros(big,1);
tic;
for ii = 1:big
    x(ii) = rand;
end
toc

x = []; 
tic; 
for ii = 1:big
    x(end+1) = rand; 
end; 
toc 

x = []; 
tic; 
for ii = 1:big
    x = [x rand]; 
end; 
toc

   Elapsed time is 0.004611 seconds.
   Elapsed time is 0.016448 seconds.
   Elapsed time is 0.034107 seconds.
   Elapsed time is 12.341434 seconds.

I got these times running in 2012b however when I ran the same code on the same computer in matlab 2010a I get

Elapsed time is 0.003044 seconds.
Elapsed time is 0.009947 seconds.
Elapsed time is 12.013875 seconds.
Elapsed time is 12.165593 seconds.

So I guess the speed advantage only applies to more recent versions of Matlab

Solution 3

As mentioned before, the use of x(end+1) = newElem has the advantage that it allows you to concatenate your vector with a scalar, regardless of whether your vector is transposed or not. Therefore it is more robust for adding scalars.

However, what should not be forgotten is that x = [x newElem] will also work when you try to add multiple elements at once. Furthermore, this generalizes a bit more naturally to the case where you want to concatenate matrices. M = [M M1 M2 M3]

---

All in all, if you want a solution that allows you to concatenate your existing vector x with newElem that may or may not be a scalar, this should do the trick:

 x(end+(1:numel(newElem)))=newElem

Comments

[Dan]

Also for the second method, x must be initialized first!

[ThijsW]

@RodyOldenhuis, no problem! I did the same, probably at the same time. @Dan, that's true, but as the question mentioned "having a vector x (of size n)", I kind of assumed n to be non-zero and the vector being initialized already :)

[Rody Oldenhuis]

+1, Edited to add the obvious as well. I'll test again on a "real" CPU (I'm on this crappy unreliable no-good APU thing now...)

[ThijsW]

@Dan, same for me, I get 0.028 for the 3rd option and 8.909 for the last

[Rody Oldenhuis]

I also think the JIT optimisation for the x(end+1) case is a pretty recent addition (R2012a or so...). I vagualey remember reading something like that in some changenotes at smoe point. I also get very different results on my APU/Matlab R2010, but I'm not sure if that's due to the Matlab version or the APU...

[Dan]

@RodyOldenhuis and ThiijsW see my recent comparison between older and newer matlab

[Rody Oldenhuis]

@Dan: Ah :) OK, NOW it's all better :)

[Rody Oldenhuis]

@ThijsW: Still, using end rather than some variable n (which might be a global for all you know!) is the more universal, robust-no-cost way to go

[Digna]

I think your last example should be: x(end+1:end+length(newElem)) = newElem

[Dennis Jaheruddin]

@Digna Thanks for finding the bug, I have updated the answer to fix the problem.

[JaBe]

With my Matlab2011b, there was also a drastic (~50x) speed improvement on vector concatenation with this method vs. the a=[a b] method.

[Robino]

@Jabe the [a b] method is concatenation. I suppose you are saying that extending the array is quicker than concatenation?

[Robino]

@Dan x must be initialised for both methods.

[Dan]

@Robino I don't think so, I don't have MATLAB to check but in octave you can certainly do x(end+1)=... without x being initialised and I'm almost certain you can do it in MATLAB as well

[Robino]

@Dan x(end+1)=123 gives Undefined function or variable 'x'.