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];
?
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 :)
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
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