Sum over part of array without knowing array size

7 vues (au cours des 30 derniers jours)
JE
JE le 9 Sep 2015
Commenté : John D'Errico le 9 Sep 2015
I currently have a function that takes a two dimensional (2D) array as one of its inputs. Part of the function code sums over all but the first column of this input array, i.e.:
function DoSomething( Array2D, OtherInputs)
...
SumOverColumns = sum(Array2D(:, 2:end), 2);
...
end
I want to expand the use of the function so that the array can be 3D or more. However, I still want to be able so sum over all columns except the first one. I can't think of a general way to do this because I think that I need to specify ":" for every other dimension.
Any help would be much appreciated.
  2 commentaires
Cedric
Cedric le 9 Sep 2015
Modifié(e) : Cedric le 9 Sep 2015
You need to define a little more precisely what happens with higher dimensions, and what it is that you want to output. In 2D you are summing over the 2nd dimension excluding the first column and outputting a column vector of sums. What happens in 3D? Do you still need to sum over the 2nd dimension only excluding the first column of each page (and hence output a 2D array of sums), or do you need to sum over dimensions 2 and 3 excluding the first, simultaneously (output a vector) or separately (output two 2D arrays)?
JE
JE le 9 Sep 2015
Hi Cedric
In the following, I've used a1 to denote the size of dimension 1 (of the input array), a2 for the size of the 2nd dimension and so on.
Yes, in the 2D example, the result is a column vector (e.g., a1 x 1).
For 3D, I want the output to be a1 x 1 x a3. For 4D, I want the output to be a1 x 1 x a3 x a4. and so on.

Connectez-vous pour commenter.

Réponse acceptée

John D'Errico
John D'Errico le 9 Sep 2015
Modifié(e) : John D'Errico le 9 Sep 2015
Or more? How many more dimensions? Really high dimensional arrays will be huge, so it makes little sense to have a 300 dimensional array. Even a 2- dimensional array actually has infinitely many trailing singleton dimensions. So a 5x6 array is actually implicitly of size
[5,6,1,1,1,1,1,1,1,1, ...]
Suppose you are willing to assume your array will have no more than 5 dimensions in the future. Then the trivial solution is:
SumOverColumns = sum(Array2D(:, 2:end, :, :, :), 2);
There are ways to do this with more potential dimensions, but why bother?
  4 commentaires
JE
JE le 9 Sep 2015
Hi John
That works with one slight change to the 2nd line:
S = size(ArrayNd);
S(2) = 1;
SumOverColumns = reshape(sum(ArrayNd(:, 2:end,:), 2),S);
Thanks very much!
John D'Errico
John D'Errico le 9 Sep 2015
Oops, yes. You were summing on that dimension. Yes.

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Logical dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by