meanX

meanX calculates the mean (a single value) for any X-dimensional matrix.

Description

Just like mean2() calculates the entire mean (a single value) of a 2D matrix, meanX calculates the entire mean for any X-dimensional matrix.

Use meanX() to replace mean2() as it is much more robust (it is not just restricted to 2D matrices). Use it on a 3D, or anyD matrix, and it will work just fine.

This is an extremely simple but incredibly useful macro which I use much more often than I would have initially expected.

• Required Products: MATLAB
• MATLAB release: MATLAB 7.13 (2011b)

[Jan Simon]

This is faster:
n = numel(A);
for i = 1:ndims(A), A = sum(A); end
A = A / n;

[dberm22]

Hmm, didn't really take speed into consideration when writing this, as my code worked fast enough for my case. Thanks for the info, though.

[dberm22]

I will update the code when I get a chance.

[Shaun]

There are faster ways to do this (as Jan pointed out), but it works well enough.

[Xianmin]

why not just mean(A(:))?

[Sheng Yue]

mean(A(:))

[dberm22]

Yes, that works, too. As one of my professors used to say, there are many ways to skin a cat, but they are all messy.

[Jan Simon]

@Xianmin: mean(A(:)) is slower than sum(A(:))/numel(A), but both do not use the multi-threaded capabilities of SUM, if it is called for an array.
Another idea to reduce the computing time is sorting the dimensions in descending order:
d = size(A); [dummy, index]=sort(d, 2, 'descend');
for i = index; A = sum(A, i); end
A = A / prod(d);
This reduces the number of elements as fast as possible. E.g. for RAND(4, 100000) this is twice as fast as the original. It is even faster than MEAN(A(:)) and SUM(A(:))/numel(A).

[Xianmin]

@Jan Simon
Never test it before. But it's good to know that. Thanks!