calculation of a mean matrix

Hi I have two matrices

a = [1 2 3; 2 3 4]

and

b = [2 3 4; 3 4 5];

I want a mean output matrix "c," whose output should be

c= [1.5 2.5 3.5; 2.5 3.5 4.5].

so basically "c" should have a mean of respective parameters and same dimension as "a" and "b". Can someone help?

Thanks, Subrat

[Yanbo]

you might just simply add a to b, and them divide the sum by 2. Or, are you looking for a specific command?

Answer 1

[Oleg Komarov]

Unfortunately your example doesn't allow to propose a unique solution, i.e.:

c1 = [mean(a); mean(b)]
c2 = squeeze(mean(cat(3,a,b),3));

c1 simply takes the vertical mean (along rows) of a and then concatenates the vertical mean of b

c2 takes the mean of row 1 from a AND b and then concatenates the mean of the second row fro the two matrices.

Which one do you want?

[Subrat kumar sahoo]

Thanks Oleg, I wanted the operation like c2 is what I was looking for. Thanks again. Subrat

[Subrat kumar sahoo]

I have a bit different requirement now: if "a" and "b" happens to be two elements of the same cell like d{1} and d{2} then is there a possibility of getting "c2" (*c2 = squeeze(mean(cat(3,a,b),3));*) with elements "a" and "b" (i.e. now d{1} and d{2}) picked thru a "for" loop? Thanks, Subrat

Answer 2

[Image Analyst]

a = [1 2 3; 2 3 4];
b = [2 3 4; 3 4 5];

c = (a+b)/2

In the command window:

c =
          1.5          2.5          3.5
          2.5          3.5          4.5

[Oleg Komarov]

This is the so much more intuitive version of my c2!

[Alfredo Scigliani]

what if you have a ridculous amount of matrices (1000) and you want to find the average? I think a for loop, but not sure how.

[Steven Lord]

what if you have a ridculous amount of matrices (1000)

Then I'd recommend you revise the code to avoid that scenario. More likely than not you dynamically created variables with numbered names like x1, x2, x3, etc.

Can you do that? Yes.

Should you do this? The general consensus is no. That Answers post explains why this is generally discouraged and offers several alternative approaches.

Answer 3

[Thomas]

a = [1 2 3; 2 3 4];
b = [2 3 4; 3 4 5];
c=[mean(a);mean(b)]

Answer 4

[Benjamin Klugah-Brown]

what if matrix a and b have different size

[Walter Roberson]

How would you like to define the results for locations that exist in one array but not in the other array?

[Benjamin Klugah-Brown]

Thinking of using zeros?

[Walter Roberson]

s=max(size(A),size(B)) ;
A1=A;
B1=B;
A1(end+1:s(1),1)=0;
B1(end+1,s(1),1)=0;
A1(1,end+1:s(2))=0;
B1(1,end+1:s(2))=0;
(A1+B1)/2

[Benjamin Klugah-Brown]

Thanks very much... by the ways does it work for NA too?

[Walter Roberson]

If by NA you mean NaN, then you would have to use

mean(cat(3, A1, B1), 3, 'omitnan')

or you would have to use something like

maskA = isnan(A1);
maskB = isnan(B1);
C1 = (A1 + B1) / 2;
C1(maskA) = B1(maskA);
C1(maskB) = A1(maskB);