Suppose you have four vectors (A, B, C, and D) with different lengths, you can aggregate them into one vector G as:

G= [A; B; C; D];

Then you can find the average of G as:

mean(G);

Or  just sum the components of the vectors and divide by total number of elements.

For example if you have three vectors X={4,5,6} , Y = {10, 11, 9} and  Z ={4, 3,5} .

Here, Sum= 57 and total components (n)= 9, Sum/n=6.3

you can also calculate as: For X vector, Sum/3=5, for Y, sum/3= 10, for z, sum/3= 4. Now add these numbers and divide by total number of vectors i.e.3, 19/3= 6.3.

thanks all for answering... I want to get a new vectors for the average , so I have to make the vectors same length first,for example: 

A={1,2,3,4,5}

B={1,2,3}

C={1}

then 

I want to get a G vector which represent the average as a vector not just one number, Please any assistance will be so appreciated...  

Hiba,

In that case scenario the code is:

G=[mean(A); mean(B); mean(C)];

Your 3 vectors are in different vector space to it dosenot makes any scene to take their average , but if project all them to same vector space then you can do so.

eq,

A=[1,2,3,4,5] '=> Ap=[1,2,3,4,5]'

B=[1,2,3]' => Bp=[1,2,3,0,0]'

C=[1]' => Cp=[1,0,0,0,0]'

mean([Ap,Bp,Cp],2)={1,4/3,2,4/3,5/3}.

use cells is the easiest way for different vector length

http://www.mathworks.com/help/matlab/ref/cell.html

Hiba,

matlab : Given

A={1,2,3,4,5}

B={1,2,3}

C={1}

x={mean(A),mean(B),mean(C)}

x =

[3] [2] [1]

where x is a cell of matrix elements

To get the matrix back  from the cell you have : mat2cell ()

http://www.mathworks.com/help/matlab/ref/cell2mat.html

PS: Any dimensions , I work with huge data elements...

you have two ways to do it

1. use cell

>> A = [1 2 3];

B = [2 3 4 2];

C = [1 2 ];

D = 1 : 5;

ce = {A,B,C,D};

av = cellfun(@mean,ce)

av =

2.0000 2.7500 1.5000 3.0000

2. use matrix

>> A = [1 2 3];

B = [2 3 4 2];

C = [1 2 ];

D = 1 : 5;

E = NaN(4,5);

E(1,1:3) = A;

E(2,1:4) = B;

E(3,1:2) = C;

E(4,1:5) = D;

nanmean(E,2)

ans =

2.0000

2.7500

1.5000

3.0000

I hope this can help you