average of range of values in a matrix

Hello,

I have a matrix d (NxN), which looks like that:

Can I use a for loop to say, calculate the average value of all lines in column 1, which have the same value in column 3?

n = amount of same values (here 3 x 11)

and give me a vektor which looks like that:

Can I tell the for loop to find automatically all multiple values in column 3? Like now it's 11, 16, 21, but maybe the next matrix has in column 3 10, 12, 14,...

Answer 1

[dpb]

[u,~,ib]=unique(x(:,3));  % get the unique values and locations for each
val=x(:,1:2);             % shorthand for the subset over which to average
mx(:,3)=u;                % preallocate for the result array & assign index values
for i=1:2                 % for the two columns
  mx(:,i)=accumarray(ib,val(:,i),[],@mean);  % compute the mean over the subsets
end

Seems like should be able to do it w/o the loop over the columns but I drew a blank at the moment as to how to since accumarray val argument must be a vector

[Image Analyst]

Yeah, I thought of accumarray too, but accumarray is such a mind-bender I have to research it in the help every time because it's kind of complicated, so sometimes I just do the brute force but intuitive approach. However, I'm sure Andrei Bobrov, master of the one-liners, will come along shortly and do the whole thing in one compact but totally cryptic and impenetrable line of code.

[dpb]

Yeah, Andrei's the guru there, fur shur... :) I agree w/ accumarray; it's capable but complex-enough to always take a couple of tries and if it's been a while since last foray.

I realize I forgot to add the OP's last wish above; namely to include the value of the indexing variable as the last column...

Could change a little; start with

mx(:,3)=u;

for the preallocation and setting the values together...

Answer 2

[Fabi Boro]

thank you very much fot the detailed and fast reply.

I'm afraid I'm a beginner in Matlab and I'm trying to understand your code but I don't get it entirely. "mx" stands for my Matrix, but I get an error message for the line

 [u,~,ib]=unique(x(:,3));

"Reference to a cleared variable x." what is the difference bewteen your first suggestion and your second to start with

   mx(:,3)=u;

?

[dpb]

x is a placeholder for your original variable--you don't say what you've named the...well, in looking again I see that I did overlook that in, sorry--use 'd' in place of 'x'.

mx stands for "mean of x" for the result; use whatever you want instead, of course.

The difference between the two solutions is that the first leaves you with a 2-column result of the two means; the latter finishes the job by storing the unique indices vector of the groups as the third column per the original request. Matlab automagically allocates full memory to the largest subscript dimension in an assignment of this kind; try, say,

newvar(3,3)=pi

at the command line to see how the assignment to a single location serves the same function as

newvar=zeros(3);
newvar(3,3)=pi;

it simply saves a step.

Answer 3

[Andrei Bobrov]

Hi friends! In four rows

A = [1234    785     11
   5678    457     11
   9101    390     11
   2467    381     16
   3573    737     16
   2354    938     16
   8467    729     21
   3477    900     21];

[a,~,c] = unique(A(:,3));
[x,y] = ndgrid(c,1:2);
A1 = A(:,1:2);
out = [accumarray([x(:),y(:)],A1(:),[],@mean),a];