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From: Ulrik Nash on 29 Jun 2010 09:41
Hi everyone,

Suppose I have

A = [1 2 3 0 0 0; 0 3 0 5 0 0]

What command lets me get

B = [2;4]

which is the mean of each row in A, disregarding zeros?

Regards,

Ulrik.
From: Adam Chapman on 29 Jun 2010 09:49
On Jun 29, 2:41 pm, "Ulrik Nash" <u...(a)sam.sdu.dk> wrote:
> Hi everyone,
>
> Suppose I have
>
> A = [1 2 3 0 0 0; 0 3 0 5 0 0]
>
> What command lets me get
>
> B = [2;4]
>
> which is the mean of each row in A, disregarding zeros?
>
> Regards,
>
> Ulrik.

I couldn't find a way to do all rows without a for loop, but here is a
solution:



for i=1:size(A,1)

%get row data
B=A(i,:);

%remove zero values:
B=B(B~=0);

get mean of row:
m(i,1)=mean(B);

end

From: us on 29 Jun 2010 09:56
"Ulrik Nash" <uwn(a)sam.sdu.dk> wrote in message <i0ct5g$ekm$1(a)fred.mathworks.com>...
> Hi everyone,
>
> Suppose I have
>
> A = [1 2 3 0 0 0; 0 3 0 5 0 0]
>
> What command lets me get
>
> B = [2;4]
>
> which is the mean of each row in A, disregarding zeros?
>
> Regards,
>
> Ulrik.

one of the solutions
- if(f) you own the stats tbx...

% the data
v=[
1 2 3 0 0 0
0 3 0 5 0 0
];
% the engine
tf=v~=0;
b=sum(tf,2);
a=sum(v,2);
r=a./b;
% the result
r
% r = 2 4

us
From: Oleg Komarov on 29 Jun 2010 10:03
"Ulrik Nash" <uwn(a)sam.sdu.dk> wrote in message <i0ct5g$ekm$1(a)fred.mathworks.com>...
> Hi everyone,
>
> Suppose I have
>
> A = [1 2 3 0 0 0; 0 3 0 5 0 0]
>
> What command lets me get
>
> B = [2;4]
>
> which is the mean of each row in A, disregarding zeros?
>
> Regards,
>
> Ulrik.

% If you don't have the Stats toolbox
[row,~,val] = find(A);
accumarray(row,val,[],@mean)

% If you have the Stats toolbox
A(A == 0) = NaN;
nanmean(A,2)
From: Matt Fig on 29 Jun 2010 10:07
"us " <us(a)neurol.unizh.ch> wrote in message
> one of the solutions
> - if(f) you own the stats tbx...
>
> % the data
> v=[
> 1 2 3 0 0 0
> 0 3 0 5 0 0
> ];
> % the engine
> tf=v~=0;
> b=sum(tf,2);
> a=sum(v,2);
> r=a./b;
> % the result
> r
> % r = 2 4
>
> us

Do I misread you us? Why does one need the stats toolbox for this? Isn't your solution the same as:

rows_mean = sum(A,2)./sum(logical(A),2)
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