Reply by dutmatlab April 16, 20052005-04-16

Hi,

two solutions :

A=[];
for n=1:10
A=[A rand(1,5)];
end

or

A=zeros(10,5);
for n=1:10
A(n,:)=rand(1,5);
end

The second one should be faster because matrix A is pre-alocated.

Je


Reply by V Ravi Chander April 15, 20052005-04-15

hi cal,
if 'b' is the 1X5 vector generated inside the for loop, and 'a' is the
10X5 matrix that you need,

for i=1:10
%some processing to obtain b
a(i,:)=b;
%or a=[a;b];
end

regards,
Ravi

>
> Hi
>
> Could somebody please help.
>
> Inside a loop I generate a (1X5)vector. At each step I want to
> include this vector as a new row of a matrix. So, for example, after
> 10 times through the loop I will have generated a (10X5) matrix.
>
> Is there a quick way of doing this?
>
> Any help would be much appreciated.
>
> Thanks
>
> Cal >


--
V Ravi Chander,
Emerging Technologies and Services Group,
Reliance Infocomm Ltd.


Reply by camdmag April 14, 20052005-04-14


Hi

Could somebody please help.

Inside a loop I generate a (1X5)vector. At each step I want to
include this vector as a new row of a matrix. So, for example, after
10 times through the loop I will have generated a (10X5) matrix.

Is there a quick way of doing this?

Any help would be much appreciated.

Thanks

Cal