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From: ryan on 27 Jun 2010 20:27
Hi!
When I run this loop I obtain 5 simultaneous 'z=' statements:
e.g.:

z=
1 2 3 4 5
z=
2 9 8 7 6

is there a way to generate one matrix 'Z' with 'n' rows and the '5' colums out of these five arrays instead?
ex:

z=
1 2 3 4 5
2 9 6 7 8
....
loop portion of code:

for k=1:5


for h=1:n
i=1:n;
j=1:n;

T=Dz(i,j)*X;

S=sum(T');

C(i);

U=(C(i)-T(i));
d;
h;
X(h)=U(h)/d(h);

end

Z=[k;X]'

for h=n;
k;
h;
X';

end
end

X=X';


Thanks in advance!
From: Faraz Afzal on 27 Jun 2010 21:52
"ryan " <ryanlendslasvegas(a)yahoo.com> wrote in message <i08q8r$mn6$1(a)fred.mathworks.com>...
> Hi!
> When I run this loop I obtain 5 simultaneous 'z=' statements:
> e.g.:
>
> z=
> 1 2 3 4 5
> z=
> 2 9 8 7 6
>
> is there a way to generate one matrix 'Z' with 'n' rows and the '5' colums out of these five arrays instead?
> ex:
>
> z=
> 1 2 3 4 5
> 2 9 6 7 8
> ...
> loop portion of code:
>
> for k=1:5
>
>
> for h=1:n
> i=1:n;
> j=1:n;
>
> T=Dz(i,j)*X;
>
> S=sum(T');
>
> C(i);
>
> U=(C(i)-T(i));
> d;
> h;
> X(h)=U(h)/d(h);
>
> end
>
> Z=[k;X]'
>
> for h=n;
> k;
> h;
> X';
>
> end
> end
>
> X=X';
>
>
> Thanks in advance!

Hi ryan..

Yes you can do this .. Just add a little bit more to your for loop..

Such as the part that I have mentioned with ">>>>" You can change it to columns using transpose or leaving transpose..

for k=1:5


for h=1:n
i=1:n;
j=1:n;

T=Dz(i,j)*X;

S=sum(T');

C(i);

U=(C(i)-T(i));
d;
h;
X(h)=U(h)/d(h);

end
>>>>> Z = zeros(n,k);
>>>>> Z(:,k)=[k;X]' ;

for h=n;
k;
h;
X';

end
end

X=X';

I hope it help..

Let me know if it worked..

Regards,
Faraz
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