Creating a loop for matrix multiplication

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Wojtek
Wojtek am 11 Mai 2014
Kommentiert: Jos (10584) am 13 Mai 2014
Hi. I am trying to create a loop to multiply two matricies 40 times such that A*B0=B1 and A*B1=B2 and so on. So far I have got this
%
a=1-0.0277;
b=(1/15)*(1-0.008);
c=0.0215*a;
d=(14/15)*(1-0.008);
A=[[a,b];[c,d]];
B0=[6820;2140];
B=[B0];
for j=1:40
B1=A*B0;
B0=B1;
B=[B;B1];
disp(round(B(j)))
end
plot(B)
It does work but the result comes up as a list of all the values together and I want it to be a in a 1x2 matrix format.
Can anyone help? Thank you
  5 Kommentare
Wojtek
Wojtek am 12 Mai 2014
Bearbeitet: Wojtek am 12 Mai 2014
Basically, this loop creates an array B which has 82 entries in one column. What I would like to have is 81 separate arrays created using elements of B.
So for example C1=B(1:2,1), C2=B(3:4,1) and so on.
I have tried to write something like this:
for k=1:41
for a=1:2:81
for b=2:2:82
C(k)=B(a:b,1)
disp(C)
end
end
end
But that doesn't work.
John D'Errico
John D'Errico am 12 Mai 2014
NO NO NO NO. Do not create 81 different named variables. Learn to use arrays of variables. Your code will be the better for it. Your mental health a bit better too, because your code will not drive you crazy.

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Antworten (1)

Wojtek
Wojtek am 13 Mai 2014
Problem solved. I have used this
x=length(B);
y=x/2;
C=[];
for i=1:41
C=[C B(2*i-1:2*i)];
end
Thank you all for your comments.
  1 Kommentar
Jos (10584)
Jos (10584) am 13 Mai 2014
for i=1:41,
C = [C B(2*i-1:2*i)]
end
is exactly the same as
C = B(1:82)

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