Anyone can help me to improve this code?

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Brwa
Brwa 2013-5-22
Dear friends; I have a code which is not suitable for big matrices or unlimited matrice. I would like to make it better because some times i have a very big matrices such as A = (100,100)
A = zeros(n,n) matrix.
R = [ 0 0 0 1 1] vector
c=2
if R(5,1) == 1
A(5,5) = c
A(5,4) = - c
A(4,5) = - c
else
A(5,5) = 0
A(5,4) = 0
A(4,5) = 0
end
if R(4,1) == 1
A(4,4) = c + A(5,5)
A(3,4) = - c
A(4,3) = - c
else
A(4,4) = 0
A(3,4) = 0
A(4,3) = 0
end
.
.
.
ans
A =
0 0 0 0 0
0 0 0 0 0
0 0 0 -2 0
0 0 -2 4 -2
0 0 0 -2 2
Thanks in advance, your help always appreciated

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2013-5-22
编辑：Andrei Bobrov 2013-5-22
[EDIT]
n = 5; %
R = [0 0 0 1 1]';% eg
c = 2; %
A1 = spdiags(-c*R,1,n,n);
A = A1 + A1' + spdiags(c*(R + [R(2:end);0]),0,n,n);
full(A)
OR
Rin(:,[1 3 2]) = c*[-[circshift(R,-1) R],R + [R(2:end);0]];
A = spdiags(Rin,-1:1,n,n);
full(A)
OR
Rin(:,[1 3 2]) = c*[-[circshift(R,-1) R],conv(R,[1;1],'same')];
A = spdiags(Rin,-1:1,n,n);
full(A)
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Brwa
Brwa 2013-5-22
still have the same mistake, the diagonal line should not be sum of
A(1,1) must not equal A(5,5)+A(4,4)+A(3,3)+A(2,2)+c
if R =[ 1 1 1 1 1]
A(5,5) must equa = c
A(4,4) must equal A(5,5) + c
A(3,3) must equal A(4,4) + c
A(2,2) must equal A(3,3) + c
A(1,1) must equal A(2,2) + c
And if
if R =[ 1 1 1 0 1]
A(5,5) must equa = c
A(4,4) must equal A(5,5) + 0
A(3,3) must equal 0 + c this means we only consider about original number of A(4,4) when R =[ 1 1 1 0 1] the fourth value is 0 so A(4,4) = 0 therefore, A(3,3) = 0 + c = c
A(2,2) must equal A(3,3) + c
A(1,1) must equal A(2,2) + c

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更多回答（2 个）

David Sanchez
David Sanchez 2013-5-22
Ii this what you need?
for k=1:n-1
if R(k,1) == 1
A(k,k)= c + A(k+1,k+1);
else
A(k,k) = A(k+1,k+1);
end
  4 个评论
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David Sanchez
David Sanchez 2013-5-22
I see, you have to start the for loop by the end.
c=2;
if R(5,1) == 1
A(5,5) = c
A(5,4) = - c
A(4,5) = - c
else
A(5,5) = 0
A(5,4) = 0
A(4,5) = 0
end
for k=(n-1):-1:1
if R(k) == 1
A(k,k)= c + A(k+1,k+1);
else
A(k,k) = A(k+1,k+1);
end
Brwa
Brwa 2013-5-22
oh, my god still not correct, but I still appreciate your help very very much.
the problem is for R= [ 1 1 1 0 1]
A(3,3) must = c not 2c because A(4,4) =0 so A(3,3) = c + 0 = c

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Brwa
Brwa 2013-5-22
Thanks for your helps, Andrei Bobrov and David Sanchez. you guys are so nice. This is my code. it works good, to be honest i learnt alot from you two. but if someone can ake it shorter that will be great.
n=5;
A = zeros(n,n);
R = [1;1;0;1;1];
c = 2;
for i = n-1 : -1 : 2;
if R(i,1) == 1 && R(i+1,1) ==1
A(i,i) = 2*c ;
A(i,i-1) = - c ;
A(i-1,i) = - c ;
elseif R(i,1) == 1 && R(i+1,1) ~=1
A(i,i) = c ;
A(i,i-1) = - c ;
A(i-1,i) = - c ;
elseif R(i,1) ~= 1 && R(i+1,1) ==1
A(i,i) = c ;
A(i,i-1) = 0 ;
A(i-1,i) = 0 ;
elseif R(i,1) ~= 1
A(i,i) = 0 ;
A(i,i-1) = 0 ;
A(i-1,i) = 0 ;
end
if R(1,1)== 1 && R(2,1) == 1
A(1,1) = 2*c ;
elseif R(1,1) == 1 || R(2,1) == 1
A(1,1) = c;
else
A(1,1) = 0;
end
if R(n,1) ==1 && R(n-1,1) ==1
A(n,n) = c ;
A(n-1,n) = -c;
A(n,n-1) = -c;
A(n-1,n-1) = 2*c;
elseif R(n,1) ==1
A(n,n) = c ;
A(n-1,n) = -c;
A(n,n-1) = -c;
A(n-1,n-1) = c;
end
end
Thank you again

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