Anyone can help me to improve this code?

조회 수: 1 (최근 30일)
Brwa
Brwa 2013년 5월 22일
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

채택된 답변

Andrei Bobrov
Andrei Bobrov 2013년 5월 22일
편집: Andrei Bobrov 2013년 5월 22일
[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)
  댓글 수: 3
Andrei Bobrov
Andrei Bobrov 2013년 5월 22일
EDITed
Brwa
Brwa 2013년 5월 22일
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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추가 답변 (2개)

David Sanchez
David Sanchez 2013년 5월 22일
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
  댓글 수: 4
David Sanchez
David Sanchez 2013년 5월 22일
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 2013년 5월 22일
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 2013년 5월 22일
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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