我编写了这段代码，使用jacobi迭代解决了带状矩阵。它是一个改进的Jacobi函数，经过优化和＃34;在带状矩阵上更快地工作。我已成功修改它只在矩阵中计算非零项，但它运行速度比原始代码慢得多。如何让它更有效地运行？
原始代码：
while (true)
iter = iter+1;
for r=1:m % looping through all rows of A
sum = 0; % setting sum equal to 0
for c=1:n % looping through all columns of A
if r~=c % if the r does not equal c, meaning the current elements are not diagonals...
sum = sum + A(r,c)*x0(c);% set sum = sum plus current element times cth element of x
end
x(r) = (-sum + b(r))/A(r,r); % if r and c are diagonals, the x value at that element is -sum + b(r) divided by that element of A
end
end
if abs(norm(x) - norm(x0)) < TOL % checking tolerance
break
end
x0 = x; % setting x0 = x before relooping
end
带状代码：
while (true)
% iter = iter+1;
for r=1:m % looping through all rows of A
for i = 1:hi % looping through the section of A at the beginning that does not have a complete "low"
sum = 0; % assigning sum = 0 for each loop
for r = i % looping through rows of first section
for c = 1:r+hi; % looping through only columns that go from the first column to hi after the diagona
if r~=c % if the r does not equal c, meaning the current elements are not diagonals...
sum = sum + A(r,c)*x0(c);% set sum = sum plus current element times cth element of x
end
x(r) = (-sum + b(r))/A(r,r); % solving the row
end
end
end
for i = 1+hi:n-hi % looping through second section (section with a complete "low" and "hi" for each diagonal
sum = 0; % setting sum = 0 for each loop
for r = i % looping through each row of second section
for c = r+low:r+hi; % looping through only columns of current row that are in the bandwidth
if r~=c % if the r does not equal c, meaning the current elements are not diagonals...
sum = sum + A(r,c)*x0(c);% set sum = sum plus current element times cth element of x
end
x(r) = (-sum + b(r))/A(r,r); % solving current row
end
end
end
for i = n+low+1:n % looping through final section of A (section with out a complete "hi" for each diagonal
sum = 0; % setting sum = 0 for each loop
for r = i % looping through each row of final section
for c = r + low:n; % looping through only columns of current row that are from "low" away from the diagonal to the end of the row
if r~=c % if the r does not equal c, meaning the current elements are not diagonals...
sum = sum + A(r,c)*x0(c);% set sum = sum plus current element times cth element of x
end
x(r) = (-sum + b(r))/A(r,r); % solving current row
end
end
end
end
if abs(norm(x) - norm(x0)) < TOL % checking tolerance
break
end
x0 = x; % assigning x0 as the new estimate from the previous attempt, x
end
sln = x0; % assigning output variable
我对其所做的更改将其分为3个部分（1个部分用于没有完整＆＃34;低＆＃34;以及＆＃34; hi＆＃34;在开始时，结尾处为1，对于中间部分为1。然后我将循环改为从低处开始并转到hi。为什么这个函数需要这么长时间才能运行？大约需要一分钟对于20x20矩阵。