Storage of a loop from negative number

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Rafael Freire
Rafael Freire 2011-10-19
How i can storage a loop from negative number For example I have the below loop
for x=-1.5:0.01:1.5
for y=-1.5:0.01:1.5
(x,y)=92.5+(32.53333333)*(x)+(40.12702079)*(y)+(-42.2)*(x^2)+(-34.93538288)*(y^2)+(-3.464203233)*((x)*(y))
end
end
Best regards

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采纳的回答

Walter Roberson
Walter Roberson 2011-10-19
xvals=-1.5:0.01:1.5;
yvals=-1.5:0.01:1.5;
for J = 1:length(xvals)
for K = 1:length(yvals)
F(J,K)=92.5+(32.53333333)*(xvals(J))+(40.12702079)*(yvals(K)+(-42.2)*(xvals(J)^2)+(-34.93538288)*(yvals(K)^2)+(-3.464203233)*((xvals(J))*(yvals(K)))
end
end
This can be greatly simplified as:
F = bsxfun(@(x,y) 92.5+(32.53333333)*.(x)+(40.12702079).*(y)+(-42.2).*(x.^2)+(-34.93538288).*(y.^2)+(-3.464203233).*((x).*(y)), (-1.5:0.01:1.5).', -1.5:0.01:1.5);

更多回答（2 个）

Andrei Bobrov
Andrei Bobrov 2011-10-19
i1 = -1.5:0.01:1.5;
f = @(x,y)92.5+32.53333333*x+40.12702079*y-42.2*x.^2-34.93538288*y.^2-3.464203233*x.*y;
Z = bsxfun(f,i1',i1);
Oleg Komarov
Oleg Komarov 2011-10-19
Loop version:
z = zeros(1,301^2);
n = 0;
for x =-1.5:0.01:1.5
for y =-1.5:0.01:1.5
n = n+1;
z(n)=92.5+32.53333333*x+40.12702079*y-42.2*x^2-34.93538288*y^2-3.464203233*x*y;
end
end

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