Storage of a loop from negative number
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
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
0 Kommentare
Akzeptierte Antwort
Walter Roberson
am 19 Okt. 2011
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);
0 Kommentare
Weitere Antworten (2)
Andrei Bobrov
am 19 Okt. 2011
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
am 19 Okt. 2011
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
0 Kommentare
Siehe auch
Kategorien
Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!