Optimizing or Simplifying code
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Hi
I have the following code:
for count = 1:1:size(PointLoads,1);
if Positions > PointLoads(count,2) % Checks if the location of PointLoad falls within posiyions
ShearForce = 0;
else
Z = (Positions<=PointLoads(count,2)); %If condition is false then Positions are multiplied by the corresponding forces
ShearForce = ShearForce + PointLoads(count,1).*Z;
end
end
%ShearForce for DistributedLoadsLoads
for count = 1:1:size(DistributedLoads,1)
if Positions > max(DistributedLoads(count,2:3))
ShearForce = ShearForce;
else
DistributedLoads1 = DistributedLoads(count,1);
A = (max(DistributedLoads(count,2:3)) - min(DistributedLoads(count,2:3)));
B = (Positions<min(DistributedLoads(count,2:3)));
C = (max(DistributedLoads(count,2:3)) - Positions);
D = (and(Positions>=min(DistributedLoads(count,2:3)),Positions<=max(DistributedLoads(count,2:3))));
ShearForce = ShearForce + DistributedLoads1.*A.*B;
ShearForce = ShearForce + DistributedLoads1.*C.*D;
end
end
I was wondering if anyone would be able to assist me in optimizing the code and perhaps making it less complicated.
Thanks
0 Kommentare
Antworten (1)
Jan
am 1 Mai 2016
Bearbeitet: Jan
am 1 Mai 2016
index = (Positions <= PointLoads(k, 2));
ShearForce = sum(PointLoads(index, 1));
%ShearForce for DistributedLoadsLoads
D = DistributedLoads; % Nicer name
maxD = max(D(:, 2:3), [], 1);
minD = min(D(:, 2:3), [], 1);
for k = 1:size(D,1)
if Positions >= maxD(k)
A = maxD(k) - minD(k);
B = (Positions < minD(k));
C = maxD(k) - Positions;
D = (and(~B, Positions <= maxD(k)));
ShearForce = ShearForce + D(k,1) .* (A .* B + C .* D);
end
end
Is "Positions" a vector? If so, are you sure that "if Positions >= maxD(k)" does, what you want?
0 Kommentare
Siehe auch
Kategorien
Mehr zu Surrogate Optimization 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!