Double for loop is a problem ??? Any help, I can't find the right results when using the second for loop!!!!!
3 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Hello, I have to use two for loops in my code to calculate the matrix c, xp and yp. The first one concerning the parameter rr and the second for xxb.
To cheek my loops if works well, I compare the values of C and xp for yp when I use only the foo loop for rr for fixed values of xxp. But for some values of xxb I don't find the right results. For exemple , for xxb=0.05, and for 0.1, and for 0.15 the resuls not the same for the two cases.
Any Idea??? Thank you in advance,
Adam
Here is below my code:
xb=-5:0.05:5;
xxb=xb';% column vector
dr=0.01;
r=0:dr:5;
rr=r;% row vector
R=0.05;
for j=1:length(xxb);
for m=1:length(rr);
a(j)=2.*xxb(j);
c(j,m)=xxb(j).^2+rr(m).^2-R.^2;
xp(j,m)=c(j,m)./a(j);
yp(j,m)=R.^2-((((2.*c(j,m))-(a(j).^2))./(2.*a(j))).^2);
end
end
the end of code:
For exemple
using two for loop For the firt values of arrow of index 102 corresponding to the xxb=0.05:
-1,82145964977565e-17
9,99999999999816e-05
0,000399999999999982
0,000899999999999982
0,00159999999999998
0,00249999999999998
0,00359999999999998
0,00489999999999998
..........
using one for loop for xxb=0.05
0
9,99999999999998e-05
0,000400000000000000
0,000900000000000000
0,00160000000000000
0,00250000000000000
0,00360000000000000
..................
3 comentarios
Bård Skaflestad
el 18 de Ag. de 2012
Also, you may want to write your code using vectorised statements, e.g., something like this:
xb = -5:0.05:5;
dr = 0.01;
r = 0:dr:5;
R = 0.05;
[xxb, rr] = ndgrid(xb, r);
a = 2 .* xxb;
c = xxb.^2 + rr.^2 - R.^2;
xp = c ./ a;
yp = R.^2 - (((2.*c - a.^2) ./ (2 .* a))).^2;
which, in addition to being easier to read, is also likely to run more efficiently--at least if the sizes of xb or r do not increase too much.
Respuestas (0)
Ver también
Categorías
Más información sobre Timing and presenting 2D and 3D stimuli en Help Center y File Exchange.
Productos
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!