how to store value from loop

3 Ansichten (letzte 30 Tage)
sazzad hossen
sazzad hossen am 26 Mai 2014
Kommentiert: sazzad hossen am 27 Mai 2014
for i=1:0.1:5
ly=i;
for j=1:0.1:5;
lx=j;
c1=co1(x1,y1,z1,lx,ly,lz,f);
c2=co2(x2,y2,z1,lx,ly,lz,f);
c3=co3(x3,y3,z1,lx,ly,lz,f);
%options = optimoptions('fsolve','Display','iter');
func = @(x)efunction(x,x1,x2,x3,y1,y2,y3,c1,c2,c3,f);
[x,fval] = fsolve(func,x0);
X=x(1,1);
Y=x(1,2);
H=x(1,3);
x_error= sprintf('%.10f',abs(X-lx));
y_error=sprintf('%.10f',abs(Y-ly));
z_error=sprintf('%.10f',abs((z1-H)-lz));
end
end
problem : i just get only one value in each parameter of x_error, y_error, z_error
but i want to get/store all value of x_error, y_error and z_error

Akzeptierte Antwort

rifat
rifat am 27 Mai 2014
Your index isnt integer.. Use the following modification. And your program isnt self sufficient for me to execute.
idx = 0;
for i=1:0.1:5
ly=i;
idx=idx+1;
idy=0;
for j=1:0.1:5;
idy=idy+1;
lx=j;
c1=co1(x1,y1,z1,lx,ly,lz,f);
c2=co2(x2,y2,z1,lx,ly,lz,f);
c3=co3(x3,y3,z1,lx,ly,lz,f);
%options = optimoptions('fsolve','Display','iter');
func = @(x)efunction(x,x1,x2,x3,y1,y2,y3,c1,c2,c3,f);
[x,fval] = fsolve(func,x0);
X=x(1,1);
Y=x(1,2);
H=x(1,3);
x_error{idx,idy}= sprintf('%.10f',abs(X-lx));
y_error{idx,idy}=sprintf('%.10f',abs(Y-ly));
z_error{idx,idy}=sprintf('%.10f',abs((z1-H)-lz));
end
end

Weitere Antworten (1)

Sara
Sara am 26 Mai 2014
Define a cell array before the loop:
x_error = cell(50,50); % Adjust num elements according to your loops
and then do:
x_error{i,j} = ....
The same for the other variables.
  1 Kommentar
sazzad hossen
sazzad hossen am 27 Mai 2014
it's not wotk
x_error = cell(1681,1); % because of two loop 41*41=1681
x_error = cell(1681,1);
x_error = cell(1681,1);
for i=1:0.1:5 ly=i;
for j=1:0.1:5;
lx=j;
c1=co1(x1,y1,z1,lx,ly,lz,f);
c2=co2(x2,y2,z1,lx,ly,lz,f);
c3=co3(x3,y3,z1,lx,ly,lz,f);
%options = optimoptions('fsolve','Display','iter');
func = @(x)efunction(x,x1,x2,x3,y1,y2,y3,c1,c2,c3,f);
[x,fval] = fsolve(func,x0);
X=x(1,1);
Y=x(1,2);
H=x(1,3);
x_error{i,j}= sprintf('%.10f',abs(X-lx));
y_error{i,j}=sprintf('%.10f',abs(Y-ly));
z_error{i,j}=sprintf('%.10f',abs((z1-H)-lz));
end
end
error result:
Subscript indices must either be real positive integers or logicals.
Error in solution2 (line 39) x_error{i,j}= sprintf('%.10f',abs(X-lx));

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Solver-Based Optimization Problem Setup finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by