how to store value from loop

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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

采纳的回答

rifat
rifat 2014-5-27
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

更多回答(1 个)

Sara
Sara 2014-5-26
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 个评论
sazzad hossen
sazzad hossen 2014-5-27
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));

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