Re: [Scilab-users] evaluate matrix in a function
A possible solution, we compare "diffcode_jacobian" and "numderivative". in this case seem to calculate with the same accuracy function val = HH(z) x=z(1); y=z(2); val=(1.25*y-sqrt(abs(x))).^2+x.^2-1; endfunction; function z=f(x) z(1,1:2)=diffcode_jacobian(HH,x) z(1,3:4)=numderivative(HH,x) endfunction xt = -1.4:0.01:1.4; function Z=g(X,Y,f) for i=1:size(X,"*") T=[X(i) Y(i)]; Z(i,:)=f(T); end endfunction //And for 280 values calculate fast F=g(xt,xt) -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] evaluate matrix in a function
Hi Philippe:
Where this error?
// The function to differentiate
function val = HH(z)
x=z(1);
y=z(2);
val=(1.25*y-sqrt(abs(x))).^2+x.^2-1;
endfunction;
// The exact gradient
function Sys=ge(x)
gx=x(:,1);
gy=x(:,2);
g1=2*gx-(gx.*(1.25*gy-sqrt(abs(gx./abs(gx).^(3/2);
g2=2.5*(1.25*gy-sqrt(abs(gx)));
Sys(:,1)=g1;
Sys(:,2)=g2;
endfunction
// Compute the approximate Jacobian
t=[3,1]; //for function f, which does not allow multiple inputs [x1,1 x2,1
x3,1; ...; x1, n x2, n x3, n]
t1=[3,1;3,1];//for function f, allow multiple inputs [x1,1 x2,1 x3,1; ...;
x1, n x2, n x3, n]
*// test function*
function z=f(x)
z=diffcode_jacobian(HH,x) endfunction
disp(f(t)',ge(t1),HH(t))
function z=feval2(x,y,g)
// eval f on the curve t->(x(t),y(t))
// and on the grid defined by x,y
deff('z=newf(k)',['a=[x(k),y(k)]';'z=g(a)'])
// x,y are global variables for the "newf" function
z=feval(1:length(x),newf)
// "newf" is a local variable for "feval2"
endfunction
// Compute the approximate Jacobian
x=-3:3;y=x;
z=feval2(x,y,f) // curve evaluation
disp(z ," feval2");
*in builtin
feval2 ( C: \Curve11.sce line 47 )
newf: Wrong size for output argument #1: A Scalar expected.
敦慶l: An error occured in '硥捥敆慶䙬' subroutine.
*
Gracias
Hermes
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Re: [Scilab-users] evaluate matrix in a function
Right, and if one follows the thread till the end, how does that apply to
diffcode_jacobian ?
-Original Message-
From: users [mailto:users-boun...@lists.scilab.org] On Behalf Of philippe
Sent: Tuesday, October 10, 2017 8:56 AM
To: users@lists.scilab.org
Subject: Re: [Scilab-users] evaluate matrix in a function
Hi,
Le 07/10/2017 à 13:05, Hermes a écrit :
> how do I declare the functions to be able to evaluate a matrix
> variable(Multiple evaluation of a function). Where the first column
> corresponds to the first variable of the function. And so on.
> is only possible within a "for" cycle? how to declare the function to be
> able to use the operator "dot"
I'm not sure of what you are asking for. It looks like you want to
evaluate a function f on the curve defined by 2 vectors x,y instead of
evaluating it on the grid defined by x,y (which is done by feval). If
this is what you are searching for you can use something like feval2
below :
function z=feval2(x,y,f)
// eval f on the curve t->(x(t),y(t))
// and on the grid defined by x,y
deff('z=newf(k)','z=f(x(k),y(k))')
// x,y are global variables for the "newf" function
z=feval(1:length(x),newf)
// "newf" is a local variable for "feval2"
endfunction
then try the example :
-->deff('z=f(x,y)','z=x.^2-y.^2') // test function
-->x=-3:3;y=x;
-->z=feval(x,y,f)// grid evaluation
z =
0.5.8.9.8.5.0.
- 5.0.3.4.3.0. - 5.
- 8. - 3.0.1.0. - 3. - 8.
- 9. - 4. - 1.0. - 1. - 4. - 9.
- 8. - 3.0.1.0. - 3. - 8.
- 5.0.3.4.3.0. - 5.
0.5.8.9.8.5.0.
-->z=feval2(x,y,f) // curve evaluation
z =
0.0.0.0.0.0.0.
Best regards
Philippe
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Re: [Scilab-users] evaluate matrix in a function
Hi,
Le 07/10/2017 à 13:05, Hermes a écrit :
> how do I declare the functions to be able to evaluate a matrix
> variable(Multiple evaluation of a function). Where the first column
> corresponds to the first variable of the function. And so on.
> is only possible within a "for" cycle? how to declare the function to be
> able to use the operator "dot"
I'm not sure of what you are asking for. It looks like you want to
evaluate a function f on the curve defined by 2 vectors x,y instead of
evaluating it on the grid defined by x,y (which is done by feval). If
this is what you are searching for you can use something like feval2
below :
function z=feval2(x,y,f)
// eval f on the curve t->(x(t),y(t))
// and on the grid defined by x,y
deff('z=newf(k)','z=f(x(k),y(k))')
// x,y are global variables for the "newf" function
z=feval(1:length(x),newf)
// "newf" is a local variable for "feval2"
endfunction
then try the example :
-->deff('z=f(x,y)','z=x.^2-y.^2') // test function
-->x=-3:3;y=x;
-->z=feval(x,y,f)// grid evaluation
z =
0.5.8.9.8.5.0.
- 5.0.3.4.3.0. - 5.
- 8. - 3.0.1.0. - 3. - 8.
- 9. - 4. - 1.0. - 1. - 4. - 9.
- 8. - 3.0.1.0. - 3. - 8.
- 5.0.3.4.3.0. - 5.
0.5.8.9.8.5.0.
-->z=feval2(x,y,f) // curve evaluation
z =
0.0.0.0.0.0.0.
Best regards
Philippe
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Re: [Scilab-users] evaluate matrix in a function
diffcode_jacobian does not seem to accept multiple input points, but only one N-dimension point X: // Calling Sequence: // J = diffcode_jacobian(f,x) // x : a n-by-1 matrix of doubles, real, the point where to compute the derivatives You cannot compare g and J, because the former is a function and the latter has type constant. However you can compare J and V but the answer is false because of tiny numerical difference: J==V ans = T T T T F T T T J-V ans = 0. 0. 0. 0. - 4.441D-160. 0. 0. Rgds, Rafael -Original Message- From: users [mailto:users-boun...@lists.scilab.org] On Behalf Of Hermes Sent: Saturday, October 07, 2017 8:10 PM To: users@lists.scilab.org Subject: Re: [Scilab-users] evaluate matrix in a function Hello The H and g functions work. But the diffcode_jacobian (H, r) evaluation does not accept the dot operator. Only redefining the H function (without dot operators) will achieve the results function val = HH(z) x=z(1); y=z(2); val=(1.25*y-sqrt(abs(x)))^2+x^2-1; // switched *.^* to *^* endfunction; t=[3. 1.] disp(diffcode_jacobian(HH,t)); -->6.2783122 -1.205127 J=[]; for k=1:size(r,"r") J(k,:)=diffcode_jacobian(HH,r(k,:)); end disp(J, "Jacobian"); -->Jacobian 6.2783122 -1.205127 7.751.25 3.3915608 11.294873 16.116117 -0.8210678 disp(and(J==g)) -->F Why? The results of J and g look the same !! g 6.2783122 -1.205127 7.75 1.25 3.3915608 11.294873 16.116117 -0.8210678 J 6.2783122 -1.205127 7.751.25 3.3915608 11.294873 16.116117 -0.8210678 Gracias -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] evaluate matrix in a function
Hello The H and g functions work. But the diffcode_jacobian (H, r) evaluation does not accept the dot operator. Only redefining the H function (without dot operators) will achieve the results function val = HH(z) x=z(1); y=z(2); val=(1.25*y-sqrt(abs(x)))^2+x^2-1; // switched *.^* to *^* endfunction; t=[3. 1.] disp(diffcode_jacobian(HH,t)); -->6.2783122 -1.205127 J=[]; for k=1:size(r,"r") J(k,:)=diffcode_jacobian(HH,r(k,:)); end disp(J, "Jacobian"); -->Jacobian 6.2783122 -1.205127 7.751.25 3.3915608 11.294873 16.116117 -0.8210678 disp(and(J==g)) -->F Why? The results of J and g look the same !! g 6.2783122 -1.205127 7.75 1.25 3.3915608 11.294873 16.116117 -0.8210678 J 6.2783122 -1.205127 7.751.25 3.3915608 11.294873 16.116117 -0.8210678 Gracias -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] evaluate matrix in a function
Hi, It seems you are missing a few dots in g1, try: g1=2*gx-(gx .*(1.25*gy-sqrt(abs(gx./abs(gx).^(3/2); Also, I believe you should also remove the transposed input in V=g(r'), and make it V=g(r); Rafael and I believe your -Original Message- From: users [mailto:users-boun...@lists.scilab.org] On Behalf Of Hermes Sent: Saturday, October 07, 2017 6:44 PM To: users@lists.scilab.org Subject: Re: [Scilab-users] evaluate matrix in a function Hello The H function works. the other two evaluations I can not solve the problem. function val = H(z) x=z(:,1); y=z(:,2); val=(1.25*y-sqrt(abs(x))).^2+x.^2-1; endfunction; function Sys=g(x) gx=x(:,1); gy=x(:,2); g1=2*gx-(gx *(1.25*gy-sqrt(abs(gx/abs(gx).^(3/2); g2=2.5*(1.25*gy-sqrt(abs(gx))); Sys(:,1)=g1; Sys(:,2)=g2; endfunction r=[3 1;4 2;3 5;8 2] disp(H(r),"H"); H 8.232373 15.25 28.411865 63.107864 V=g(r'); disp(V,"g"); J = diffcode_jacobian(H,r); disp(J, "Jacobian"); at line 4 of function g ( E:\Heart Curve8.sce line 28 ) at line36 of executed file E:\Heart Curve8.sce *Inconsistent row/column dimensions. * Gracias ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] evaluate matrix in a function
Hello The H function works. the other two evaluations I can not solve the problem. function val = H(z) x=z(:,1); y=z(:,2); val=(1.25*y-sqrt(abs(x))).^2+x.^2-1; endfunction; function Sys=g(x) gx=x(:,1); gy=x(:,2); g1=2*gx-(gx *(1.25*gy-sqrt(abs(gx/abs(gx).^(3/2); g2=2.5*(1.25*gy-sqrt(abs(gx))); Sys(:,1)=g1; Sys(:,2)=g2; endfunction r=[3 1;4 2;3 5;8 2] disp(H(r),"H"); H 8.232373 15.25 28.411865 63.107864 V=g(r'); disp(V,"g"); J = diffcode_jacobian(H,r); disp(J, "Jacobian"); at line 4 of function g ( E:\Heart Curve8.sce line 28 ) at line36 of executed file E:\Heart Curve8.sce *Inconsistent row/column dimensions. * Gracias -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] evaluate matrix in a function
Hello The H function works. the other two evaluations did not solve the problem. function val = H(z) x=z(:,1); y=z(:,2); val=(1.25*y-sqrt(abs(x))).^2+x.^2-1; // switched ^ to .^ to handle vectors endfunction; function Sys=g(x) gx=x(:,1); gy=x(:,2); g1=2*gx-(gx *(1.25*gy-sqrt(abs(gx/abs(gx).^(3/2); g2=2.5*(1.25*gy-sqrt(abs(gx))); Sys(:,1)=g1; Sys(:,2)=g2; endfunction r=[3 1;4 2;3 5;8 2] disp(H(r),"H"); H 8.232373 15.25 28.411865 63.107864 V=g(r'); disp(V,"g"); J = diffcode_jacobian(H,r); disp(J, "Jacobian"); at line 4 of function g ( E:\Heart Curve8.sce line 28 ) at line36 of executed file E:\Heart Curve8.sce *Inconsistent row/column dimensions. * Gracias -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] evaluate matrix in a function
Your code has the solution (for the matrix function output). Do the same for input: r=[3 1;4 2;3 5;8 2]; function val=H(z) x=z(:,1); y=z(:,2); val=(1.25*y-sqrt(abs(x))).^2 + x.^2 - 1; endfunction; Regards, Rafael -Original Message- From: users [mailto:users-boun...@lists.scilab.org] On Behalf Of Hermes Sent: Saturday, October 07, 2017 1:05 PM To: users@lists.scilab.org Subject: [Scilab-users] evaluate matrix in a function how do I declare the functions to be able to evaluate a matrix variable(Multiple evaluation of a function). Where the first column corresponds to the first variable of the function. And so on. is only possible within a "for" cycle? how to declare the function to be able to use the operator "dot" r=[3 1;4 2;3 5;8 2]; function val = H(z) x=z(1); y=z(2); val=(1.25*y-sqrt(abs(x)))^2+x^2-1; endfunction; function Sys=g(x) gx=x(1); gy=x(2); g1=2*gx-(gx*(1.25*gy-sqrt(abs(gx/abs(gx)^(3/2); g2=2.5*(1.25*gy-sqrt(abs(gx))); Sys(:,1)=g1; Sys(:,2)=g2; endfunction V=g(r)'; disp(V,"g"); J = diffcode_jacobian(H,r); disp(J, "Jacobian"); Gracias Hermes ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
[Scilab-users] evaluate matrix in a function
how do I declare the functions to be able to evaluate a matrix variable(Multiple evaluation of a function). Where the first column corresponds to the first variable of the function. And so on. is only possible within a "for" cycle? how to declare the function to be able to use the operator "dot" r=[3 1;4 2;3 5;8 2]; function val = H(z) x=z(1); y=z(2); val=(1.25*y-sqrt(abs(x)))^2+x^2-1; endfunction; function Sys=g(x) gx=x(1); gy=x(2); g1=2*gx-(gx*(1.25*gy-sqrt(abs(gx/abs(gx)^(3/2); g2=2.5*(1.25*gy-sqrt(abs(gx))); Sys(:,1)=g1; Sys(:,2)=g2; endfunction V=g(r)'; disp(V,"g"); J = diffcode_jacobian(H,r); disp(J, "Jacobian"); Gracias Hermes -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
[Scilab-users] evaluate matrix in a function
how do I declare the functions to be able to evaluate a matrix variable(Multiple evaluation of a function). Where the first column corresponds to the first variable of the function. And so on.is only possible within a "for" cycle? how to declare the function to be able to use the operator "dot"r=[3 1;4 2;3 5;8 2];function val = H(z)x=z(1); y=z(2); val=(1.25*y-sqrt(abs(x)))^2+x^2-1; endfunction;function Sys=g(x) gx=x(1); gy=x(2); g1=2*gx-(gx*(1.25*gy-sqrt(abs(gx/abs(gx)^(3/2);g2=2.5*(1.25*gy-sqrt(abs(gx))); Sys(:,1)=g1; Sys(:,2)=g2;endfunctionV=g(r)';disp(V,"g");J = diffcode_jacobian(H,r);disp(J, "Jacobian");GraciasHermes -- Sent from: http://mailinglists.scilab.org/Scilab-users-Mailing-Lists-Archives-f2602246.html___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
