필터 지우기
필터 지우기

Help vectors size doesn't match...why?

조회 수: 5 (최근 30일)
cakey
cakey 2014년 10월 23일
댓글: cakey 2014년 11월 16일
This is my code to evaluate Lagrange where x_in are the given interpolation points and x_out is suppose to be f(x_in), but it keeps saying x_in and x_out don;t have the same vector length
x = -1:50;
y = 1.0 ./ (1+9*x.^2);
x_in=linspace(-1,1,100);
x_out=(1.0) ./ (1+9*(x_in).^2);
y_out=make_Ln(x,x_in,x_out);
plot(x,y,':',x_in,y_out,'*');
title('Lagrange Interpolation:x_in=linspace(-1,1,100)')
legend('y','Interpolation');
figure;
  댓글 수: 6
cakey
cakey 2014년 10월 23일
Error using plot Vectors must be the same lengths.
cakey
cakey 2014년 10월 23일
편집: Andrei Bobrov 2014년 10월 24일
Sorry it is this one:
y_out=make_Ln(x,x_in,x_out)
n=size(x_in,2);
L=ones(n,size(x,2));
if (size(x_in,2)~=size(x_out,2))
fprintf(1,'x_in and x_out must have same elements');
y=NaN;
else
for i=1:n
for j=1:n
if (i~=j)
L(i,:)=L(i,:).*(x-x_in(j))/(x_in(i)-x_in(j));
end
end
end
y=0;
for i=1:n
y_out=y_out+x_out(i)*L(i,:);
end
end

댓글을 달려면 로그인하십시오.

채택된 답변

Julia
Julia 2014년 10월 23일
Hi,
x_in and x_out have the same length (100).
x and y are shorter (52), but I don't know if this bothers your function make_Ln().
I suppose y_out and x_in do not match, but without knowing your function make_Ln() I cannot say more.
To plot x_in and x_out is no problem
plot(x,y,':',x_in,x_out,'*')
  댓글 수: 7
Patrik Ek
Patrik Ek 2014년 10월 24일
Cakey i think that you should take a look at the debugger. You can easily set breakpoints in the code and check dimensions, values, etc. It is also possible to set a breakpoint if error. Both under the editor menu under breakpoints, or simply by typing dbstop error. This can be of help in the future and also to make it easier to find solutions of problems.
cakey
cakey 2014년 11월 16일
Thank you!

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Creating and Concatenating Matrices에 대해 자세히 알아보기

제품

Community Treasure Hunt

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

Start Hunting!

Translated by