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Error using reshape Number of elements must not change. Use [] as one of the size inputs to automatically calculate the appropriate size for that dimension.
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Kutlu Yigitturk
on 28 Dec 2020
Commented: Image Analyst
on 28 Dec 2020
% dening data points, vectors x and y
x_value = [0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5];
y_value = [0 0.2474665 0.8813736 1.550158 2.0947125 2.5320681 2.893444 3.2003349 3.466711 3.7019111 3.9124228];
% dening x in range 0 to 5 with total 100 values
xx = linspace(0,5,100);
plot(x_value, y_value,'O')
%calculation of coeffiecients
for i=1:xx
for j=1:xx
c.c(i,j)=(x_value(1,i)-x_value(1,j));
end
end
c=c.*c;
q=size(c);
c=c';
c(c==0)=[];%deleting the zero values from the matrix
c=reshape(c,q(2)-1,q(1));
c=reshape(c,q(2)-1,q(1)); -> Error in Project (line 19)
c=c';
j=1;
for i=1:xx
d.d(i,j)=prod(c(i,:));
end
d=d.d;
d=d';
for i=1:xx
coff.coff(i,j)=y_value(1,i)./d(1,i);
end
coff=coff.coff;
coff=coff';%coefficients
for i=1:xx
for j=1:xx
p.p(i,j)=(x-x_value(1,j));
end
end
p=p.p;
m=size(p);
p=p';
p(1:xx+1:end)=[];%deleting the diagonal elements
p=reshape(p,m(2)-1,m(1));
p=prod(p);
pol=coff.*p;
pol=sum(pol);%Lagrange polynomial
l=input('Enter a value to compute the b value:');
output=subs(pol,l);
%Displaying output
disp('');
fprintf('Value of input x:=%f\n',double(output));
if l>max(x_value)
x=min(x_value):0.1:l;
y=subs(pol,x);
plot(x,y,'r');
xlabel('X')
ylabel('Y')
else
x=min(x_value):0.1:max(x_value);
y=subs(pol,x);
plot(x,y,'r');
xlabel('X')
ylabel('Y')
end
At line 19, the 'Error using reshape Number of elements must not change. Use [] as one of the size inputs to automatically calculate the appropriate size for that dimension. ' I am getting this error. How can I solve this problem. Thank you for your help.
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Accepted Answer
Image Analyst
on 28 Dec 2020
When you say this:
c.c(i,j)=(x_value(1,i)-x_value(1,j));
you're saying that you're assigning the right hand side to a field called "c" that is a member of a structure also (unfortunately) called "c". Why are you doing that?
Then you do this
c=c.*c;
q=size(c);
c=c';
c(c==0)=[];%deleting the zero values from the matrix
c=reshape(c,q(2)-1,q(1));
The first line will not even work since you can't square a structure. But anyway, you get q based on the size of c but then delete some elements from c so the size of c changes. However you never updated q. Why didn't you? And why do you want c to be a 2D array? What is the overall intent of this code? Some kind of filtering, convolution, or gradient?
The whole code is a mess and I think you'll discover the error if you do two things:
3 Comments
Image Analyst
on 28 Dec 2020
For spline, you can use interp1() or spline(). See attached demo.
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