Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.

1 次查看(过去 30 天)
Hello,
farfieldvelocity = [-100:.01:100] % velocity est.
lockingdepth = [-100:.01:100] % locking depth est.
for i = 1:length(lockingdepth);
for j = 1:length(farfieldvelocity);
v_e_est = (farfieldvelocity(j)./ pi).* atan(dist./lockingdepth(i));
e = v_est - v_e;
CF(j,i) = e'*e;
end
end
Here I always get the error "Unable to perform assignment because the indices on the left side are not compatible with the size of the right side." for the line CF(j,i) = e'*e

回答(2 个)

James Tursa
James Tursa 2019-3-22
What is size(dist) and size(v_e)? If these are row vectors, then maybe you need to do e*e' instead to get a scalar result.
  4 个评论
James Tursa
James Tursa 2019-3-22
Sorry I misread your variables. What is size(v_est)? And what is the variable v_e_est supposed to be used for? Is this what was meant:
e = v_e_est.' - v_e; % make e a column vector

请先登录,再进行评论。


Sevil Cansu Yildirim
for i = 1:length(lockingdepth);
for j = 1:length(farfieldvelocity);
v_e_est = (farfieldvelocity(j)./ pi).* atan(dist./lockingdepth(i));
e = v_e_est - v_e;
CF(j,i) = e.*e';
end
end
I am sorry.
Here, e is the error whic is the difference between v_e_est (v_e estimated) and (v_e observed).
cf is the cross function, which minimize the error. (e'*e is for that, leat squares method)

类别

Help CenterFile Exchange 中查找有关 Tables 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by