Filter löschen
Filter löschen

Same code I can successfully run in R2021b but in R2016a it is showing Error using - Matrix dimensions must agree. What should I change to run this in R2016a?

1 Ansicht (letzte 30 Tage)
function [facilities,R,Dij,E,C]=a(r1,r2,d1,d2,i,j)
tic
facilities=(1:i)';
R = randi([r1 r2],[i 2]);
R(:,2)=(R(:,1))+(R(:,2));
Dij=randi([d1 d2],[i,j]);
E=zeros(i,1);
E(:)=10^-6;
for m=1:i
for n=1:j
C = max((R(:,2)-Dij(:,:)),0)./max(max((R(:,2)-R(:,1)),(R(:,2)-Dij(:,:))),E(:,1));
n=n+1;
end
m=m+1;
end
toc
end
% Same code I can successfully run in R2021b but in R2016a it is showing Error using - Matrix dimensions must agree. What should I change to run this in R2016a?
  3 Kommentare
Stephen23
Stephen23 am 8 Dez. 2022
Bearbeitet: Stephen23 am 8 Dez. 2022
You can read more about implicit array expansion here:
The code throws an error on R2016a because it does not have implicit expansion.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

DGM
DGM am 8 Dez. 2022
Bearbeitet: DGM am 8 Dez. 2022
R2016b introduced implicit array expansion. You're relying on that. The loops are also unnecessary.
i = 10;
j = 5;
r1 = 1;
r2 = 10;
d1 = 1;
d2 = 10;
facilities=(1:i)';
R = randi([r1 r2],[i 2]);
R(:,2)=(R(:,1))+(R(:,2));
Dij=randi([d1 d2],[i,j]);
E=zeros(i,1);
E(:)=10^-6;
num = max(bsxfun(@minus,R(:,2),Dij),0);
den = max((R(:,2)-R(:,1)),E(:,1)); % these terms agree
den = bsxfun(@max,bsxfun(@minus,R(:,2),Dij),den);
C = num./den;
I tested this in R2015b

Weitere Antworten (0)

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by