indexing must appear last in an index expression ERROR

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Hamid on 12 Dec 2014
Commented: Hamid on 12 Dec 2014
Hi everybody,
I get this error : ()-indexing must appear last in an index expression. Line: available(b,6)... Column: 17
also I have this warning at that line : can not call or index into a temporary array.
for i=1:ne
Le(i)=sqrt((Ex(i,1)-Ex(i,2))^2+(Ey(i,1)-Ey(i,2))^2+(Ez(i,1)-Ez(i,2))^2);
if Le(i)==0
j=+j1;
else
u=u+1;
end
Le(i)/200<available(i,6);
available(b,6)=available(i,6)(min(find(available(i,6)*200>Le(i)==1)));
Ep(u,:)=available(b,:);
end
I will be very thankful for any suggestion.

Guillaume on 12 Dec 2014
In matlab you can't chain indexing and as the error message tells you, you can't index temporaries. So you can't do
m(SomeIndexingOperation)(SomeOtherIndexingOperation)
m(SomeIndexingOperation) is a temporary, and you can't index it further without assigning it to a variabe. So to resolve your problem, you first need to assign available(i,6) to a variable before ou can index into it:
temp = available(i, 6)
available(b, 6) = temp(min(...
That's one issue with your code. There are unfortunately more:
1.
min(find(something))
is the same as
find(something, 1) %but this is much faster
2. The code inside your find is suspicious. You usually don't have operators > and == in the same expression. As it is since the expression before the == is either 0 or 1 the == 1 does not do anything, so your expression is the same as
available(i,6)*200>Le(i)
3. The line
Le(i)/200<available(i,6);
does not do or assign anything since the result is not assigned to anything.
Hamid on 12 Dec 2014
Beautiful, thanks a lot

Azzi Abdelmalek on 12 Dec 2014
You probably forgot an operator, maybe a prod *
available(b,6)=available(i,6)*(min(find(available(i,6)*200>Le(i)==1)));
Hamid on 12 Dec 2014
Thank you Azzi,
available is a 128*6 matrix, I have available(i,6)<Le(i)/200 and I want available(i,6) to be minimum.
after that I want to assign available(i,6) minimums to available(b,6) , while in this try I got that error.
Thanks.