indexing must appear last in an index expression ERROR

1 view (last 30 days)
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.

Accepted Answer

Guillaume
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.
  10 Comments
Guillaume
Guillaume on 12 Dec 2014
Use the second return value of min, which will be the row index where the minimum is found
[~, minidx] = min(available(available(:, 6) > Le(i)/200, 6));
minrow = available(minidx, :)

Sign in to comment.

More Answers (1)

Azzi Abdelmalek
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)));
  1 Comment
Hamid
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.

Sign in to comment.

Tags

Community Treasure Hunt

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

Start Hunting!