indexing must appear last in an index expression ERROR
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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.
0 件のコメント
採用された回答
Guillaume
2014 年 12 月 12 日
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 件のコメント
Guillaume
2014 年 12 月 12 日
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, :)
その他の回答 (1 件)
Azzi Abdelmalek
2014 年 12 月 12 日
You probably forgot an operator, maybe a prod *
available(b,6)=available(i,6)*(min(find(available(i,6)*200>Le(i)==1)));
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