2 views (last 30 days)
Triveni on 4 Aug 2016
Commented: Triveni on 4 Aug 2016
for l = 1:6
c11 (l)=opt11{l};
end
output = c11 =
Columns 1 through 5
89.0475 98.7932 98.7932 98.7932 98.7932
Column 6
98.7932

Guillaume on 4 Aug 2016
Edited: Guillaume on 4 Aug 2016
for ...
... do something
if condition
break; %stop the loop
end
end
for l = 1:6
c11(l) = opt11{l}
if l>1 && c11(l) <= c11(l-1)
break;
end
end
Of course, the above could be just written without the loop:
c11 = cell2mat(opt11(1:6));
c11 = c11(logical(cumprod(diff(c11) > 0)));
##### 3 CommentsShow 1 older commentHide 1 older comment
Stephen23 on 4 Aug 2016
@Triveni: it sounds like you are really trying achieve some task which you have not told us about. Rather than getting us to fix some broken code, it would be better if you actually described what you are trying to achieve:
Triveni on 4 Aug 2016
Actually if i post whole code then you need to pay attention at-least for 5 to 10 minutes to analyze my code. My question is not so difficult, i have to stop only loop in different circumstances which i have told you.
[93.6568] [98.9636] [93.4562] [103.8119]
Columns 5 through 8
[101.4730] [103.8119]
In column 1, 2, 3, 4, and 5 value increase and then decrease, then value is continuously increase and decrease. If increase and decrease is steady then i have to break my loop. I have to find out maximum value from c11. When value is not maximizing and it's going steady then i need not to solve it.

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