4 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)));
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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