Can I use if ..break...else inside the same loop?

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Bee
Bee 2015-10-21
编辑： Bee 2015-10-26
My apology if it seems very trivial to all, but I have looked into the answers and could not find one single example which used if..break...else in the same loop. In my code, I want the value of t, where any of the variables x(t) or y(t) or z(t).... equals M (some pre set value) for the first time and then discontinue the iteration; otherwise it will return 0. So my code snippet looks like:
for i = 1:T
...
...
if x(t) == M || y(t) == M || z(t) == M || a(t) == M || b(t) == M
target = t;
break
else
target = 0;
end
end
I got some suspicious answers, so I was thinking whether the problem is in the use of break or the random probabilities I used in my code. Thanks.
  2 个评论
per isakson
per isakson 2015-10-21
编辑：per isakson 2015-10-21
  • The use of break looks ok!
  • "I got some suspicious answers". How come "suspicious" ?
Bee
Bee 2015-10-22
编辑：Bee 2015-10-22
Dear _ _per isakson,__thanks for your comments. The word 'suspicious' has a story - I was looking into the occurrence of consensus in a scale-free network and got some reasonable points of consensus; but after some tweaks here and there to change the simulation scenario, I am getting consensus even with a large population having varying opinions - this is 'suspicious' to me, because reaching consensus in a large network is fairly impossible.
However, I am happy that I used if..break..else correctly, so the 'suspicious' results must have something to do with the probabilities of opinion updates or network structure.

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Robert
Robert 2015-10-21
编辑：Robert 2015-10-22
Your code looks fine. However, you can simplify the structure by adding target = 0; before your loop and omitting the else statement.
You might also consider using find(X==M,1), replacing X with each of your variables and then using min to find the index of the first occurrence of M, equivalent to your iterator i when you hit the break statement.
In any case, your use of if...break...else is not incorrect.
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Guillaume
Guillaume 2015-10-22
A possibly more efficient way of achieving the same result:
target = find(any([x;y;z;a;b] == M), 1); %assuming x,y,etc. are row vectors.
Less to type anyway.
Bee
Bee 2015-10-26
编辑：Bee 2015-10-26
Thanks a bunch Guillaume, I was looking for something like this; I didn't know how to use 'any' in the code. Your suggestion is awesome :)

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