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

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Bee on 21 Oct 2015
Edited: Bee on 26 Oct 2015
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;
target = 0;
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.


per isakson
per isakson on 21 Oct 2015
  • The use of break looks ok!
  • "I got some suspicious answers". How come "suspicious" ?
Bee on 22 Oct 2015
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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Accepted Answer

Robert on 21 Oct 2015
Edited: Robert on 22 Oct 2015
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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Robert on 22 Oct 2015
You are welcome. In your question, you iterate over i = 1:T. Was that supposed to be t = 1:T? If so, your comment above is close, but not quite there. Also, I now see that your comment matches my suggestion, which also needs to be tweaked. I will update it so I don't mislead other readers who might not check the comments.
The find function looks for anything that evaluates to true (in other words anything but zero). If it is important to your code, check doc find to see what it does with nan. So to find values of N, we need to use
target = min([find(x==N,1), find(y==N,1), find(z==N,1)])
Guillaume on 22 Oct 2015
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 on 26 Oct 2015
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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