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

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.

[per isakson]

• The use of break looks ok!
• "I got some suspicious answers". How come "suspicious" ?

[Bee]

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.

Answer 1

[Robert]

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.

[Bee]

Thank you very much Robert. The suggestions are great, I will definitely try them out. I am not much of a coder, just wondering, about your second suggestion, will it look like:

target = min([find(x,N,1), find(y,N,1), find(z,N,1)])

Thanks again for your wonderful comments and prompt answer :)

[Robert]

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]

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]

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 :)