Getting values from a for loop when specific condition met

Ali Tawfik (view profile)

on 25 Aug 2019
Latest activity Edited by dpb

on 26 Aug 2019

dpb (view profile)

I am doing a loop, and want the iteration stops when condition of x = 5 however, the code terminates before x=5 ?
clear all;
clc;
x=[1 2 3 4 5 6 7 8 9 10];
for n=1:10
if x(n)>= 5;
break
end
disp('how many iteration done now');
final(n,:)=x(n)
z(n,:)=x(n)+2
end

Rik

Rik (view profile)

on 25 Aug 2019
It doesn't calcluate anything for n=5, because you tell Matlab to stop before that. What is your question?

dpb (view profile)

on 25 Aug 2019

break is like drawing the "Go To Jail" card in Monopoly--you "go directly to jail, do not pass GO, do not collect \$200". Here when you get to the conditional and it is satisfied, then the code immediately goes to the next line of code after the terminating end of the for loop and so the other code isn't executed that last iteration--the loop did do five iterations here because it took until then for x(n) to be >=5; just that you didn't do anything at all in the loop other than the test.
Perhaps your expectations could be met simply by reordering:
...
for n=1:10
disp('how many iteration done now');
final(n,:)=x(n)
z(n,:)=x(n)+2
if x(n)>= 5, break, end
end
...
Then again, that assumes that what happens now is what you want to happen for the loop for all iterations that it does do...

dpb

dpb (view profile)

on 25 Aug 2019
You left off the second (and optional, but needed here) argument to find
Instead of me spoon-feeding, look at the documentation for find and see if you can't learn the difference between the two invocations yourself.
Going forward, it will be much more productive if you learn to read and use the doc instead of waiting for somebody here to answer a question that can be resolved thru the input descriptions.
Ali Tawfik

Ali Tawfik (view profile)

on 26 Aug 2019
However, I don't agree with you in what you said about optional, or even your last code in general
(x>=5,1) 1 here means 1 which means first element that met the condition,
second, why you add the values to 2 ?
I meant, I need from x array all the numbers that met the conditions written.
Thanks,
dpb

dpb (view profile)

on 26 Aug 2019
"Optional" in that it is an optional input to find(), not that it is optional to use it in this case to find the first location in x that satisfies the condition which is what your break logic is doing.
It produced identically the same results as your code here...
>> x=1:10;
>> for n=1:10
disp('how many iteration done now');
final(n,:)=x(n);
z(n,:)=x(n)+2;
if x(n)>= 5, break, end
end
how many iteration done now
how many iteration done now
how many iteration done now
how many iteration done now
how many iteration done now
>> [final z(:,5)]
ans =
1 3
2 4
3 5
4 6
5 7
>>
Your code above w/o the intermediate results while mine produced:
>> final=find(x>=5,1);
>> z=x(1:final)+2;
>> final
final =
5
>> z.'
ans =
3
4
5
6
7
>>
with the assumption as before that there really wasn't any need for final to be an array and the desired result was that for z when the condition is met.
Now, if that isn't the result wanted, then the actual problem definition needs clarification. So, if the problem is really the statement "I meant, I need from x array all the numbers that met the conditions written." then that's just
z=x(x>=5);
using logical addressing/indexing, one of the most powerful of Matlab features. That isn't what your code actually does, but maybe that was the real question underneath the one you asked.