“if” statement using “or” operator.
3,509 views (last 30 days)
I have a very simple question....and I have been working on it for some time but cannot figure it out. This is essentially what I would LIKE to say:
for r = 1:length(FreqSec)-1
if FreqSec(1,r+1) > FreqSec(r)*1.01 "OR" FreqSec(1,r+1) <FreqSec(r)*0.99
LagStart = [FreqSec(1,r) r];
FreqSec is a vector with lots and lots of values....generally within the range of 0.99 and 1.01, except for in a certain interval. I want to detect the exact index point at which the values start changing from the 0.99 to 1.01 range.
Thanks for the help in advance :)
Walter Roberson on 3 Feb 2012
if FreqSec(1,r+1) > FreqSec(r)*1.01 | FreqSec(1,r+1) <FreqSec(r)*0.99
if FreqSec(1,r+1) > FreqSec(r)*1.01 || FreqSec(1,r+1) <FreqSec(r)*0.99
The first of these is more general. The second of these, , is the short-circuiting OR that does not bother to evaluate the second expression if it already knows the final result after the first operation. The operator can only be used between expressions that produce scalar outputs.
More Answers (2)
Geoff on 3 Feb 2012
So you want the last index within the valid range?
I don't know why you are multiplying by 1.01 and 0.99. Perhaps you have described the problem incorrectly or that is the cause of your difficulties. What I think you are trying to do is this:
idx = find( FreqSec >= 0.99 & FreqSec <= 1.01, 1, 'last' )
LagStart = [FreqSec(1,idx) idx];
Or, since it's symmetric:
idx = find( abs(FreqSec-1) <= 0.01, 1, 'last' )
If instead you want the index of the first out-of-range value, use:
idx = find( abs(FreqSec-1) > 0.01, 1, 'first' )
The parameter 'first' is optional, but good for clarity.
Rehman Tabasum on 30 Apr 2021
this is my code for the start button gui for countdown timer so anyone know how to stop coutdown timer when it is runnnig