# “if” statement using “or” operator.

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Sarah 2012 年 2 月 2 日

Hello everyone,
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];
break;
end
end
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 :)

#### 1 件のコメント

Aaron Eldridge 2016 年 4 月 28 日
The "||" counts as an or on matlab

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### 採用された回答

Walter Roberson 2012 年 2 月 3 日
if FreqSec(1,r+1) > FreqSec(r)*1.01 | FreqSec(1,r+1) <FreqSec(r)*0.99
or
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.

#### 1 件のコメント

Harry MacDowel 2013 年 11 月 23 日
good clarification on the difference between | and !

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### その他の回答 (1 件)

Geoff 2012 年 2 月 3 日
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.
-g-

#### 1 件のコメント

Geoff 2012 年 2 月 3 日
Oops, those statements with 'last' are incorrect. You are probably better off doing the 'first' statement and subtracting 1.

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