# if statement is not working

조회 수: 2 (최근 30일)
댓글: Abolfazl Chaman Motlagh 2021년 12월 9일
Hi all,
I have the follwoing code that runs fine but no output of the if statement for some reason:
theta = [-90.0000 -63.4746 -52.1364 -43.1736 -35.3765 -28.2737 -21.6183 ...
-15.2575 -9.0847 -3.0170 3.0170 9.0847 15.2575 21.6183 28.2737 ...
35.3765 43.1736 52.1364 63.4746 90.0000];
x0 = 0.1250;
y0 = 0;
if theta >= 41.19 & theta <= 90
inter_r = -(1/sin(theta))*[-sin(theta) cos(theta)*(x0-1)-y0.*sin(theta)];
end
An error of (Unrecognized function or variable 'inter_r') occurs.
Any help would be appreicted.
Thanks.

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### 채택된 답변

Jon 2021년 12월 8일
편집: Jon 2021년 12월 8일
The problem is that your if statement will only be true if all of the elements of theta satisfy the condition.
If you only want to assign inte_r for elements where the condition is met you could do something like this:
theta = [-90.0000 -63.4746 -52.1364 -43.1736 -35.3765 -28.2737 -21.6183 ...
-15.2575 -9.0847 -3.0170 3.0170 9.0847 15.2575 21.6183 28.2737 ...
35.3765 43.1736 52.1364 63.4746 90.0000];
x0 = 0.1250;
y0 = 0;
% assign logical vector which is true for elements that are in range
inRange = theta >= 41.19 & theta <= 90
% select elements of theta that are in range
thetaInRange = theta(inRange)
% assign function values just for elements that are in range
inter_r = -(1.0 ./sin(thetaInRange)).*...
##### 댓글 수: 2없음 표시없음 숨기기
Jon 2021년 12월 8일
You also seem to have another problem in your function if theta is a vector. First of all you aren't consistent with your use of element by element multiplication and division. You use it in the final multiplication, but if theta is a vector then you would also need it for
-(1/sin(theta))
and also the next multiply.
You would also have a problem, even with element by element multiplication of
-(1./sin(theta)).*[-sin(theta) cos(theta)*(x0-1)-y0.*sin(theta)]
because 1 ./ sin(theta) has as many element as theta, but you multiply it by a new vector, defined by your square brackets that has twice as many elements.
편집: Lama Hamadeh 2021년 12월 9일
Thanks Jon. That did solve the problem.

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### 추가 답변 (1개)

Abolfazl Chaman Motlagh 2021년 12월 8일
the condition theta >= 41.19 & theta <= 90 is a logical vector with 20 values. if you want to calculate such a formula for those theta that satisfy the condition. you can use for loop or vectorize code.
theta = [-90.0000 -63.4746 -52.1364 -43.1736 -35.3765 -28.2737 -21.6183 ...
-15.2575 -9.0847 -3.0170 3.0170 9.0847 15.2575 21.6183 28.2737 ...
35.3765 43.1736 52.1364 63.4746 90.0000];
x0 = 0.1250;
y0 = 0;
for i=1:numel(theta)
if theta(i)>= 41.19 & theta(i)<= 90
inter_r(i,1:2) = -(1/sin(theta(i)))*[-sin(theta(i)) cos(theta(i))*(x0-1)-y0.*sin(theta(i))];
else
inter_r(i,1:2) = nan; % for example
end
end
or vectorize
Condition = theta >= 41.19 & theta <= 90;
inter_r(Condition,1:2)=repmat(-(1./sin(theta(Condition)))',[1 2]).* ...
[-sin(theta(Condition))' (cos(theta(Condition))*(x0-1)-y0.*sin(theta(Condition)))'];
inter_r(~Condition,:)=nan;
##### 댓글 수: 2없음 표시없음 숨기기
Thanks for the reply. I tried the for loop method (which I find it very clear to use) but I am getting wrong output, I'm afraid.
Abolfazl Chaman Motlagh 2021년 12월 9일
i see you accept another answer. it's good to hear it solves your problem. in case you still want to tell me what exactly is wrong in this answer, i can gladly help.

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