Substituting a number for NaN in anonymous function

조회 수: 12(최근 30일)
I'm trying unsuccessfully to substitute a number for NaN in anonymous function. Here it's an example of the problem. Bear with it's silliness please:
clc;
clear all;
f=@(x) x*1/x;
g=@(x) (~isnan(f(x))).*f(x) + (isnan(f(x))).*1;
I'd expect that g(0)=1, but it's still NaN. What is wrong in the way that I defined g?
  댓글 수: 2
Guillaume
Guillaume 2015년 1월 13일
It's getting a bit messy. You should have accepted the answer that helped you the most and started a new question. Don't pile questions on top of questions, because now there's no appropriate place to answer your new question.
Anyway, the reason why it does not work in the second case is that isnan(fval) == 1 (which is just the same as isnan(fval)) is a logical vector equal to
1 0 0 0 0 0 0 0 0 0 0
If you pass a vector to if it will only evaluate to true if and only if all elements are not zeros. Therefore your expression is false.
To fix this:
if any(isnan(fval))
out = 1;
else
out = fval;
end
Or if you just want to replace the Nans by 1:
fval(isnan(fval)) = 1; %no need for if

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채택된 답변

Alfonso Nieto-Castanon
Alfonso Nieto-Castanon 2015년 1월 13일
편집: Alfonso Nieto-Castanon 2015년 1월 13일
There may be "cleaner" ways to do this but one possibility would be:
g = @(x) [f(x) 1]*sparse(1+isnan(f(x)),1,1,2,1);
EDIT: and as others have pointed out the problem with your original g function is that, when f(x) is NaN, you get 0*NaN + 1*1 which still evaluates to NaN...

추가 답변(2개)

John Petersen
John Petersen 2015년 1월 13일
You have .*f(x) in g(x), which is still giving you a NaN

Star Strider
Star Strider 2015년 1월 13일
If you want to define L’Hospital’s rule, you have to define it specifically. In the IEEE standard that MATLAB implements, 0/0 is NaN.
See if this works in your application:
n = @(x) 2.*x; % Numerator Function
d = @(x) x; % Denominator Function
Lh = @(n,d,x) (n(x+1E-12)-n(x)) ./ (d((x+1E-12)-d(x))); % L’Hospital’s RUle
Lh0 = Lh(n,d,0) % Evaluating AT 0
Lh2 = Lh(n,d,2) % Evalutaing At 2

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