# Skip analysis when you have NaN or 0

3 ビュー (過去 30 日間)
Luccas S. 2021 年 12 月 26 日
コメント済み: Luccas S. 2021 年 12 月 26 日
I think I am having problems with my analysis result due to the presence of NaN and 0. How to skip analysis when I have NaN or 0?
p_fix = [];
for n = 4:size(t,1)
X = [Ia(n-1,1) Ia(n-2,1) ; Ia(n-2,1) Ia(n-3,1)];
future = [Ia(n,1) ; Ia(n-1,1)];
C = X\future;
Ia_future(n,1) = C(1,1)*Ia(n,1)+C(2,1)*Ia(n-1,1);
PE(n,1)=Ia(n,1)+Ia_future(n,1);
p(n,1)=(1+0.2)*max(PE(n-1,1));
if isempty(p_fix) && PE(n-1,1)>p(n,1)
p_fix = p(n,1);
if (mod(N,5)==0) && (PE(n-1,1)>p_fix)
fprintf('Accuses IC\n')
else
fprintf('Does not accuse IC\n')
p_fix = [];
end
end
Basically I wanted this part to be ignored when (PE=NaN or 0) and (p = NaN or 0) happened. And perform the PE and p calculations again.
if isempty(p_fix) && PE(n-1,1)>p(n,1)
p_fix = p(n,1);
if (mod(N,5)==0) && (PE(n-1,1)>p_fix)
fprintf('Accuses IC\n')
else
fprintf('Does not accuse IC\n')
p_fix = [];

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### 採用された回答

Image Analyst 2021 年 12 月 26 日
Do you want to break out of the loop and go somewhere else,
PE(n,1)=Ia(n,1)+Ia_future(n,1);
p(n,1)=(1+0.2)*max(PE(n-1,1));
if (isnan(PE(n, 1)) || PE(n, 1) == 0) && (isnan(p(n, 1)) || p(n, 1) == 0)
break; % Quit the loop totally
end
or do you want to continue with the next iteration of the loop?
PE(n,1)=Ia(n,1)+Ia_future(n,1);
p(n,1)=(1+0.2)*max(PE(n-1,1));
if (isnan(PE(n, 1)) || PE(n, 1) == 0) && (isnan(p(n, 1)) || p(n, 1) == 0)
continue; % Skip to bottom of loop and continue the loop with the next iteration.
end
##### 2 件のコメント表示非表示 1 件の古いコメント
Luccas S. 2021 年 12 月 26 日
Oh sorry, I ended up deleting the comment because I had found my mistake. That's right, it was the position of the end.

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### その他の回答 (1 件)

William Rose 2021 年 12 月 26 日
a=1; %try replacing "1" with "0" or "NaN"
if ~(isnan(a) || a==0)
disp('a is not NaN and is not 0.')
else
disp(a)
end
a is not NaN and is not 0.
Try.

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