# I get index error, but the correct result. How can I avoid this?

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Alicia Cuber Caballero on 1 Aug 2019
Commented: KALYAN ACHARJYA on 1 Aug 2019
I am trying to solve this Cody problem where I need to remove all rows that contain NaN-s.
My code looks like this:
[m,n]=size(A)
b = abs(sum(A'))
b1=length(b)
if n==1
iter2 = 0
for l=1:m
if isnan(A(l,:))==1
A(l,:)=[]
B=A
elseif isnan(A(l))==0
end
end
elseif n>1
iter1 = 0
for i=1:b1
if isnan(b(i))==1
elseif isnan(b(i))==0
iter1 = iter1+1
B(iter1,:) = A(i,:)
end
end
end
where, in the overarching if, the if part works for column matrices, and the elseif for any other matrices. (I know there are probably better ways to solve it, but I'm trying to figure them out on my own).
It works. However, I get the alert that in
if isnan(A(l,:))==1
"Index exceeds the number of array elements (4)" for the column vector A = [ 1 3 6 NaN 3 NaN]'. I am assuming it is becauyse for l=5, I still get a four element vector. However, the result is correct, and if it weren't for that error message, I believe the Cody problem would accept my answer.

KALYAN ACHARJYA on 1 Aug 2019
Edited: KALYAN ACHARJYA on 1 Aug 2019
"I am trying to solve this Cody problem where I need to remove all rows that contain NaN-s".
A =[1 3 6 NaN 3 NaN]'
idx=find(isnan(A));
A(idx,:)=[]
Result:
A =
1
3
6
3
##### 2 CommentsShowHide 1 older comment
KALYAN ACHARJYA on 1 Aug 2019
A(isnan(A),:)=[]
I am sure you have more simpler one.
Thanks Always

Alicia Cuber Caballero on 1 Aug 2019
Thank you! That's a simpler solution :)