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

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.

Thanks in advance!

Answer 1

[KALYAN ACHARJYA]

"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

[madhan ravi]

find() can be omitted here

[KALYAN ACHARJYA]

Agreed @Madhan

A(isnan(A),:)=[]

I am sure you have more simpler one.

Thanks Always

Answer 2

[Alicia Cuber Caballero]

Thank you! That's a simpler solution :)