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

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Alicia Cuber Caballero il 1 Ago 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.
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### Risposta accettata

KALYAN ACHARJYA il 1 Ago 2019
Modificato: KALYAN ACHARJYA il 1 Ago 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 CommentiMostra NessunoNascondi Nessuno
madhan ravi il 1 Ago 2019
find() can be omitted here
KALYAN ACHARJYA il 1 Ago 2019
A(isnan(A),:)=[]
I am sure you have more simpler one.
Thanks Always

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### Più risposte (1)

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