Can any one explain me why there is list index out of range error in this code

Question

def remove_odd(list1):
for index in range(len(list1)):
if list1[index]%2==0:
list1.remove(list1[index])
list1=[1,2,3,4,5]
remove_odd(list1)
print(list1)

Answer 1

Because you are working with len(original) from a list that you are already deleting elements from. I think that's it.

I just noticed that also, you would be deleting the even ones, not the odd ones haha
Maybe this can help you:
i=0
while i<=len(arr):
if arr[i]%2!=0:
del arr[i]
i+=1

Answer 2

Thanks for anwers and thanks Cristian Gabriel for your explanation i completely understand it

Answer 3

Instead of removing items from the original `list` you have an option to filter the `list` instead. You can use filter() function or `list` comprehension to do this 👍

P.S. I think the function name 'remove_odd()' doesn't match what it does. The condition in the loop appears to be targeting even numbers instead of the odd ones : )

Also, as a future reference, use appropriate tags

https://code.sololearn.com/W3uiji9X28C1/?ref=app