List index out of range - Python

Question

def Task4():
 import random
 t = True
 list1 = []
 list2 = []
 rand = random.randint(1,6)
 list1.append(rand)
 print(list1)
 for x in range(0,5):
 if list1[0] == (rand):
 list1.pop(0)
 else:
 list2.append(rand)
 print(list2)
 list1.pop(0)

Can't workout why if list1[0] == (rand): keeps coming up with a list index out of range.

• grc: You need to check that the first element exists (i.e. the list isn't empty) before indexing it -> if list1 and list1[0] == rand
• Maroun: Debug your code and you'll know why. You can debug it using a pencil and a paper.
• Andrew: You appended to list1 only one element but in the loop you tried to pop 5 elements.

Answer 1

Let’s see what happens to list1:

list1 = []

Ok, the list is created, and it’s empty.

rand = random.randint(1,6)
list1.append(rand)

rand is added to the list, so at this point list1 = [rand].

for x in range(0,5):

This will loop four times; so let’s take a look at the first iteration:

if list1[0] == (rand):

Since list1 = [rand], list1[0] is rand. So this condition is true.

list1.pop(0)

The list element at index 0 is removed; since list1 only contained one element (rand), it’s now empty again.

for x in range(0,5):

Second iteration of the loop, x is 1.

if list1[0] == (rand):

list1 is still empty, so there is no index 0 in the list. So this crashes with an exception.

---

At this point, I would really like to tell you how to solve the task in a better way, but you didn’t specify what you are trying to do, so I can only give you a hint:

When you remove items from a list, only iterate for as often as the list contains elements. You can either do that using a while len(list1) loop (which will loop until the list is empty), or by explicitely looping over the indexes for i in len(list1). Of course, you can also avoid removing elements from the list and just loop over the items directly using for x in list1.

Answer 2

• when x is 0: -> you pop the only existing list item, thus the list is empty afterwards

• when x is 1: -> you try to access list1[0], which doesn't exist -> list index out of range error

• DonCristobal: Dear Downvoter, I'm interested in improving my answers, thus: What's so bad about the answer that you downvoted it? Yes, it could certainly be more complete, but given that it's a simple question, I thought a simple answer would do.
• Stefan Pochmann: I think your answer is perfect. Curious about the downvote as well. Anyway, have my upvote.

Answer 3

truth is evident with debugging with print and try except:
def Task4():
 import random
 t = True
 list1 = []
 list2 = []
 rand = random.randint(1,6)
 list1.append(rand)
 for x in range(0,5):
 try:
 print("list1 before if and pop",list1)
 if list1[0] == (rand):
 list1.pop(0)
 print("list1 after pop",list1)
 else:
 list2.append(rand)
 print("list2 {}".format(list2))
 list1.pop(0)
 except IndexError as e:
 print("Index Error {}".format(e))
 break
Task4()
list1 before if and pop [3]
list1 after pop []
list1 before if and pop []
Index Error list index out of range