how do i solve "list index out of range" error in python? (IndexError)

Question

I tried to create a sorting algorithm but it didn’t work.

Here’s my code:

from random import *

sort = [9, 7, 4, 5, 8, 3, 2, 1, 6, 10]

def sort_array(array):
    for i in range(len(array)):
        if array[i + 1] < array[i + 2]:
            array[i + 1], array[i + 2] = array[i + 2], array[i + 1]
        else:
            pass

    print(array)

sort_array(sort)

I get this error when I’m trying to run it:

if array[i + 1] < array[i + 2]: IndexError: list index out of range

I was expecting it to just sort the array.

Asked By: kdkj

||

Source

Answer 1

You should try an iteration until the array length – 2 since i+2 will go out of bound throwing an error try with for i in range(len(array)-2).

Anyway i don’t think your code will work.

Read about Sorting algorithms or if you’re too lazy there’s a spoiler:


sort = [9, 7, 4, 5, 8, 3, 2, 1, 6, 10]


def sort_array(arr):
    for i in range(1, len(arr)):
        key = arr[i]
        j = i - 1
        while j >= 0 and key < arr[j]:
            arr[j + 1] = arr[j]
            j -= 1
        arr[j + 1] = key
    print(arr)


sort_array(sort)

Answered By: Matt

Answer 2

Well, if you print(i) inside of your loop (it is better to use log), you will see that it reaches 8 and then breaks. You have a list of length 10, thus its maximum index value can be 9, while you are calling array[8 + 2] which doesn’t exist and it is out of your list’s bounds.

Moreover, your current sorting algorithm doesn’t move the zero element (in your case "9"). FYI, I don’t think that your array will be sorted in one loop, so you may want to modify your algorithm.

P.S.

Don’t import this way from random import *. It is a bad practice.
Check enumerate(array) instead of range(len(array)). It is a better practice

Answered By: Thoughtful_monkey

how do i solve "list index out of range" error in python? (IndexError)

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