[SOLVED] List Index Out of Range Meaning, How to Fix | Python Program
[SOLVED] List Index Out of Range Meaning, How to Fix | Python Program.
If you are frequent in Python programming, you will definitely come across this error message, “indexerror: list index out of range“. You might want to ask, what does list index out of range mean? and how to fix the list index out of range in python?
Well, in this article, I will share with you everything you need to know about this common list assignment index out of range error in Python. Here, you will follow this guide and fic the error once and for all. After reading this post, you will find yourself at a position where you can even teach others on how to fix this error message.
Why do you have “List Index Out of Range” Error?
Table of Contents
If you write a simple python program, as follow;
l=[1,2,3,0,0,1] for i in range(0,len(l)): if l[i]==0: l.pop(i)
This will give you an error message ‘list index out of range’ on line
And you will observe that
i is getting incremented and the list is getting reduced.
However, you may have a loop termination condition
i < len(l).
See other explanations here:
check out the following code in python
- >>> even = [ 2, 4, 6, 8 ]
- >>> even
- >>> even
- >>> even
- Traceback (most recent call last):
- File “<pyshell#7>”, line 1, in <module>
- IndexError: list index out of range
So in the above code, we have defined an even list that contains 4 elements with starting index 0 and last index 3.
clearly, this means you can’t access any elements in even list if you go beyond your last index value.
And if u try, then u will get a list index out of range error.
List Index Out of Range Meaning
List index out of range means that you are providing an index for which a list element does not exist. Index out of range means that the code is trying to access a matrix or vector outside of its bounds.
In python list is mutable, so the size is not fixed.
Due to size is not fixed, the available index is greater than the assigned index for a list(available index > assigned index).
2 Ways to Fix List Index Out of Range Error
See how to handle indexerror list index out of range in python here. You are reducing the length of your list
l as you iterate over it, so as you approach the end of your indices in the range statement, some of those indices are no longer valid.
It looks like what you want to do is:
l = [x for x in l if x != 0]
which will return a copy of
l without any of the elements that were zero (that operation is called a list comprehension, by the way). You could even shorten that last part to just
if x, since non-zero numbers evaluate to
There is no such thing as a loop termination condition of
i < len(l), in the way you’ve written the code, because
len(l) is precalculated before the loop, not re-evaluated on each iteration. You could write it in such a way, however:
i = 0 while i < len(l): if l[i] == 0: l.pop(i) else: i += 1
If you try to access the empty or None element by pointing available index of list, Then you will get the List index out of range error.
open python and dirty your hands with this below code, to have a better understanding.
>> x = list(‘1234’)
[‘1’ , ‘2’ , ‘3’, ‘4’]
>> length = len(x)
Index Error: list index out of range
Final Answer [SOLVED]: If you have a list with 53 items, the last one is
thelist because indexing starts at 0 (zero).
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