Python:List Index Out of Range Error While Iteratively Popping Elements

python : list index out of range error while iteratively popping elements

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l as you iterate over it, so as you approach the end of your indices in the range statement, some of those indices are no longer valid.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 True.

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

(Python) List index out of range - iteration

You're modifying the list you're iterating over. If you do that, the size of the list shrinks, so eventually lst[i] will point beyond the list's boundaries.

>>> lst = [1,2,3]
>>> lst[2]
3
>>> lst.remove(1)
>>> lst[1]
3
>>> lst[2]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list index out of range

It's safer to construct a new list:

return [item for item in lst if item[0]!=1 and item[1]!=1]

IndexError: list index out of range in very simple 3 lines of Python code

It's not clear what exactly you're trying to do. When you iterate over an iterable like a list with

for i in my_list:

each i is each member of the list, not the index of the member of the list. So, in your case, if you want to print each member of the list, use

for i in my_list:
    print(i)

Think about it: what if the 3rd member of the list was 9, for example? Your code would be trying to print my_list[9], which doesn't exist.

How to Fix List Index out of range in python?

You may try this instead:

def getMoneySpent(keyboards, drives, b):
    q = []
    for k in keyboards :
        for d in drives :
            price = k + d
            if price <= b :
                q.append(price) 

    return max(q) if q else -1

Instead of removing elements, just don't add them in the first place...

Unexpected error: IndexError: list index out of range when trying to access a list within a function

You pass empty list : myMax([]) to the function and you point to the first element of this empty list L[0]. You need to check if the list L is not empty.

Adding the below as the first lines of the function will help you to protect against None or empty list

if L is None or len(L) == 0:
  raise ValueError('Input list can not be None or empty')

Why do I get List index out of range when trying to add consecutive numbers in a list using for i in list?

1. In your for loop, you're iterating through the elements of a list a. But in the body of the loop, you're using those items to index that list, when you actually want indexes.
   
   Imagine if the list a would contain 5 items, a number 100 would be among them and the for loop would reach it. You will essentially attempt to retrieve the 100th element of the list a, which obviously is not there. This will give you an IndexError.

   We can fix this issue by iterating over a range of indexes instead:

   for i in range(len(a))
   

   and access the a's items like that: a[i]. This won't give any errors.

2. In the loop's body, you're indexing not only a[i], but also a[i+1]. This is also a place for a potential error. If your list contains 5 items and you're iterating over it like I've shown in the point 1, you'll get an IndexError. Why? Because range(5) is essentially 0 1 2 3 4, so when the loop reaches 4, you will attempt to get the a[5] item. Since indexing in Python starts with 0 and your list contains 5 items, the last item would have an index 4, so getting the a[5] would mean getting the sixth element which does not exist.

   To fix that, you should subtract 1 from len(a) in order to get a range sequence 0 1 2 3. Since you're using an index i+1, you'll still get the last element, but this way you will avoid the error.

3. There are many different ways to accomplish what you're trying to do here. Some of them are quite elegant and more "pythonic", like list comprehensions:

   b = [a[i] + a[i+1] for i in range(len(a) - 1)]
   

   This does the job in only one line.

Python 3 : IndexError: list index out of range

When you start to remove items from a list, it changes in size. So, the ith index may no longer exist after certain removals:

>>> x = ['a', 'b', 'c', 'd', 'e']
>>> x[4]
'e'
>>> x.pop()
'e'
>>> x[4]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list index out of range

A simpler way to remove duplicate items is to convert your list to a set, which can only contain unique items. If you must have it as a list, you can convert it back to a list: list(set(X)). However, order is not preserved here.

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If you want to remove consecutive duplicates, consider using a new array to store items that are not duplicates:

unique_x = []
for i in range(len(x) - 1):
    if x[i] != x[i+1]:
        unique_x.append(x[i])
unique_x.append(x[-1])

Note that our range bound is len(x) - 1 because otherwise, we would exceed the array bounds when using x[i+1].

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