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I have written a simple python program

l=[1,2,3,0,0,1]
for i in range(0,len(l)):
       if l[i]==0:
           l.pop(i)

This gives me error 'list index out of range' on line if l[i]==0:

After debugging I could figure out that i is getting incremented and list is getting reduced. However, I have loop termination condition i < len(l). Then why I am getting such error?

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l as you iterate over it, so as you approach the end of your indices in the range statement, some of those indices are no longer valid.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 error, 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  But in case you mention an index in your code that is outside the range of the list, you will encounter an IndexError. “ List index out of range ” error occurs in Python when we try to access an undefined element from the list. The only way to avoid this error is to mention the indexes of list elements properly.

The expression len(l) is evaluated only one time, at the moment the range() builtin is evaluated. The range object constructed at that time does not change; it can't possibly know anything about the object l.

P.S. l is a lousy name for a value! It looks like the numeral 1, or the capital letter I.

Python Loop: List Index Out of Range, Traceback (most recent call last): File “python”, line 7, in The error occurs as soon as it hits x[3] as this is out of range, as are the other possibilities. In essence i  Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Learn more Python 3 : IndexError: list index out of range [duplicate]

You're changing the size of the list while iterating over it, which is probably not what you want and is the cause of your error.

Edit: As others have answered and commented, list comprehensions are better as a first choice and especially so in response to this question. I offered this as an alternative for that reason, and while not the best answer, it still solves the problem.

So on that note, you could also use filter, which allows you to call a function to evaluate the items in the list you don't want.

Example:

>>> l = [1,2,3,0,0,1]
>>> filter(lambda x: x > 0, l)
[1, 2, 3]

Live and learn. Simple is better, except when you need things to be complex.

IndexError: list index out of range, How to fix the list index out of range. In python “list index out of range” error occurs when we try to access an undefined element from the list. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Learn more Python: IndexError: list index out of range

What Mark Rushakoff said is true, but if you iterate in the opposite direction, it is possible to remove elements from the list in the for-loop as well. E.g.,

x = [1,2,3,0,0,1]
for i in range(len(x)-1, -1, -1):
    if x[i] == 0:
        x.pop(i)

It's like a tall building that falls from top to bottom: even if it is in the middle of collapse, you still can "enter" into it and visit yet-to-be-collapsed floors.

Indexerror: list Index Out of Range in Python, To get rid of this error you can try the following ways:- if guess in word: print("\​nYes!", guess, "is in the word!") new = "". i = 0. Generally it means that you are providing an index for which a list element does not exist. E.g, if your list was [1, 3, 5, 7], and you asked for the element at index 10, you would be well out of bounds and receive an error, as only elements 0 through 3 exist. share. Share a link to this answer.

I think the best way to solve this problem is:

l = [1, 2, 3, 0, 0, 1]
while 0 in l:
    l.remove(0)

Instead of iterating over list I remove 0 until there aren't any 0 in list

Python error: string index out of range python, In this post, we will see about Indexerror: list Index Out of Range in Python. You will get this error when you are trying to access an index that  IndexError: list assignment index out of range List elements can be modified and assigned new value by accessing the index of that element. But if you try to assign a value to a list index that is out of the list’s range, there will be an error. You will encounter an IndexError list assignment index out of range.

Fix for Python Error: List Index Out Of Range, I am trying to execute the following python code: def construct(s, k, a): index = 0 # the following error: IndexError: list index out of range. The way Python indexing works is that it starts at 0, so the first number of your list would be. You would have to print, as the starting index is 0 and therefore line 53 is. Subtract 1 from the value and you should be fine.

Indexerror: list Index Out of Range, IndexError: list index out of range. Ps: i tried to install this module on bitnami VM and it works whithout any problem but when i tried to installe it  I think I have my program completed, but it doesn't work. I'm trying to write a == winning_numbers[i]: IndexError: list index out of range

Python error "IndexError: list index out of range", Enter list of element:2,3,4,5,6 Enter search element:5 results: C:\Python36\python.exe M:/python/ProvingGround/newStuff/FindIndex1.py Enter list of element:2,3,4,5,6

Comments
  • "I have loop termination condition i < len(l)" Why do you say that? Where in your code do you see that?
  • @ S. Lott , i in range(0,len()) means 'i will go upto len(l)-1'
  • Another python tip - you could have just written range(len(l)), as 0 is the default starting value.
  • From PEP 8: Never use the characters l (lowercase letter el), O (uppercase letter oh), or I (uppercase letter eye) as single character variable names. python.org/dev/peps/pep-0008
  • @atv: What makes you think range(0,len(l)) has a result that varies when l is changed? Why do you think that? Where did you read it?
  • You don't even need the lambda, since 0 evaluates to False. filter(None, l)
  • @Steve Losh - This is what I love about SO... Learning simple little tricks like that to save me keystrokes in the longrun! Thanks!
  • -1: Why do you point a beginner to this legacy way? List comprehensions are now the preferred way to do this.
  • -1 for filter or map. You should always use a list comprehension if it will do the job.
  • You guys are right, but filter is still an important part of the toolkit for when you need more advanced evaluation.
  • From Review: Hi, please don't answer just with source code. Try to provide a nice description about how your solution works. See: How do I write a good answer?. Thanks
  • Not only is this wrong, it's addressing a different problem. OP has an issue where they themselves are removing elements of the list as they iterate over the list. Also, i starts at 0 in OP's question. Your example would have OP start iterating at the last element of the array, and then wrapping back around into the first.
  • Try this (since this is a comment, you're on your own to figure out the formatting in your IDE): python l=[1,2,3,0,0,1] for i in range(0,len(l)): print("element:" + str(i - 1)) print(l[i-1])