How can I check if a list index exists?

Question

Seems as though

if not mylist[1]:
    return False

Doesn’t work.

Asked By: Sundrah

||

Source

Answer 1

In the case of integer-indexed lists, I’d simply do

if 1 < len(mylist):
  ...

For dicts, you can of course do

if key in mydict:
  ...

Answered By: csl

Answer 2

You just have to check if the index you want is in the range of 0 and the length of the list, like this

if 0 <= index < len(list):

it is actually internally evaluated as

if (0 <= index) and (index < len(list)):

So, that condition checks if the index is within the range [0, length of list).

Note: Python supports negative indexing. Quoting Python documentation,

If i or j is negative, the index is relative to the end of the string: len(s) + i or len(s) + j is substituted. But note that -0 is still 0.

It means that whenever you use negative indexing, the value will be added to the length of the list and the result will be used. So, list[-1] would be giving you the element list[-1 + len(list)].

So, if you want to allow negative indexes, then you can simply check if the index doesn’t exceed the length of the list, like this

if index < len(list):

Another way to do this is, excepting IndexError, like this

a = []
try:
    a[0]
except IndexError:
    return False
return True

When you are trying to access an element at an invalid index, an IndexError is raised. So, this method works.

Note: The method you mentioned in the question has a problem.

if not mylist[1]:

Lets say 1 is a valid index for mylist, and if it returns a Falsy value. Then not will negate it so the if condition would be evaluated to be Truthy. So, it will return False, even though an element actually present in the list.

Answered By: thefourtheye

Answer 3

In the EAFP style of Python:

try:
    mylist[1]
except IndexError:
    print "Index doesn't exist!"

Answered By: Ben

Answer 4

Alternative (but somewhat slower) way of doing it:

if index not in range(len(myList)):
    return False

It gets a bit more verbose when accounting for negative indices:

if index not in range(-len(myList), len(myList)):
    return False

Answered By: Zlatko Karakaš

Answer 5

assert len(mylist) >= abs(index) + int(index >= 0), "Index out of range"

or

assert len(mylist) > abs(index) - int(index < 0), "Index out of range"

Answered By: Ni.Go.

Answer 6

Or you can do:

if index in dict(enumerate(mylist)):
    return True

Although is will probably be even less efficient than range(len(mylist)). Maybe someone should propose a keys() method for lists that returns the range of the keys in PEP.

Answered By: Roho

Answer 7

The following approach returns the True result for index == 0 and for negative index values (if such an index is valid for the list, for example listIn[-2] for [0, 1, 2]):

def isInListRange(listIn, index):
  """ Description: Function to detect if list index out of range
    Import: from shared.isInListRange import isInListRange
    Test: python -m shared.isInListRange
  """

  try:
    return True if (index == 0 and len(listIn) > 0) or listIn[index] else False
  except:
    return False

Answered By: Roman

How can I check if a list index exists?

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