Python: How to avoid 'list assignment index out of range'

Question

Tried a search but couldn't locate anything. I'm looking for the simplest to replace

if myList and len(myList)>4

with something shorter:

myList = [1,2,3,4,5]

if myList and len(myList)>4: 
    print myList[4], len(myList)

since

if not myList[4]

doesn't do job.

if len(myList) > 4: is shorter (and equivalent as long as myList is actually a list) ... — mgilson
– mgilson, Commented Apr 11, 2014 at 5:21
Normally you could just do if len(myList) > 4. If you need the if myList, it's probbaly because of something else you're not showing here. — BrenBarn
– BrenBarn, Commented Apr 11, 2014 at 5:22
@Sputnix be explicit then. if myList is not None and len(myList)> 4: Frankly I'd test for None-ness first and handle that case separately than just if len(myList) <= 4 — Adam Smith
– Adam Smith, Commented Apr 11, 2014 at 5:24
if not myList Falses on an empty list too which works in my situation well. But thanks! — alphanumeric
– alphanumeric, Commented Apr 11, 2014 at 5:25

Adam Smith · Accepted Answer · 2014-04-11 05:21:45Z

4

Not really sure what's being asked, but I'll hazard a guess that EAFP is the right concept here.

try:
    myList[4]
except IndexError:
    # handle it

Basically, try to do the thing you want to do with index 4. If you can't, and it's because of an IndexError, handle it the way you'd handle it if your if len(myList) > 4 failed.

answered Apr 11, 2014 at 5:21

Adam Smith

54.6k13 gold badges85 silver badges120 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

alphanumeric Over a year ago

Good suggestion too. Wishing for shorter!

rubayeet · Accepted Answer · 2014-04-11 05:40:49Z

3

You can try extending the built-in type list and override __getitem__() to get your intended behavior.

class MyList(list):

    def __getitem__(self, index):
        try:
            return super(MyList, self).__getitem__(index)
        except IndexError:
            return None

mylist = MyList([1,2,3,4,5])
print mylist[4] #prints 5
print mylist[6] is None #prints True

Personally, I'd go for Adam Smith's suggestion.

answered Apr 11, 2014 at 5:40

rubayeet

9,4009 gold badges48 silver badges55 bronze badges

1 Comment

alphanumeric Over a year ago

I actually like yours! Thanks!

Nishant Nawarkhede · Accepted Answer · 2014-04-11 06:00:36Z

0

Try this,

>>> if len(myList)> 4 : print myList[4]

Update

>>> if len(myList)> 4 and isinstance(myList,list) : print myList[4]

edited Apr 11, 2014 at 6:00

answered Apr 11, 2014 at 5:27

Nishant Nawarkhede

8,41813 gold badges61 silver badges83 bronze badges

4 Comments

alphanumeric Over a year ago

len(None) throws TypeError: object of type 'NoneType' has no len()

alphanumeric Over a year ago

well, in "a real life" there will be no way to anticipate what myList is.

alphanumeric Over a year ago

Thanks! A check for None needs to proceed len() otherwise things get screwed really fast! :)

Adam Smith Over a year ago

You should also use isinstance(myList,list) rather than type(myList) is list to preserve inheritance.

Collectives™ on Stack Overflow

Python: How to avoid 'list assignment index out of range'

3 Answers 3

1 Comment

1 Comment

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

1 Comment

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related