Getting a default value on index out of range in Python [duplicate]

Answer 1

If you are looking for a quick hack for reducing the code length characterwise, you can try this.

a=['123','2',4]
a.append('sss') #Default value
n=5 #Index you want to access
max_index=len(a)-1
b=a[min(max_index, n)]
print(b)

But this trick is only useful when you no longer want further modification to the list

Answer 2

Using try/catch?

try:
    b=a[4]
except IndexError:
    b='sss'

Answer 3

Since this is a top google hit, it's probably also worth mentioning that the standard "collections" package has a "defaultdict" which provides a more flexible solution to this problem.

You can do neat things, for example:

twodee = collections.defaultdict(dict)
twodee["the horizontal"]["the vertical"] = "we control"

Read more: http://docs.python.org/2/library/collections.html

Answer 4

another way:

b = (a[4:]+['sss'])[0]

Answer 5

You could create your own list-class:

class MyList(list):
    def get(self, index, default=None):
        return self[index] if len(self) > index else default

You can use it like this:

>>> l = MyList(['a', 'b', 'c'])
>>> l.get(1)
'b'
>>> l.get(9, 'no')
'no'

Answer 6

In the Python spirit of beautiful is better than ugly

Code golf method, using slice and unpacking (not sure if this was valid 4 years ago, but it is in python 2.7 + 3.3)

b,=a[4:5] or ['sss']

Nicer than a wrapper function or try-catch IMHO, but intimidating for beginners. Personally I find tuple unpacking to be way sexier than list[#]

using slicing without unpacking:

b = a[4] if a[4:] else 'sss'

or, if you have to do this often, and don't mind making a dictionary

d = dict(enumerate(a))
b=d.get(4,'sss')