Default return values for out-of-bounds list item

Tim Chase python.list at tim.thechases.com
Fri Jan 22 04:38:33 CET 2010

```MRAB wrote:
> gburdell1 at gmail.com wrote:
>> Is there a built-in method in python that lets you specify a "default"
>> value that will be returned whenever you try to access a list item
>> that is out of bounds? Basically, it would be a function like this:
>>
>> def item(x,index,default):
>>    try:
>>       return x[index]
>>    except IndexError:
>>       return default
>>
>> So if a=[0,1,2,3], then item(a,0,44)=0, item(a,1,44)=1, and item(a,
>> 1000,44)=44, item(a,-1000,44)=44
>>
>> What I want to know is whether there is a built-in method or notation
>> for this.
>  >
> There's no built-in method or notation for that.
>
>  > What if, for example, we could do something like a [1000,44] ?
>
> That's actually using a tuple (1000, 44) as the index. Such tuples can
> be used as keys in a dict and are used in numpy for indexing
> multi-dimensional arrays, so it's definitely a bad idea.
>
> If such a method were added to the 'list' class then the best option (ie
> most consistent with other classes) would be get(index, default=None).

But there's nothing stopping you from creating your own subclass
of "list" that allows for defaults:

class DefaultList(list):
def __init__(self, default, *args, **kwargs):
list.__init__(self, *args, **kwargs)
self.default = default
def __getitem__(self, index):
try:
return list.__getitem__(self, index)
except IndexError:
return self.default
ml = DefaultList(42, [1,2,3,4])
for i in range(-5,5):
print i, ml[i]

(yeah, there's likely a "proper" way of using super() to do the
above instead of "list.__<whatever>__" but the above worked for a
quick 3-minute example composed in "ed").  The output of the
above is

-5 42
-4 1
-3 2
-2 3
-1 4
0 1
1 2
2 3
3 4
4 42