Default return values for out-of-bounds list item
__peter__ at web.de
Fri Jan 22 09:21:48 CET 2010
Steven D'Aprano wrote:
> On Thu, 21 Jan 2010 17:56:00 -0800, 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):
>> return x[index]
>> except IndexError:
>> return default
> That's probably the best way to do it.
>> 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. What if, for example, we could do something like a [1000,44] ?
> You can use slicing instead:
> and then detect the empty list and use default:
> def item(x, index, default):
> a = x[index:index+1]
> return a if a else default
You need to special-case -1:
>>> for i in range(-4, 4):
... print "x[%d] -> %r" % (i, item("abc", i, "default"))
x[-4] -> 'default'
x[-3] -> 'a'
x[-2] -> 'b'
x[-1] -> 'default'
x -> 'a'
x -> 'b'
x -> 'c'
x -> 'default'
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