default value in a list

[Peter Otten]

Peter Otten wrote:

> Paul McGuire wrote:
> 
>>> Is there an elegant way to assign to a list from a list of unknown
>>> size?  For example, how could you do something like:
>>>
>>> >>>  a, b, c = (line.split(':'))
>>> if line could have less than three fields?
> 
>> I asked a very similar question a few weeks ago, and from the various
>> suggestions, I came up with this:
>> 
>> line = "AAAA:BBB"
>> expand = lambda lst,default,minlen : (lst + [default]*minlen)[0:minlen]
>> a,b,c = expand( line.split(":"), "", 3 )
> 
> Here is an expand() variant that is not restricted to lists but works with
> arbitrary iterables:
> 
> from itertools import chain, repeat, islice
> 
> def expand(iterable, length, default=None):
>     return islice(chain(iterable, repeat(default)), length)

Also nice, IMHO, is allowing individual defaults for different positions in
the tuple:

>>> def expand(items, defaults):
...     if len(items) >= len(defaults):
...             return items[:len(defaults)]
...     return items + defaults[len(items):]
...
>>> expand((1, 2, 3), (10, 20, 30, 40))
(1, 2, 3, 40)
>>> expand((1, 2, 3), (10, 20))
(1, 2)

Peter