Better solution

[Mark McEahern]

> Well, I want to throw away a _same_ garbage from a list with less 
> of coding.
> This is current code, sure not the best ;-) Is any better solutions?
> 
> --------------8<------------------------
> lst = ['', 'a', '', 'b', 'c', '', 'd']
> map(lambda z:lst.pop(lst.index('')), range(0, lst.count('')))
> --------------8<------------------------
> 
> Now lst equals to ['a', 'b', 'c', 'd'].

These would also do the trick and be less cryptic:

  filter(lambda x: x, lst)

  [x for x in lst if x]

Cheers,

// m

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