Better solution

Bo M. Maryniuck b.maryniuk at
Tue Aug 20 16:51:18 CEST 2002

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?

lst = ['', 'a', '', 'b', 'c', '', 'd']
map(lambda z:lst.pop(lst.index('')), range(0, lst.count('')))

Now lst equals to ['a', 'b', 'c', 'd'].

Regards, Bogdan

If you are good, you will be assigned all the work.  If you are real
good, you will get out of it.

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