q at q.nl
Mon Jan 23 18:41:55 CET 2006
Op 19 jan 2006 vond "markscala at gmail.com" :
> another approach:
> ref = [2,2,4,1,1]
> lis = [2,2,5,2,4]
> len([ref.pop(ref.index(x)) for x in lis if x in ref])
This is the type of solution I was hoping to see: one-liners, with no use
of local variables. As Tim Chase already wrote, it has only one less
elegant side: it alters the original ref list.
Thanks for your suggestion.
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