reduce expression to test sublist
tjreedy at udel.edu
Sat Jan 5 22:55:50 CET 2013
On 1/5/2013 1:58 PM, Dave Angel wrote:
> If you're trying to make a faster loop, then I suggest you look into set
> differences. Turn both lists into sets, and subtract them. Something
> like (untested):
> result = not bool( set(lst1) - set(lst2) )
This does not return False as soon as an item in set1 is found that is
not in set2.
set(lst1) < set(lst2)
will, and directly return False/True. The OP is trying to compute the
lst1 < lst2, where lst1 and lst2 are interpreted as sets, rather than as
sequences with the lexicographic ordering default.
Terry Jan Reedy
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