On 15 Nov 2005 04:51:05 -0800, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote:
> >Simon Brunning wrote: >> On 15/11/05, Ben Bush <[EMAIL PROTECTED]> wrote: >> > I found I named the following python file as sets.py, which brought the >> > problem (is that right?). i changed it to other name and it works. >> > But the logic output is wrong. >> > from sets import Set as set >> > lisA=[1,2,5,9] >> > lisB=[9,5,0,2] >> > lisC=[9,5,0,1] >> > def two(sequence1, sequence2): >> > set1, set2 = set(sequence1), set(sequence2) >> > return len(set1.intersection(set2)) == 2 >> > print two(lisA,lisB) >> > False(should be true!) >> >> It looks to me like A and B have three members in common - 2, 5 and 9. >> >Any chance that while "two" performs correctly for what is intended, it >is not what the OP wants ? > >Seems that he only thins the "5,9"/"9,5" are the common case, thus >expect 2 ? > That's what I thought, hence: >>> lisA=[1,2,3,4,5,6,9] >>> lisB=[1,6,5] >>> lisC=[5,6,3] >>> lisD=[11,14,12,15] >>> def pairs(seq): ... it = iter(seq) ... curr = it.next() ... while True: ... curr, last = it.next(), curr ... yield curr, last ... if curr != last: ... yield last, curr ... >>> list(pairs(lisA)) [(2, 1), (1, 2), (3, 2), (2, 3), (4, 3), (3, 4), (5, 4), (4, 5), (6, 5), (5, 6), (9, 6), (6, 9)] >>> list(pairs(lisB)) [(6, 1), (1, 6), (5, 6), (6, 5)] >>> set(pairs(lisA)).intersection(set(pairs(lisB))) set([(5, 6), (6, 5)]) >>> len(set(pairs(lisA)).intersection(set(pairs(lisB)))) == 2 True I guess the OP meant exactly one shared order-insensitive pair, so that last test is insufficient. Maybe (untested beyond what you see ;-) >>> def lisCheck(L1, L2): ... common = set(pairs(L1)).intersection(set(pairs(L2))) ... if len(common) == 1: return True # [(x,y)] not possible since pairs makes (y,x) ... elif len(common) == 2: # guard against [(x,x), (y,y)] ... t0, t1 = common ... return t0[0]==t1[1] and t0[1]==t1[0] and t0[0]!=t0[1] ... else: return False ... >>> lisCheck(lisA, lisB) True >>> lisCheck(lisA, lisC) True >>> lisCheck(lisB, lisC) True >>> lisCheck(lisB, [5,3,6]) False >>> lisCheck([1,2,2,3], [2,2]) True >>> lisCheck([1,2,2,3,3], [2,2]) True >>> lisCheck([1,2,2,3,3], [2,2,3]) False >>> lisCheck([1,2,2,3,3], [2,2,0, 3]) True >>> lisCheck([1,2,2,3,3], [2,2,0,3,3]) False Also, too lazy to make pairs produce less redundant pairs than eliminating equal flips. If I did it again, I guess I'd also yield last,curr before curr,last for better left-to-right scan mnemonics, though it doesn't affect the end result. Regards, Bengt Richter -- http://mail.python.org/mailman/listinfo/python-list
