On Sat, 08 Nov 2008 22:53:14 -0800, Kay Schluehr wrote: >> How often do you care about equality ignoring order for lists >> containing arbitrary, heterogeneous types? > > A few times. Why do you care, Steven?
I'm a very caring kind of guy. >> In any case, the above doesn't work now, since either L1 or L2 might >> contain complex numbers. >> The sorted() trick only works because you're making an assumption about >> the kinds of things in the lists. If you want to be completely general, >> the above solution isn't guaranteed to work. > > You are right. I never used complex numbers in Python so problems were > not visible. Otherwise the following comp function in Python 2.X does > the job: [...] > Not sure how to transform it into a search key that is efficient and > reliable. Yes, that's a general problem. It's not always easy to convert a sort comparison function into a key-based sort. I know that 99% of the time key is the right way to do custom sorts, but what about the other 1%? > [...] > >> Here is a way to >> solve the problem assuming only that the items support equality: >> >> def unordered_equals2(L1, L2): >> if len(L1) != len(L2): >> return False >> for item in L1: >> if L1.count(item) != L2.count(item): >> return False >> return True > > Which is O(n**2) as you might have noted. Yes, I did notice. If you can't assume even ordering, then you need to do a lot more work. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
