Steven Bethard <[EMAIL PROTECTED]> writes: > Cameron Walsh wrote: > > Which brings me to the question, would this solution: > > B = set(B) > > A = B + list(x for x in A if x not in B) > > be faster than this solution: > > B = set(B) > > A.sort(key=B.__contains__, reverse=True) > [timings deleted] > That said, I'd probably still use the first solution -- it's more > immediately obvious why that one works.
Wait a minute, the first example looks wrong, B has gotten replaced by a set and then it's added to a list. Anyway how about timing C = set(A) - set(B) A = B + filter(C.__contains__, A) This scans A twice, but it does more of the work in native code, without sorting. -- http://mail.python.org/mailman/listinfo/python-list
