Tom Fitzhenry wrote: > On Fri, Aug 10, 2007 at 02:54:44AM -0700, Jaggo wrote: >> Can anyone think of any better way? > > If SmallList and BigList are sorted (in order), there is a faster method: > > def IsAPartOfList(SmallList,BigList): > for i in BigList: > for j in SmallList: > if i==j: > return true > if i>j: > break > return false > > (I'm not sure how encouraged using break statements are, so wait for a tutor > to > answer)
break is fine! If the list you are searching is sorted you can use the bisect module to do a binary search instead of the linear search above. > If one list is already sorted but the other isn't, it may still be faster to > sort the unsorted list then use the method above. I don't think BigList has to be sorted in the above algorithm. If both lists are sorted I suppose you could write it like a merge sort, walking along both lists looking for a match. Kent _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
