Thanks Michael, this worked great, time to read up on collections! On Jan 24, 2008 3:36 AM, Michael Langford <[EMAIL PROTECTED]> wrote:
> You use a variation on bucket sort do to this: > http://en.wikipedia.org/wiki/Bucket_sort > > You make a dict where the keys are the values in the lists, and a name > for each list is in the problem. > > So you get something that looks like: > > one: list1, list2 > two: list1, list2, list3 > etc > > Doing this with collections.defaultdict is a breeze: > import collections > > dd = collections.defaultdict(list) > for eachlist in lists: > for each in eachlist: > dd[each].append(getListName(eachlist)) > > Then to find the repeated elements you filter where the list is of length > > 1. > > for each in dd: > print "%s: %s" % (each, dd[each]) > > I'd provide code, but I'm not sure what an appropriate naming function > is for you, nor am I sure what you're doing with this when you're > done. > > --Michael > > > > On Jan 24, 2008 3:15 AM, Fiyawerx <[EMAIL PROTECTED]> wrote: > > I have been able to find a few articles on comparing 2 lists, but I have > 4 > > lists that I need to compare. I need to find any repeated elements and > the > > list that they came from. For example, > > > > list1 = ['one', 'two', 'three'] > > list2 = ['one', 'two', 'four', 'five'] > > list3 = ['two', 'three', 'six', 'seven'] > > list4 = ['three', 'five', 'six'] > > https://mail.google.com/mail/#label/Pythontutor/117aadf8364dbf3b > Gmail - [Tutor] Comparing more than 2 lists - [EMAIL PROTECTED] > > > > I need to be able to get along the lines of output: > > Element 'one' contained in list1 and list2 > > Element 'two' contained in list1 and list2 and list3 > > ... > > Element 'five' contained in list2 and list4 > > > > etc.. and I can't quite figure out how to go about it > > > > > > _______________________________________________ > > Tutor maillist - Tutor@python.org > > http://mail.python.org/mailman/listinfo/tutor > > > > > > > > -- > Michael Langford > Phone: 404-386-0495 > Consulting: http://www.RowdyLabs.com >
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