hiro wrote: > bare in mind that I have a little over 10 million objects in my list > (l2) and l1 contains around 4 thousand > objects.. (i have enough ram in my computer so memory is not a > problem)
Glad to see you solved the problem with the trailing space. Just one minor point, I did say > or probably less efficiently ... As far as i know, my suggestion's running time is proportional to len(l1)*len(l2), which gets quite big for your case where l1 and l2 are large lists. If I understand how python dictionaries work, Paul Rubin's suggestion > from itertools import izip, count > pos = map(dict(izip(l2, count())).__getitem__, l1) or the (I think) approximately equivalent from itertools import izip, count d = dict(izip(l2,count())) pos = [ d[i] for i in l1 ] or the more memory intensive d = dict(zip(l2,range(len(l2)))) pos = [ d[i] for i in l1 ] should all take take running time proportional to (len(l1)+len(l2))*log(len(l2)) For len(l1)=4,000 and len(l2)=10,000,000 Paul's suggestion is likely to take about 1/100th of the time to run, ie be about 100 times as fast. I was trying to point out a somewhat clearer and simpler (but slower) alternative. Charles -- http://mail.python.org/mailman/listinfo/python-list
