Charles Sanders <[EMAIL PROTECTED]> writes: > 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 ]
If you're itertools-phobic you could alternatively write d = dict((x,i) for i,x in enumerate(l2)) pos = [ d[i] for i in l1 ] dict access and update is supposed to take approximately constant time, btw. They are implemented as hash tables. -- http://mail.python.org/mailman/listinfo/python-list
