On Jan 21, 4:23 pm, Scott David Daniels <scott.dani...@acm.org> wrote: > prueba...@latinmail.com wrote: > > ... If you have duplicates this will not work. You will have to do > > something like this instead: > > >>>> o=[] > >>>> i=0 > >>>> ln=len(l) > >>>> while i<ln: > > if l[i]['title']=='ti': > > o.append(l.pop(i)) > > ln-=1 > > else: > > i+=1 > > Or the following: > indices = [i for i,d in enumerate(l) if d['title']=='ti'] > for i in reversed(indices): # so del doesn't affect later positions > del l[i] > > --Scott David Daniels > scott.dani...@acm.org
Cool. How come I didn't think of that! That means I can create an evil one liner now >:-). replacecount=len([o.append(l.pop(i)) for i in reversed(xrange(len(l))) if l[i]['title']=='ti']) -- http://mail.python.org/mailman/listinfo/python-list
