On Tue, Feb 9, 2010 at 10:22 AM, Srinivas Reddy Thatiparthy <srinivas_thatipar...@akebonosoft.com> wrote: > I have written four solutions to a problem(self explanatory) ,out of > them ,which one is the pythonic way of doing and > is there any other ways of solving it? > > > 1.sum([i for i in range(1000) if i%3==0 or i%5==0]) > > 2.gen=(i for i in range(1000)) > sum([i for i in gen if i%3==0 or i%5==0]) > > 3.sum(filter(lambda a:a%3==0 or a%5==0,range(1000))) > > 4.def generator(m): > count=0 > while count<m: > if count%3==0 or count%5==0: > yield count > count+=1 > sum([i for i in generator(1000)])
#1. You can even remove the square brackets. sum(i for i in range(1000) if i%3==0 or i%5==0) _______________________________________________ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers
