Will McGugan wrote: > Neal Becker wrote: > >> I can do this with a generator: >> >> def integers(): >> x = 1 >> while (True): >> yield x >> x += 1 >> >> for i in integers(): >> Is there a more elegant/concise way? >> > > import itertools > for i in itertools.count(): > print i >
Actualy itertools.count(1) to start at 1.. Will McGugan -- http://www.willmcgugan.com "".join({'*':'@','^':'.'}.get(c,0) or chr(97+(ord(c)-84)%26) for c in "jvyy*jvyyzpthtna^pbz") -- http://mail.python.org/mailman/listinfo/python-list
