On Thu, 13 Oct 2005 13:31:13 -0400, 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?
Others have given answers involving xrange() and itertools.count(), but I
thought I'd just mention that in my opinion, what you have written is
pretty elegant and concise and best of all, doesn't have the same problems
xrange() and itertools.count() have when then hit maxint. Not everything
needs to be a one-liner or a mysterious blackbox.
You could even modify your function to take start and step arguments:
def integers(start=1, step=1):
x = start
while True:
yield x
x += step
for odd in integers(step=2):
print odd
for even in integers(0, 2):
print even
etc.
--
Steven.
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