Thanks to both Dan and William. However, Dan's result puzzled me.
Aren't they suggested that the for loop is faster?
Here is what I got:
sage: timeit('list(2*vector([random() for j in range(10)]))')
625 loops, best of 3: 332 µs per loop
sage: timeit('[2*random() for j in range(10)]')
125 loops, best of 3: 8.3 ms per loop
which suggested otherwise. Also how does timeit determine how many
loops to perform in a test?
On Oct 21, 4:09 pm, Dan Drake <[EMAIL PROTECTED]> wrote:
> On Tue, 21 Oct 2008 at 07:50AM -0700, pong wrote:
> > I wrote two list tests:
>
> > def test1():
> > for k in range(10^3):
> > v=[2*random() for j in range(10)]
> > else: pass
>
> > def test2():
> > for k in range(10^3):
> > v=list(2*vector([random() for j in range(10)]))
> > else: pass
>
> > It turns out that
> > time test1() gives around 8s CUP time while time test2() shows only
> > 0.35s CUP time is used. Wow!
>
> Also, instead of doing your own loops, you can use %timeit:
>
> sage: %timeit [2*random() for j in range(10)]
> 100000 loops, best of 3: 6.87 µs per loop
> sage: %timeit list(2*vector([random() for j in range(10)]))
> 10000 loops, best of 3: 112 µs per loop
>
> Using list comprehensions is almost always a good choice in Sage/Python;
> as I understand (and as we see above), they're very well-optimized.
>
> Dan
>
> --
> --- Dan Drake <[EMAIL PROTECTED]>
> ----- KAIST Department of Mathematical Sciences
> ------- http://mathsci.kaist.ac.kr/~drake
>
> signature.asc
> < 1KViewDownload
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