Emille,
Thank you for the example of list splicing. Do you know if this is
faster than a more conventional loop statement as in my code for
primearray which is in my original post (reprinted here)
The code is as follows:
def BuildSieve(itemsin):
TheSieve=list()
TheSieve = range(0,itemsin+1)
TheSieve[1]=0
for i in range(2,itemsin+1):
if (TheSieve[i] > 0):
j = i + i
while (j <= itemsin):
TheSieve[j] = 0
j+=i
return TheSieve
It is called with PrimaryList = BuildSieve(1000000)
Again, thank you for your help.
Robert
On Wed, 2009-06-17 at 17:01 -0700, Emile van Sebille wrote:
> On 6/17/2009 4:48 PM Robert Berman said...
> > Emile,
> >
> > Thank your for your comments. I do have a list running from 0-1000001.
> > Yes, it is true, I only needed 0 - 100000 and yes I will change it.
> > However, if you use primearray
>
> you haven't posted the primearray code...
>
> <snip>
> >
> > However, for the time being, can you perhaps share some suggestions on
> > list splicing?
>
> So, this part of your original post--
> > Out[53]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
> >
> > Since 3 is a prime, we can eliminate all multiples of 3. Within l1,
> > these are expressed as
> >
> > In [52]: l1[n+n:len(l1):n]
> > Out[52]: [6, 9, 12]
> >
> > when n = 3. ( do know 12 would have been eliminated by the prime
> > number 2)
> >
> > It would be great if I could say l1[n+n:len(l1):n] = 0
>
> but you can say:
>
> for ii in l1[n+n:len(l1):n]: l1[ii] = 0
>
> Is something like that what you're after?
>
> Emile
>
>
>
> > but obviously
> >
> > that will fail for obvious reasons. I am looking for the right hand
> > side of the statement to set a list within the list to all zeros.
> ---
>
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