Robert P. J. Day said:
>    the ubiquitous sieve of eratosthenes requires you to pre-specify
>  your maximum value, after which -- once the sieve completes -- all you
>  know is that you have all of the prime numbers up to n.  whether
>  you'll have 1000 of them isn't clear, which means that you might have
>  to start all over with a larger maximum value.  (being able to
>  directly determine the n'th prime number would solve a *lot* of prime
>  number problems. :-)
> 
>    
In April of this year, members of this forum helped me to devise a 
prime-number iterator [1]. So here's a simple solution for the OP:

#---------------
from itertools import ifilter, count

# create iterator
prime_iter = ifilter(
    lambda n, P=[]: all(n%p for p in P) and not P.append(n),
                    count(2))

# throw away lots of primes
for i in range(999):   
    prime_iter.next()
   
# here's the one we want
print prime_iter.next()      #7919
#---------------

I don't think this is a solution that a course instructor would expect, 
though!

As mentioned in [1], optimizations of this algorithm (using 
itertools.takewhile() and/or caching previously-computed squares) are 
still at cl1p.net [2].

-John

[1]  http://mail.python.org/pipermail/python-list/2009-April/177415.html

[2]  http://www.cl1p.net/python_prime_generators


 <http://cl1p.net/python_prime_generators/> 



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