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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