Generators/iterators, Pythonicity, and primes
Miles
semanticist at gmail.com
Sun Apr 5 06:28:17 CEST 2009
On Sat, Apr 4, 2009 at 2:50 PM, John Posner wrote:
> Inspired by recent threads (and recalling my first message to Python
> edu-sig), I did some Internet searching on producing prime numbers using
> Python generators. Most algorithms I found don't go for the infinite,
> contenting themselves with "list all the primes below a given number".
>
> Here's a very Pythonic (IMHO) implementation that keeps going and going and
> going ...:
>
> from itertools import count
> from math import sqrt
>
> def prime_gen():
> """
> Generate all prime numbers
> """
> primes = []
> for n in count(2):
> if all(n%p for p in primes if p < sqrt(n)):
> primes.append(n)
> yield n
>
> The use of all() is particularly nifty (see
> http://code.activestate.com/recipes/576640/). And so is the way in which the
> list comprehension easily incorporates the sqrt(n) optimization.
>
> Question: Is there a way to implement this algorithm using generator
> expressions only -- no "yield" statements allowed?
def prime_gen():
primes = []
return (primes.append(n) or n for n in count(2) if all(n%p for p
in primes if p<=sqrt(n)))
That version is only marginally faster than your original version.
The biggest performance penalty is that the (... for p in primes ...)
generator isn't aborted once p > sqrt(n). Here's a less nifty but much
more efficient version:
def prime_gen():
prime_list = [2]
for p in prime_list: yield p
for n in itertools.count(prime_list[-1] + 1):
for p in prime_list:
if p * p > n:
prime_list.append(n)
yield n
break
elif n % p == 0:
break
else:
raise Exception("Shouldn't have run out of primes!")
When generating the first 1000 primes, this version's approximately 20
times faster; for the first 10,000 primes, ~80x (but still much slower
than a simple Sieve of Eratosthenes). To make repeated calls faster,
move prime_list = [2] outside the function.
-Miles
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