Python- Sieve of Eratosthenes- Compact Python

Question

This is my code for finding primes using the Sieve of Eratosthenes.

list = [i for i in range(2, int(raw_input(\"Compute primes up to what number? \"))+1)]  
         


        

Answer 1

The following one-liner is not related at all to your code:

def primes(n):
    return set(range(2,n))-{c for i in range(2,n) for c in range(2*i,n,i)}

Like your code, this is still not really the Sieve of Eratosthenes because, for example, it will futilely try to cross off multiples of 6 and 9 etc. Nevertheless it still runs significantly faster than most other Sieve look-alikes for values less than a million or more, since for small N there are "about as many" primes as non-primes (the fraction of numbers < N that are prime is 1/log(N)).

Heavily modified from source, possibly less efficient than original: http://codeblog.dhananjaynene.com/2011/06/10-python-one-liners-to-impress-your-friends/

Answer 2

Here is the most compact true sieve I have come up with so far. This performs surprisingly well.

def pgen(n): # Sieve of Eratosthenes generator
    np = set() # keeps track of composite (not prime) numbers
    for q in xrange(2, n+1):
        if q not in np:
            yield q
            np.update(range(q*q, n+1, q))

>>> list(pgen(100))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73,
79, 83, 89, 97]            

This slightly more complex version is the fastest I have seen:

def pgen(n): # Sieve of Eratosthenes generator by Dan Salmonsen
    yield 2
    np = set()
    for q in xrange(3, n+1, 2):
        if q not in np:
            yield q
            np.update(range(q*q, n+1, q+q))            

Here is a true sieve as a list comprehension:

def primes(n):
    sieve = set(sum([range(q*q, n+1, q+q) for q in odds], []))
    return [2] + [p for p in odds if p not in sieve]

>>> primes(100)
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73,
79, 83, 89, 97]