求N之下的所有素数

No.1

f=lambda n: [x for x in range(1,n) if not [y for y in range(2,x) if x%y ==0 ]]

No.2

def prime(n):
    for i in range(1,n+1):
        for j in range(2,int(i**0.5+1)):
            if i % j == 0:
                break
        else:
                yield i

NO.3（还没有搞懂、最高效的一种）

def rwh_primes2(n = 10**6):
    """ Input n>=6, Returns a list of primes, 2 <= p < n """

    correction = (n%6>1)
    n = {0:n,1:n-1,2:n+4,3:n+3,4:n+2,5:n+1}[n%6]
    sieve = [True] * (n/3)
    sieve[0] = False
    for i in xrange(int(n**0.5)/3+1):
      if sieve[i]:
        k=3*i+1|1
        sieve[      ((k*k)/3)      ::2*k]=[False]*((n/6-(k*k)/6-1)/k+1)
        sieve[(k*k+4*k-2*k*(i&1))/3::2*k]=[False]*((n/6-(k*k+4*k-2*k*(i&1))/6-1)/k+1)
    return [2,3] + [3*i+1|1 for i in xrange(1,n/3-correction) if sieve[i]]