Sieve of Eratosthenes - Finding Primes Python

Question

Just to clarify, this is not a homework problem :)

I wanted to find primes for a math application I am building & came across Sieve of Eratosthenes approach.

Answer 1

Much faster:

import time
def get_primes(n):
  m = n+1
  #numbers = [True for i in range(m)]
  numbers = [True] * m #EDIT: faster
  for i in range(2, int(n**0.5 + 1)):
    if numbers[i]:
      for j in range(i*i, m, i):
        numbers[j] = False
  primes = []
  for i in range(2, m):
    if numbers[i]:
      primes.append(i)
  return primes

start = time.time()
primes = get_primes(10000)
print(time.time() - start)
print(get_primes(100))

Answer 2

not sure if my code is efficeient, anyone care to comment?

from math import isqrt

def isPrime(n):
    if n >= 2: # cheating the 2, is 2 even prime?
        for i in range(3, int(n / 2 + 1),2): # dont waste time with even numbers
            if n % i == 0:
                return False
    return True

def primesTo(n): 
    x = [2] if n >= 2 else [] # cheat the only even prime
    if n >= 2:
        for i in range(3, n + 1,2): # dont waste time with even numbers
            if isPrime(i):
                x.append(i)  
    return x

def primes2(n): # trying to do this using set methods and the "Sieve of Eratosthenes"
    base = {2} # again cheating the 2
    base.update(set(range(3, n + 1, 2))) # build the base of odd numbers
    for i in range(3, isqrt(n) + 1, 2): # apply the sieve
        base.difference_update(set(range(2 * i, n + 1 , i)))
    return list(base)

print(primesTo(10000)) # 2 different methods for comparison
print(primes2(10000))

Answer 3

I figured it must be possible to simply use the empty list as the terminating condition for the loop and came up with this:

limit = 100
ints = list(range(2, limit))   # Will end up empty

while len(ints) > 0:
    prime = ints[0]
    print prime
    ints.remove(prime)
    i = 2
    multiple = prime * i
    while multiple <= limit:
        if multiple in ints:
            ints.remove(multiple)
        i += 1
        multiple = prime * i

Answer 4

My implementation:

import math
n = 100
marked = {}
for i in range(2, int(math.sqrt(n))):
    if not marked.get(i):
        for x in range(i * i, n, i):
            marked[x] = True

for i in range(2, n):
    if not marked.get(i):
        print i

Answer 5

i think this is shortest code for finding primes with eratosthenes method

def prime(r):
    n = range(2,r)
    while len(n)>0:
        yield n[0]
        n = [x for x in n if x not in range(n[0],r,n[0])]


print(list(prime(r)))

Answer 6

def eratosthenes(n):
    multiples = []
    for i in range(2, n+1):
        if i not in multiples:
            print (i)
            for j in range(i*i, n+1, i):
                multiples.append(j)

eratosthenes(100)