Project Euler 10 "Sum of Prime Numbers"

The sum of prime numbers less than or equal to 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all prime numbers below 2 million. http://odz.sakura.ne.jp/projecteuler/index.php?cmd=read&page=Problem%2010

I wrote it quickly using mymath I made earlier. (296ms) http://qiita.com/cof/items/45d3823c3d71e7e22920

The algorithm for finding prime numbers is written in mymath, but it is as follows.

Using the Sieve of Eratostenes, make a list of booleans whose prime number is True and non-prime number is False.
Create a list of prime numbers from the list in list comprehension notation.

Then, sum the created list of prime numbers (pri ['list'] below) with sum ().

import mymath
import mytime
import math
def cof():
    target = 2 * (10**6)
    pri = mymath.get_primes(target)
    print sum(pri['list'])

I wondered if the following algorithm was faster, but it was slower. (420ms) It seems that the processing speed differs depending on whether the list is Boolean or a number.

Make a list of numbers. [0,0,2,3,4,5,6,7,8,9,10, ‥](1 is set to 0 in advance.
Loop on prime numbers and set the multiples of the prime numbers contained in the digit string to 0. [0,0,2,3,0,5,0,7,0,0,0, ‥]
Take the sum () of the remaining list.

def cof2():
    target = 2 * (10**6)
    L = [0,0]+range(2,target+1)
    L[4::2] = [0] * (len(L[4::2]))
    p=3
    p_max = int(math.sqrt(target))
    while p<=p_max:
        if L[p]:
            L[p*2::p] = [0] * (len(L[p*2::p]))
        p+=2
    L.remove(0)
    print sum(L)

When I looked at what other people were writing on the correct answer bulletin board, Lucy_Hedgehog's code posted on Fri, 3 May 2013, 19:14 was amazing. When I ran it at home, it took 15ms. I couldn't understand his explanation and code well, but he seemed to use an algorithm different from Eratosthenes (similar to the prime number counting algorithm).