素数代码的优化

Question

This is my code in python for calculation of sum of prime numbers less than a given number.
What more can I do to optimize it?

import math
primes = [2,]                      #primes store the prime numbers



for i in xrange(3,20000,2):                    #i is the test number
    x = math.sqrt(i)
    isprime = True
    for j in primes:               #j is the devider. only primes are used as deviders
        if j <= x:
            if i%j == 0:
                    isprime = False
                    break


    if isprime:
        primes.append(i,)


print sum (primes,)

Answer 1

You can use a different algorithm called the Sieve of Eratosthenes which will be faster but take more memory. Keep an array of flags, signifying whether each number is a prime or not, and for each new prime set it to zero for all multiples of that prime.

N = 10000

# initialize an array of flags
is_prime = [1 for num in xrange(N)]
is_prime[0] = 0 # this is because indexing starts at zero
is_prime[1] = 0 # one is not a prime, but don't mark all of its multiples!

def set_prime(num):
    "num is a prime; set all of its multiples in is_prime to zero"
    for x in xrange(num*2, N, num):
        is_prime[x] = 0

# iterate over all integers up to N and update the is_prime array accordingly
for num in xrange(N):
    if is_prime[num] == 1:
        set_prime(num)

primes = [num for num in xrange(N) if is_prime[num]]

You can actually do this for pretty large N if you use an efficient bit array, such as in this example (scroll down on the page and you'll find a Sieve of Eratosthenes example).

Answer 2

Another thing you could optimize is move the sqrt computation outside the inner loop. After all, i stays constant through it, so there's no need to recompute sqrt(i) every time.

Answer 3

primes = primes + (i,) is very expensive. It copies every element on every pass of the loop, converting your elegant dynamic programming solution into an O(N ² ) algorithm. Use lists instead:

primes = [2]
...
    primes.append(i)

Also, exit the loop early after passing sqrt(i). And, since you are guaranteed to pass sqrt(i) before running off the end of the list of primes, update the list in-place rather than storing isprime for later consumption:

...
if j > math.sqrt(i):
    primes.append(i)
    break
if i%j == 0:
    break
...

Finally, though this has nothing to do with performance, it is more Pythonic to use range instead of while:

for i in range(3, 10000, 2):
    ...

Answer 4

Just another code without using any imports:

#This will check n, if it is prime, it will return n, if not, it will return 0
def get_primes(n):
    if n < 2:
        return 0
    i = 2
    while True:
        if i * i > n:
            return n
        if n % i == 0:
            return 0
        i += 1

#this will sum up every prime number up to n
def sum_primes(n):
    if n < 2:
        return 0
    i, s = 2, 0
    while i < n:
        s += get_primes(i)
        i += 1
    return s

n = 1000000
print sum_primes(n)

Answer 5

EDIT: removed some silliness while under influence

All brute-force type algorithms for finding prime numbers, no matter how efficient, will become drastically expensive as the upper bound increases. A heuristic approach to testing for primeness can actually save a lot of computation. Established divisibility rules can eliminate most non-primes "at-a-glance".