Python：找到第n个质数

Question

I'm trying to find the nth prime number using the sieve of erathostenes. Yes, I saw similar posts, but I have a problem with this piece of code. I want to stop the algorithm once it finds the nth prime. This is what I wrote:

def nth_prime(n):
    limit = 10**2
    pn = 1                     #keeps track of how many prime numbers we have found
    sieve = range(3, limit, 2)
    top = len(sieve)
    for si in sieve:
        if si:
            pn += 1
            print pn, si     #used to check while coding
            if pn == n:
                return si    #loop breaks when the nth prime is found
            else:   
                    bottom = (si*si - 3)/2
                    if bottom >= top:
                        break
                    sieve[bottom::si] = [0] * -((bottom-top)//si)   

print nth_prime(11)

It doesn't work though. At least not as I want to. If I add return filter(None, sieve)[n-2] it works fine. But I want it to stop computing at the nth prime. This is the output instead:

2 3
3 5
4 7
5 11
None

While I would expect it to continue until:

...
11 31 

If the function is able to calculate all the sieve up to the limit correctly, why does the output behave like that?

Answer 1

The Python break command breaks out of loops, not out of tests ( if - else ). I get it to work by reworking the logic to eliminate the break command. That is,

    if bottom < top:
        sieve[bottom::si] = [0] * -((bottom-top)//si)