找到第1000个素数（python）

Question

I've tried working on this question using this code:

def nthprime(n):

    cnt=1
    count=0
    while(cnt==n):
        for i in range(3, ):
            for j in range(2, i):
                if i % j != 0:
                    count = count + 1
            if count == (i - 2):
                    cnt = cnt + 1
            return I

print(nthprime(1000))

Can anyone please tell me what's wrong with my code? It only returns "None" all the time.

Answer 1

Your code would be easier to understand if you split it in two functions

def is_prime(i):   
  if i == 2: return True  # 2 is a prime number
  for j in range(2,i):  # you could stop sooner (optimize)
    if i % j == 0: return False
  return True

def nthprime(n):
  i = 1
  count = 0
  while count < n:
    i += 1
    if is_prime(i): count += 1
  return i

for n in (1,2,3,4,5,6,7,8,9,10,100,1000):
  print( n, nthprime(n) )

When I run it, I get (in python2, add from __future__ import print_function )

1 2
2 3
3 5
4 7
5 11
6 13
7 17
8 19
9 23
10 29
100 541
1000 7919    

Answer 2

You code has a few issues, as arif pointed out I is not initialized, the while loop will never run as cnt == n is not going to be valid etc. Rather than perfect your code i've got a slightly modified example of how to do what you want:

def nth_prime_number(n):
    prime_list = [2]
    num = 3
    while len(prime_list) < n:
        for p in prime_list:
            if num % p == 0:
                break
        else:
            prime_list.append(num)
        num += 2
    return prime_list[-1]

print(nth_prime_number(1000))

You could look at the differences between this example and your own code to understand where you went wrong.