如何使代码更短、更高效？

Question

import time
start_time = time.time()
a = [2]
inp = 2
while inp < 60000:
    div = 2
    inp += 1
    while div <= (inp / 2 + 1):
        prime = inp / div
        if prime.is_integer() == True:
            break
        else:
            if div >= (inp / 2):
                a.append(str(inp))
                break
            else:
                div += 1

print(a)
print(len(a))
print("Process finished --- %s seconds ---" % (time.time()-start_time))

我编写了自己的第一个程序来计算 0 到 60000 之间的每个素数。我正在寻找如何更好地格式化或改进它的技巧。

Answer 1

import math
n = 60000
primes = []
for k in range(2,n):
    if k % 2 == 0 and k > 2: 
        pass
    elif all(k % i for i in range(3, int(math.sqrt(k)) + 1, 2)):
        print(k)

编辑

由于偶数除了2之外都不是素数，所以我再次修改了答案

import math
primes2 = [2]
for k in range(3,n,2):
    if all(k % i for i in range(3, int(math.sqrt(k)) + 1, 2)):
        primes2.append(k)

或衬垫（它也将有助于性能）

from math import sqrt # Most efficient way to use external libraries
primes = [k for k in range(3,n+1,2) if all(k % i for i in range(3, int(sqrt(k)) + 1, 2))]
primes.insert(0, 2)
# To print
print(*[k for k in range(3,n+1,2) if all(k % i for i in range(3, int(sqrt(k)) + 1, 2))],sep='\n')

Answer 2

这就是我会做的事情

def isPrime(n) :
    if (n <= 1) :
        return False
    if (n <= 3) :
        return True
    if (n % 2 == 0 or n % 3 == 0) :
        return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

def printPrime(n):
    for i in range(2, n + 1):
        if isPrime(i):
            print (i, end =" ")

n = 60000  <--- you can change n by any value you want        
printPrime(n)