python判断是否为素数

import math 

def isPrime(n): 

    if n <= 1: 

    return False

    for i in range(2, int(math.sqrt(n)) + 1): 

    if n % i == 0: 

        return False

    return True

from math import sqrt 

N = 100

[ p for p in   range(2, N) if 0 not in [ p% d for d in range(2, int(sqrt(p))+1)] ]

from itertools import count 

def isPrime(n):  www.2cto.com

    if n <= 1: 

        return False

    for i in count(2): 

        if i * i > n: 

            return True

        if n % i == 0: 

            return False

def isPrime(n): 

    if n <= 1: 

        return False

    i = 2

    while i*i <= n: 

        if n % i == 0: 

            return False

        i += 1

    return True

def isPrime(n): 

    if n <= 1: 

        return False

    if n == 2: 

        return True

    if n % 2 == 0: 

        return False

    i = 3

    while i * i <= n: 

        if n % i == 0: 

            return False

        i += 2

    return True