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How to judge prime numbers in python: first use python’s mathematical functions; then scan the prime numbers in a single line program, the code is [[ p for p in range(2, N) if 0 not in [ p% d for d in range(2,int(sqrt(p)].
The operating environment of this tutorial: windows7 system, python3.9 version, DELL G3 computer.
Python prime number judgment method:
1. Use python’s mathematical functions
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
2. Single-line program to scan prime numbers
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)] ]
Using python’s itertools module
from itertools import count def isPrime(n): www.jb51.net if n <= 1: return False for i in count(2): if i * i > n: return True if n % i == 0: return False
3. Two methods without using modules
Method 1:
def isPrime(n): if n <= 1: return False i = 2 while i*i <= n: if n % i == 0: return False i += 1 return True
Method 2:
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
eg: Find the prime number (prime number) between 20001 and 40001
Since it can only be digitized by 1 or by yourself, then It shows that only when the remainder is 0 twice, the code is as follows:
#!/usr/bin/python L1=[] for x in xrange(20001,40001): n = 0 for y in xrange(1,x+1): if x % y == 0: n = n + 1 if n == 2 : print x L1.append(x) print L1
The result is as follows:
20011 20021 20023 20029 20047 20051 20063 20071 20089 20101 20107 20113 20117 20123 20129 20143 20147 20149 20161 20173 ….
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