输出第10000个素数：
from math import sqrt
def is_prim(number):
if number < 2:
return False
for i in range(2, int(sqrt(number)) + 1):
if number % i == 0:
return False
return True
def get_prim(nth):
if nth==1:return 2
x = 1
odd = 3
while True:
if is_prim(odd):x += 1
if x==nth:return odd
odd+=2 #只算奇数
nth = int(input("Input nth prime： "))
print(get_prim(nth))****************************************分割线****************************************
边读边写：逗号语法用来嵌套
with open('read me',encoding='utf8') as r,open('write me.log','r+',encoding='utf8') as w:
for line in r:
if '的' in line:
w.write(line) # line [:-1]，不读取末尾的\n相较【 for line in open(filepath).readlines()】 ，【 with open(filepath) as f】 的优点：
with负责处理open和close文件，包括抛出内部异常。而for line in f，将文件对象f当做迭代对象，将自动处理IO缓冲和内存管理，这样就无需担心几个G的 大文件的处理了。
****************************************分割线****************************************
以0~9这10个数字生产5位数，各5位数内无重复数字：
t=list(range(10))
data=[int('%s'*3 %(x,y,z)) for x in t for y in t for z in t if (x!=0 and x!=y and x!=z and y!=z)] #3位数
t=[str(x) for x in t]
import itertools
data1=[int(''.join(x)) for x in itertools.permutations(t,5) if x[0]!='0'] #元素内无相同字符，5位数
data2=[''.join(x) for x in itertools.combinations_with_replacement(t,5) if x[0]!='0']
data3=[''.join(x) for x in itertools.combinations(t,5) if x[0]!='0']
data4=[a+b+c+d+e for a,b,c,d,e in itertools.product(t,t,t,t,t) if a!='0'] #笛卡尔积
print(len(data1),data1)*********** ****** *分割线 ************ ******
由[1,6,3]，得到[[1, 6, 3], [1, 6, 3], [1, 6, 3], [1, 6, 3], [1, 6, 3]]：
import itertools
ns = itertools.repeat([1,6,3], 5)
print(list(ns))*********** ****** *分割线 ************ ******
以元素的出现次数降序排序，取前4项：
words = ['look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes', 'the', 'eyes', 'the', 'eyes',
'the', 'eyes', 'not', 'around', 'the', 'my', 'eyes', "you're", 'under', 'eyes', "don't",
'look', 'around', 'the', 'eyes', 'look', 'into' ]
from collections import Counter
print(Counter(words).most_common(4))
