9.29作业

http://www.runoob.com/python3/python3-date-time.html 时间和日期,菜鸟教程

问题:

  • 数列:
  • a = a1,a2,a3,·····,an
  • b = b1,b2,b3,·····,bn
  • 求:
  • c = a12+b13,a22+b23,a32+b33,·····+an2+bn3

1.用列表+循环实现,并包装成函数

2.用numpy实现,并包装成函数

3.对比两种方法实现的效率,给定一个较大的参数n,用运行函数前后的timedelta表示。

a = list(range(10))
b = list(range(0,50,5))
c=[]
for i in range(len(a)):
    c.append(a[i]+b[i])
print(a,b,c)

def pySum(n):
    d=list(range(n))
    e=list(range(0,5*n,5))
    f=[]
    for i in range(len(d)):
        f.append(d[i]**2+e[i]**3)
    return (f)
print(pySum(8))

import numpy
g = numpy.arange(10)
h = numpy.arange(0,50,5)
i = g + h
print(g,h,i)

def pySum(n):
    j=numpy.arange(n)
    k=numpy.arange(0,5*n,5)
    l=j**2+k**3
    return (l)
print(pySum(8))

运行结果:

import datetime;
#.datetime  .date  .time  .timedelta

#from datetime import datetime  #datetime模块包含一个datetime类

sj = datetime.datetime.today()
sjj = datetime.datetime.now() # 获取当前datetime
print(sj,sjj)

dt = datetime.datetime(2017, 5, 23, 12, 20)   # 用指定日期时间创建datetime
print(dt)
print(dt.year)    #
print(dt.day)     #
print(dt.month)    #
print(dt.hour)    #
print(dt.minute)    #
print(dt.second)    #
print(dt.microsecond)    #毫秒
timedelta = datetime.timedelta(30,60,547282)
now = datetime.datetime.now()
rq = datetime.datetime(2019,1,1)
#now + timedelta(hours=100)
#now - timedelta(days=100)
print(now + timedelta(days=100, hours=12))
#print(now-=timedelta(days=99))
原文地址:https://www.cnblogs.com/Tlzlykc/p/9722957.html