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) #年 print(dt.day) #日 print(dt.month) #月 print(dt.hour) #时 print(dt.minute) #分 print(dt.second) #秒 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))