作业5

# 数列：a1,a2,a3 ,·····,an/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
def pySum(n):
    a1 = list(range(n))
    b1 = list(range(0,3*n,3))
    c1 = []
    for i in range(len(a1)):
        c1.append(a1[i]**2+b1[i]**3)
    return (c1)
print(pySum(99))
# 2.用numpy实现，并包装成函数
import  numpy
def nySum(n):
    a = numpy.arange(n)
    b = numpy.arange(n)
    c = a**2 + b**3
    return(c)
# 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。
from datetime import datetime,timedelta
start1 = datetime.now()
pySum(100000)
delta1 = datetime.now()-start1
print(delta1)

import  numpy
start2 = datetime.now()
nySum(100000)
delta2 = datetime.now()-start2
print(delta2)