第五次作业

• 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}

def pySum(n):
    a = list(range(10))
    b = list(range(0, 60, 6))
    c = []
    for i in range(len(a)):
        c.append(a[i] ** 2 + b[i] ** 3)
    return(c)

print(pySum(10))

^{2.用numpy实现，并包装成函数}

import numpy
def npSum(n):
    a = numpy.arange(10)
    b = numpy.arange(0, 60, 6)
    c = a + b
    return(c)

print(npSum(10))

 

^{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。}

 

def pySum(n):
    a=list(range(n))
    b=list(range(0,5*n,5))
    c=[]
    for i in range(len(a)):
        c.append(a[i]**2+b[i]**3)
    return(c)
print(pySum(10))

def npSum(n):
    a=list(range(n))
    b=list(range(0,5*n,5))
    c=[]
    for i in range(len(a)):
        c.append(a[i]**2+b[i]**3)
    return(c)

print(pySum(10))

from datetime import datetime
start = datetime.now()
pySum(2000000)
delta=datetime.now()-start
print(delta)

start= datetime.now()
npSum(2000000)
delta=datetime.now()-start

print(delta)