第五次作业

numpy数组及处理：效率对比

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

#用列表+循环实现，并包装成函数
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))

#用numpy实现，并包装成函数
import numpy

def mynumpy(n):
    a = numpy.arange(n)
    b = numpy.arange(n)
    c = a**2 + b**3
    return c

print(mynumpy(10))

#对比两种方法实现的效率
from datetime import datetime
start1 = datetime.now()
pySum(1000000)
timedelta1 = datetime.now() - start1
print(timedelta1)

start2 = datetime.now()
mynumpy(1000000)
timedelta2 = datetime.now() - start2
print(timedelta2)

timedelta3 = timedelta1-timedelta2
print(timedelta3)

运行结果