python numpy 计算1-10000 平方 立方 执行效率

import sys
from datetime import datetime
import numpy as np
import matplotlib.pyplot as plt
#使用NumPy计算
def numpysum(n) :
    a = np.arange(n)**2
    b = np.arange (n)** 3
    c =a + b
    return c
#使用Python计算
#并这里由于源码为Python 2的，python 3中range的用法有变,不再直接返回列表#所以强制转化列表
def pythonsum(n) :
    a=list(range(n))
    b =list(range(n))
    c = []
    for i in range(len(a)):
        a[i]=i**2
        b[i]=i**3
        c.append(a[i] +b[i])
    return c
# prt表示是否打印结果
def printest (func,size, prt=True):
    start = datetime.now()
    c=func(size)
    delta = datetime.now() - start
    if prt==True:
        print ( "The last 2 elements of the sum ",c[-2: ])
        print ( 'Elapsed time in microsecondas ' , delta.microseconds)
    return delta.microseconds
#用于作n-time图
def timeplot () :
    pts = [ ]
    for i in range (100,10000,100):
        t_numpy = printest (numpysum,i, prt=False)
        t_python = printest (pythonsum,i,prt=False)
        pts.append ([t_numpy,t_python])
    plt.plot(pts)
    plt.legend ( [ ' Numpy' , 'Python' ])
    plt.show ( )
if __name__=='__main__':
    size = int(sys.argv[1])
    print(' Numpysum. . .')
    printest(numpysum,size)
    print ( 'Pythonsum. . . ')
    printest(pythonsum,size)
    timeplot()