python 中的numpy处理数据的效率

import numpy as np
import time

x1 = np.random.rand(100000000)
x2 = x1
#使用循环计算向量点积
tic = time.process_time()
dot = 0
for i in range(len(x1)):
    dot += x1[i] * x2[i]
toc = time.process_time()
time1 = toc-tic
print("dot=" + str(dot) +"\n for loop----computation time = " + str(1000*(toc-tic)) + 'ms')

#使用numpy函数求点积
tic = time.process_time()
dot = 0
dot = np.dot(x1,x2)
toc = time.process_time()
time2 = toc-tic
print("dot=" + str(dot) +"\n verctor version----computation time = " + str(1000*(toc-tic)) + 'ms')

import numpy  as np 
import time
import math
x= [i * 0.001 for i in np.arange(1000000)]
start = time.perf_counter()
for i,t in enumerate(x):
    x[i] = math.sin(t)
math_time = time.perf_counter()-start
print("math.sin:",math_time)  

x =np.array(x)
start = time.perf_counter()
np.sin(x)
numpy_time = time.perf_counter()-start
print("numpy.sin:",numpy_time)
print("math.time 所花的时间是numpy.sin的"+ str(math_time/numpy_time) +"倍。")