为什么numpy list访问速度比vanilla python慢​​？

Question

我的印象是numpy对于列表操作会更快,但下面的例子似乎表明不是这样:

import numpy as np
import time

def ver1():
    a = [i for i in range(40)]
    b = [0 for i in range(40)]
    for i in range(1000000):
        for j in range(40):
            b[j]=a[j]

def ver2():
    a = np.array([i for i in range(40)])
    b = np.array([0 for i in range(40)])
    for i in range(1000000):
        for j in range(40):
            b[j]=a[j]

t0 = time.time()
ver1()
t1 = time.time()
ver2()
t2 = time.time()

print(t1-t0)
print(t2-t1)

输出是:

4.872278928756714
9.120521068572998

(我在Windows 7中运行64位Python 3.4.3,在i7 920上运行)

我知道这不是复制列表的最快方法,但我试图找出我是否正在使用numpy错误.或者是这种操作的numpy速度较慢,而且在更复杂的操作中效率更高？

编辑:

我也试过以下,只是通过b [:] = a直接复制,而numpy仍然是两倍慢:

import numpy as np
import time

def ver6():
    a = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
    b = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
    for i in range(1000000):
        b[:] = a

def ver7():
    a = np.array([0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0])
    b = np.array([0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0])
    for i in range(1000000):
        b[:] = a

t0 = time.time()
ver6()
t1 = time.time()
ver7()
t2 = time.time()

print(t1-t0)
print(t2-t1)

输出是:

0.36202096939086914
0.6750380992889404

Answer 1

你使用NumPy错了.NumPy的效率依赖于在C级循环中尽可能多地完成工作而不是解释代码.当你这样做

for j in range(40):
    b[j]=a[j]

这是一个解释循环,具有所有内在解释器开销和更多,因为NumPy的索引逻辑比列表索引更复杂,并且NumPy需要在每个元素检索上创建一个新的元素包装器对象.当您编写这样的代码时,您无法获得NumPy的任何好处.

你需要编写代码,以便在C中完成工作:

b[:] = a

这也可以提高列表操作的效率,但对NumPy来说更重要.