我认为差异是因为您在中构造列表和数组，e而在中仅构造列表d。考虑：
import numpy as np
def d():
a = [1,2,3,4,5]
b = [10,20,30,40,50]
c = [i*j for i,j in zip(a,b)]
return c
def e():
a = np.array([1,2,3,4,5])
b = np.array([10,20,30,40,50])
c = a*b
return c
#Warning: Functions with mutable default arguments are below.
# This code is only for testing and would be bad practice in production!
def f(a=[1,2,3,4,5],b=[10,20,30,40,50]):
c = [i*j for i,j in zip(a,b)]
return c
def g(a=np.array([1,2,3,4,5]),b=np.array([10,20,30,40,50])):
c = a*b
return c
import timeit
print timeit.timeit('d()','from __main__ import d')
print timeit.timeit('e()','from __main__ import e')
print timeit.timeit('f()','from __main__ import f')
print timeit.timeit('g()','from __main__ import g')
这里的功能f和g周围的每一次避免重新创建列表/阵列，我们得到了非常相似的性能：
1.53083586693
15.8963699341
1.33564996719
1.69556999207
请注意，list-comp +zip仍然获胜。但是，如果我们使数组足够大，则numpy会赢得成功：
t1 = [1,2,3,4,5] * 100
t2 = [10,20,30,40,50] * 100
t3 = np.array(t1)
t4 = np.array(t2)
print timeit.timeit('f(t1,t2)','from __main__ import f,t1,t2',number=10000)
print timeit.timeit('g(t3,t4)','from __main__ import g,t3,t4',number=10000)
我的结果是：
0.602419137955
0.0263929367065
602392714
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