在Python 3中，更改列表中值的最有效方法是什么？

Question

I'm learning Python, and I was trying to change a list in different ways. For instance, if I have list called names like this:

names = ["David", "Jake", "Alex"]

and I want to add the name "Carter" into the list, what is the most efficient way to accomplish this? Here's some of the things I can do:

names.append("Carter")

names = names + ["Carter"]

names += ["Carter"]

Answer 1

append is the fastest. Here is how you build a small profile using the timeit module

import timeit
a = (timeit.timeit("l.append('Cheese')", setup="l=['Meat', 'Milk']"))
b = (timeit.timeit("l+=['Cheese']", setup="l=['Meat', 'Milk']"))
c = (timeit.timeit("append('Cheese')", setup="l=['Meat', 'Milk'];append = l.append"))
print ('a', a)
print ('b', b)
print ('c', c)
print ("==> " , (c < a < b))

As you can see, In python the access to the method append takes half of the time as the l.append itself...

a 0.08502503100316972

b 0.1582659209962003

c 0.041991976962890476

==> True

Answer 2

You can use the timeit package as shown in this blog post .

Here is a complete code running the tests 20000 times each test:

import timeit
t = 20000

print( "Addition (lst = lst + [4, 5, 6])" )
print( timeit.Timer("lst = lst + [4, 5, 6]", "lst = [1, 2, 3]").timeit(t) )
print( "Addition (lst += [4, 5, 6])" )
print( timeit.Timer("lst += [4, 5, 6]", "lst = [1, 2, 3]").timeit(t) )
print( "Extend (lst.extend([4, 5, 6]))" )
print( timeit.Timer("lst.extend([4, 5, 6])", "lst = [1, 2, 3]").timeit(t) )
print( "Append loop (lst.append([4, 5, 6]))" )
print( timeit.Timer("for i in [4,5,6]: lst.append(i)", "lst = [1,2,3]").timeit(t) )
print( "Append loop, no dot (a(i))" )
# a.b does a lookup, we don't want that, it is slower. Instead use b = a.b 
# then use b.
print( timeit.Timer("""a = lst.append
for i in [4,5,6]: a(i)""", "lst = [1,2,3]").timeit(t) )

And the results (Python 3.4.4) are:

Addition (lst = lst + [4, 5, 6])
1.947201736000352
Addition (lst += [4, 5, 6])
0.0015889199999037373
Extend (lst.extend([4, 5, 6]))
0.0020685689996753354
Append loop (lst.append([4, 5, 6]))
0.0047527769997941505
Append loop, no dot (a(i))
0.003853704999983165