3

如果我可以选择

''.join( ['a', 'b'] )

''.join( ('a', 'b') )

我应该使用哪一个(哪一个更快)?有关系吗?

4

1 回答 1

5

它们几乎是等价的,您始终可以使用timeit模块对代码进行计时:

In [145]: small_lis,small_tup = ['a','b']*10, ('a','b')*10

In [146]: avg_lis,avg_tup = ['a','b']*1000, ('a','b')*1000

In [147]: huge_lis,huge_tup = ['a','b']*10**6, ('a','b')*10**6

项目数为 20 时的计时结果:

>>> %timeit ''.join(small_lis)
1000000 loops, best of 3: 987 ns per loop

>>> %timeit ''.join(small_tup)
1000000 loops, best of 3: 1 us per loop

平均大小(2000 项):

>>> %timeit ''.join(avg_lis)
10000 loops, best of 3: 71.5 us per loop

>>> %timeit ''.join(avg_tup)
10000 loops, best of 3: 72.8 us per loop

超大尺寸(2* 10**6 件):

>>> %timeit ''.join(huge_lis)
1 loops, best of 3: 79.9 ms per loop

>>> %timeit ''.join(huge_tup)
1 loops, best of 3: 77.5 ms per loop
于 2013-04-26T19:39:40.047 回答