3.x中"加入"速度较慢吗？

Question

当我遇到这个怪癖时,我只是乱搞.我想确保我不疯狂.

以下代码(适用于2.x和3.x):

from timeit import timeit
print ('gen: %s' % timeit('"-".join(str(n) for n in range(1000))', number=10000))
print ('list: %s' % timeit('"-".join([str(n) for n in range(1000)])', number=10000))

在每个版本,同一台机器上运行3次.

注意:我将时间分组在同一行,以节省空间.

在我的Python 2.7.5上:

gen: 2.37875941643, 2.44095773486, 2.41718937347
list: 2.1132466183, 2.12248106441, 2.11737128131

在我的Python 3.3.2上:

gen: 3.8801268438439718, 3.9939604983350185, 4.166233972077624
list: 2.976764740845537, 3.0062614747229555, 3.0734980312273894

我想知道为什么这是....它可能与如何实现字符串有关？

---

编辑:我没有使用它再次这样做,range()因为它也从2.x略有变化到3.x而是我使用下面的新代码:

from timeit import timeit
print ('gen: %s' % timeit('"-".join(str(n) for n in (1, 2, 3))', number=1000000))
print ('list: %s' % timeit('"-".join([str(n) for n in (1, 2, 3)])', number=1000000))

Python 2.7.5的时间安排:

gen: 2.13911803683, 2.16418448199, 2.13403650485
list: 0.797961223325,  0.767758578433, 0.803272800119

Python 3.3.2的时间安排:

gen: 2.8188347625218486, 2.882846655874985, 3.0317612259663718
list: 1.3590610502957934, 1.4878876089869366, 1.4978070529462615

---

编辑2:似乎有更多的事情摆脱了计算,所以我尝试把它降到最低限度.

新守则:

from timeit import timeit
print ('gen: %s' % timeit('"".join(n for n in ("1", "2", "3"))', number=1000000))
print ('list: %s' % timeit('"".join([n for n in ("1", "2", "3")])', number=1000000))

时间Python 2.7.5:

gen: 1.47699698704, 1.46120314534, 1.48290697384
list: 0.323474182882, 0.301259632897, 0.323756694047

时间Python 3.3.2:

gen: 1.633002954259608, 1.6049987598860562, 1.6109927662465935
list: 0.5621341113519589, 0.5789849850819431, 0.5619928557696119

---

区别很明显,2.x更快,3.x更慢.我很好奇为什么......

Answer 1

我还没有在python3.3上工作过.我在下面说的所有这些都是基于观察.

我使用python反汇编程序来跟踪python 3.3和python 2.7.3中的代码.

s = """
''.join([n for n in ('1', '2', '3')])
"""

我发现上传码有变化.

Python 2.7.3

Python 2.7.3 (default, Apr 10 2012, 23:31:26) [MSC v.1500 32 bit (Intel)] on win
32
Type "help", "copyright", "credits" or "license" for more information.
>>> import dis
>>> from timeit import timeit
>>> s = """
... ''.join([n for n in ('1', '2', '3')])
... """
>>> timeit(s, number=100000)
0.08443676085287867
>>>
>>>
>>> c = compile(s, '<string>', 'exec')
>>> dis.dis(c)
  2           0 LOAD_CONST               0 ('')
              3 LOAD_ATTR                0 (join)
              6 BUILD_LIST               0
              9 LOAD_CONST               5 (('1', '2', '3'))
             12 GET_ITER
        >>   13 FOR_ITER                12 (to 28)
             16 STORE_NAME               1 (n)
             19 LOAD_NAME                1 (n)
             22 LIST_APPEND              2
             25 JUMP_ABSOLUTE           13
        >>   28 CALL_FUNCTION            1
             31 POP_TOP
             32 LOAD_CONST               4 (None)
             35 RETURN_VALUE
>>>

python 3.3

Python 3.3.0 (v3.3.0:bd8afb90ebf2, Sep 29 2012, 10:55:48) [MSC v.1600 32 bit (In
tel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import dis
>>> from timeit import timeit
>>> s = """
... ''.join([n for n in ('1', '2', '3')])
... """
>>> timeit(s, number=100000)
0.13603410021487614
>>>
>>>
>>> c = compile(s, '<string>', 'exec')
>>> dis.dis(c)
  2           0 LOAD_CONST               0 ('')
              3 LOAD_ATTR                0 (join)
              6 LOAD_CONST               1 (<code object <listcomp> at 0x01F70BB
0, file "<string>", line 2>)
              9 LOAD_CONST               2 ('<listcomp>')
             12 MAKE_FUNCTION            0
             15 LOAD_CONST               7 (('1', '2', '3'))
             18 GET_ITER
             19 CALL_FUNCTION            1 (1 positional, 0 keyword pair)
             22 CALL_FUNCTION            1 (1 positional, 0 keyword pair)
             25 POP_TOP
             26 LOAD_CONST               6 (None)
             29 RETURN_VALUE
>>>

从上传代码我得到的是列表理解已经改变所以我检查了两个版本中的列表理解

Python 2.7.3

Python 2.7.3 (default, Apr 10 2012, 23:31:26) [MSC v.1500 32 bit (Intel)] on win
32
Type "help", "copyright", "credits" or "license" for more information.
>>>
>>>
>>>
>>> import dis
>>> from timeit import timeit
>>> s = """
... [i for i in ('1', '2', '3')]
... """
>>> timeit(s, number=100000)
0.059500395456104374
>>> c = compile(s, '<string>', 'exec')
>>> dis.dis(c)
  2           0 BUILD_LIST               0
              3 LOAD_CONST               4 (('1', '2', '3'))
              6 GET_ITER
        >>    7 FOR_ITER                12 (to 22)
             10 STORE_NAME               0 (i)
             13 LOAD_NAME                0 (i)
             16 LIST_APPEND              2
             19 JUMP_ABSOLUTE            7
        >>   22 POP_TOP
             23 LOAD_CONST               3 (None)
             26 RETURN_VALUE
>>>

python 3.3

Python 3.3.0 (v3.3.0:bd8afb90ebf2, Sep 29 2012, 10:55:48) [MSC v.1600 32 bit (In
tel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>>
>>>
>>>
>>> import dis
>>> from timeit import timeit
>>> s = """
... [i for i in ('1', '2', '3')]
... """
>>> timeit(s, number=100000)
0.09876976988887567
>>> c = compile(s, '<string>', 'exec')
>>> dis.dis(c)
  2           0 LOAD_CONST               0 (<code object <listcomp> at 0x01FF0BB
0, file "<string>", line 2>)
              3 LOAD_CONST               1 ('<listcomp>')
              6 MAKE_FUNCTION            0
              9 LOAD_CONST               6 (('1', '2', '3'))
             12 GET_ITER
             13 CALL_FUNCTION            1 (1 positional, 0 keyword pair)
             16 POP_TOP
             17 LOAD_CONST               5 (None)
             20 RETURN_VALUE
>>>

我没有使用python3或检查更改.似乎列表理解实现已经改变.在python3.3中有MAKE_FUNCTION和CALL_FUNCTION.(现在在python2.7中调用一个函数代价很高.我不确定在python3.3中调用函数是否仍然代价高昂.如果是这样的话那么可能会增加一些时间.)