当else做得最多时，最有效的方式来执行if-elif-elif-else语句？

我有一条if-elif-elif-else语句，其中99％的时间执行else语句：

if something == 'this':
    doThis()
elif something == 'that':
    doThat()
elif something == 'there':
    doThere()
else:
    doThisMostOfTheTime()

这个构造完成了很多，但是由于它在碰到其他情况之前会遍历所有条件，所以我觉得这不是很有效，更不用说Pythonic了。另一方面，它确实需要知道是否满足任何这些条件，因此无论如何都应该对其进行测试。

是否有人知道是否以及如何可以更有效地完成此工作，或者这仅仅是做到这一点的最佳方法？

代码...

options.get(something, doThisMostOfTheTime)()

...看起来应该是快，但它实际上是慢于if... elif... else结构，因为它要调用一个函数，它可以在一个紧密的循环一个显著的性能开销。

考虑这些例子...

1.py

something = 'something'

for i in xrange(1000000):
    if something == 'this':
        the_thing = 1
    elif something == 'that':
        the_thing = 2
    elif something == 'there':
        the_thing = 3
    else:
        the_thing = 4

2.py

something = 'something'
options = {'this': 1, 'that': 2, 'there': 3}

for i in xrange(1000000):
    the_thing = options.get(something, 4)

3.py

something = 'something'
options = {'this': 1, 'that': 2, 'there': 3}

for i in xrange(1000000):
    if something in options:
        the_thing = options[something]
    else:
        the_thing = 4

4.py

from collections import defaultdict

something = 'something'
options = defaultdict(lambda: 4, {'this': 1, 'that': 2, 'there': 3})

for i in xrange(1000000):
    the_thing = options[something]

...并注意他们使用的CPU时间...

1.py: 160ms
2.py: 170ms
3.py: 110ms
4.py: 100ms

...使用来自的用户时间time(1)。

选项＃4确实有额外的内存开销，需要为每个不同的键缺失添加一个新项，因此，如果您期望数量众多的不同的键缺失，我会选择方法＃3，它在原始构造。

我要创建一个字典：

options = {'this': doThis,'that' :doThat, 'there':doThere}

现在只使用：

options.get(something, doThisMostOfTheTime)()

如果something在optionsdict中找不到，dict.get则将返回默认值doThisMostOfTheTime

一些时间比较：

脚本：

from random import shuffle
def doThis():pass
def doThat():pass
def doThere():pass
def doSomethingElse():pass
options = {'this':doThis, 'that':doThat, 'there':doThere}
lis = range(10**4) + options.keys()*100
shuffle(lis)

def get():
    for x in lis:
        options.get(x, doSomethingElse)()

def key_in_dic():
    for x in lis:
        if x in options:
            options[x]()
        else:
            doSomethingElse()

def if_else():
    for x in lis:
        if x == 'this':
            doThis()
        elif x == 'that':
            doThat()
        elif x == 'there':
            doThere()
        else:
            doSomethingElse()

结果：

>>> from so import *
>>> %timeit get()
100 loops, best of 3: 5.06 ms per loop
>>> %timeit key_in_dic()
100 loops, best of 3: 3.55 ms per loop
>>> %timeit if_else()
100 loops, best of 3: 6.42 ms per loop

对于10**5不存在的密钥和100个有效密钥：

>>> %timeit get()
10 loops, best of 3: 84.4 ms per loop
>>> %timeit key_in_dic()
10 loops, best of 3: 50.4 ms per loop
>>> %timeit if_else()
10 loops, best of 3: 104 ms per loop

因此，对于普通字典而言，在key in options这里使用键是最有效的方法：

if key in options:
   options[key]()
else:
   doSomethingElse()

可以使用pypy吗？

保留原始代码，但在pypy上运行可使我的速度提高50倍。

CPython：

matt$ python
Python 2.6.8 (unknown, Nov 26 2012, 10:25:03)
[GCC 4.2.1 Compatible Apple Clang 3.0 (tags/Apple/clang-211.12)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>>
>>> from timeit import timeit
>>> timeit("""
... if something == 'this': pass
... elif something == 'that': pass
... elif something == 'there': pass
... else: pass
... """, "something='foo'", number=10000000)
1.728302001953125

pypy：

matt$ pypy
Python 2.7.3 (daf4a1b651e0, Dec 07 2012, 23:00:16)
[PyPy 2.0.0-beta1 with GCC 4.2.1] on darwin
Type "help", "copyright", "credits" or "license" for more information.
And now for something completely different: ``a 10th of forever is 1h45''
>>>>
>>>> from timeit import timeit
>>>> timeit("""
.... if something == 'this': pass
.... elif something == 'that': pass
.... elif something == 'there': pass
.... else: pass
.... """, "something='foo'", number=10000000)
0.03306388854980469

这里是将动态条件转换为字典的if的示例。

selector = {lambda d: datetime(2014, 12, 31) >= d : 'before2015',
            lambda d: datetime(2015, 1, 1) <= d < datetime(2016, 1, 1): 'year2015',
            lambda d: datetime(2016, 1, 1) <= d < datetime(2016, 12, 31): 'year2016'}

def select_by_date(date, selector=selector):
    selected = [selector[x] for x in selector if x(date)] or ['after2016']
    return selected[0]

这是一种方法，但可能不是最Python的方法，因为对于不熟练使用Python的人来说可读性较差。

人们exec出于安全原因发出警告，但这是一个理想的案例。
这是一个简单的状态机。

Codes = {}
Codes [0] = compile('blah blah 0; nextcode = 1')
Codes [1] = compile('blah blah 1; nextcode = 2')
Codes [2] = compile('blah blah 2; nextcode = 0')

nextcode = 0
While True:
    exec(Codes[nextcode])