115

我有一个 in 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()

这个结构做了很多，但是因为它在遇到 else 之前遍历了每个条件，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这些条件中的任何一个，因此无论如何它都应该对其进行测试。

有谁知道这是否以及如何更有效地完成，或者这仅仅是最好的方法吗？

4

9 回答 9

116

编码...

options.get(something, doThisMostOfTheTime)()

...看起来它应该更快，但它实际上比if... elif...else构造慢，因为它必须调用一个函数，这在紧密循环中可能是一个显着的性能开销。

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 时间量...

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

...使用用户时间从time(1).

选项 #4 确实具有为每个不同的键未命中添加一个新项目的额外内存开销，因此如果您期望不同的键未命中的数量不受限制，我会选择选项 #3，这仍然是一个显着的改进原来的构造。

于 2013-06-18T10:29:24.827 回答
84

我会创建一个字典：

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

现在只使用：

options.get(something, doThisMostOfTheTime)()

如果在字典something中找不到，则返回默认值optionsdict.getdoThisMostOfTheTime

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，检查键 usingkey in options是这里最有效的方法：

if key in options:
   options[key]()
else:
   doSomethingElse()
于 2013-06-18T10:07:53.850 回答
9

你能用pypy吗？

保留您的原始代码但在 pypy 上运行它可以为我提供 50 倍的加速。

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

皮皮：

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
于 2013-06-18T16:02:41.513 回答
4

这是一个将动态条件转换为字典的 if 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]

这是一种方式，但可能不是最 Pythonic 的方式，因为对于不熟悉 Python 的人来说可读性较差。

于 2017-09-18T20:51:59.987 回答
1

我尝试使用 python 3.10 中引入的 match 语句：

5.py

something = 'something'
for i in range(10000000):
    match something:
        case "this":
            the_thing = 1
        case "that":
            the_thing = 2
        case "there":
            the_thing = 3
        case _:
            the_thing = 4

这是我用 3.10.0 得到的结果：
1.py: 1.4s
2.py: 0.9s
3.py:
0.7s 4.py:
0.7s 5.py: 1.0s
我想我会得到类似于 1 的结果.py 但它更快。

于 2021-10-12T12:21:22.517 回答
0

最近我遇到了一种替代“嵌套 if else”的方法，它将我的函数的运行时间从 2.5 小时减少到 ~2 分钟..Baam！让我们开始：

早期代码
bin = lambda x:"Unknown" if x==0 else("High" if x>75 else("Medium" if x>50 and x<=75 else("Medium_Low" if x>25 and x< =50 否则“低”）））

col.apply(bin) 时间 ~ 2.5hrs

优化代码

定义字典替代嵌套 if else 
 def dict_function(*args):
'Pass in a list of tuples, which will be key/value pairs'
ret = {}
for k,v in args:
    for i in k:
        ret[i] = v
return ret
Dict = dict_function(([0],"Unknown"),(range(1,25),"Low"),(range(25,50),"Medium_Low"),(range(50,75),"Medium"),(range(75,100),"High"))

col.apply(lambda x:Dict[x])

dict_function 为给定范围创建多个 key_value 对。时间~2分钟

于 2020-11-23T10:58:03.547 回答
0

我最近遇到了同样的问题，虽然不是关于性能，但我不喜欢创建函数并将它们手动添加到字典的“API”。我想要一个类似于 的 API functools.singledispatch，但基于值而不是类型进行调度。所以 ...

def value_dispatch(func):
    """value-dispatch function decorator.
    Transforms a function into a function, that dispatches its calls based on the
    value of the first argument.
    """
    funcname = getattr(func, '__name__')
    registry = {}

    def dispatch(arg):
        """return the function that matches the argument"""
        return registry.get(arg, func)

    def register(arg):
        def wrapper(func):
            """register a function"""
            registry[arg] = func
            return func
        return wrapper

    def wrapper(*args, **kwargs):
        if not args:
            raise ValueError(f'{funcname} requires at least 1 positional argument')
        return dispatch(args[0])(*args, **kwargs)

    wrapper.register = register
    wrapper.dispatch = dispatch
    wrapper.registry = registry
    return wrapper

像这样使用：

@value_dispatch
def handle_something():
    print("default")

@handle_something.register(1)
def handle_one():
    print("one")

handle_something(1)
handle_something(2)

PS：我在Gitlab上创建了一个片段供参考

于 2020-12-06T16:36:25.227 回答
0

人们exec出于安全原因发出警告，但这是一个理想的情况。
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])
于 2020-05-10T21:13:23.393 回答
0

你可以用 switch-case 类型模仿 if-elif-else，比如使用字典和 lambda 函数

例如：

x = 5
y = 5
operator = 'add'

def operation(operator, x, y): 
 return {
   'add': lambda: x+y,
   'sub': lambda: x-y,
   'mul': lambda: x*y,
   'div': lambda: x/y
 }.get(operator, lambda: None)()

result = operation(operator, x, y)
print(result)
于 2021-10-12T14:54:34.250 回答