处理大量常量的最有效方法

Question

我正在编写一个程序，该程序根据 Excel 表中的某些值进行 API 调用。将检查表中的 2 个条件：

• 语言
• 提供者

根据这两个值，API 调用需要一组不同的常量：

def run_workflow(provider, language, workflow):

    if provider == 'xxxx' and language == 0:

    wf_ready = provider_ready
    wf_unverified = provider_unverified
    wf_active = provider_active
    wf_another = provider_another
    wf_closed = provider_closed
    wf_wrongid = provider_wrongid

    elif provider == 'yyyy' and language == 0:

    wf_ready = provider_ready
    wf_unverified = provider_unverified
    wf_active = provider_active
    wf_another = provider_another
    wf_closed = provider_closed
    wf_wrongid = provider_wrongid

    elif ...


    if workflow == 'ready':
    response = requests.post(API + wf_ready),headers=header, data=json.dumps(conversation))

    elif workflow == 'unverified':
    response = requests.post(API + wf_unverified),headers=header, data=json.dumps(conversation))

    elif ...

有 2 种提供者和 7 种不同的语言，我试图找出处理这种情况的最有效（和 Pythonic 方式）并想出为每种语言创建一个类：

class Workflow_Language():

  def english(self):

    self.provider_unverified = 1112
    self.provider_ready = 1113
    self.provider_active = 1114
    self.provider_vip = 1115

  def russian(self):

    self.provider_unverified = 1116
    self.provider_ready = 1117
    self.provider_active = 1118
    self.provider_vip = 1119

  def ...

...

有没有更好的方法来处理这个问题？

Answer 1

一种方法是将常量映射到适当的处理程序：

class LanguageData:
    def __init__(self, unverified, ready, active, vip):
        self.unverified = unverified
        self.ready = ready
        self.active = active
        self.vip = vip

def english():
    return LanguageData(1,2,3,4)

def russian():
    return LanguageData(5,6,7,8)

LANGUAGE_MAP = {'en': english, 'ru': russian}

为了清楚起见，我编造了'en', 'ru' 值。你的情况似乎是0？另请注意，english 和 russian 是独立函数。最后，LanguageData 类不是强制性的，您可以简单地从这些函数中返回一个字典。但是使用属性而不是字符串键似乎更容易维护。

然后在代码中：

def run_workflow(provider, language, workflow):
    lang_data = LANGUAGE_MAP[language]()
    if workflow == 'ready':
        url = API + data.ready
    elif workflow == 'unverified':
        url = API + data.unverified
    response = requests.post(url, headers=header, data=json.dumps(conversation))

当然，workflow 如果有超过 2 个可能的值，也可以用类似的方式包装。

类似于provider。除非动作同时依赖于provider 和language，在这种情况下你需要一个双映射：

LANG_PROV_MAP = {
    ('en', 'xxxx'): first,
    ('ru', 'yyyy'): second,
}
def run_workflow(provider, language, workflow):
    data = LANG_PROV_MAP[(provider, language)]()
    ...

原始代码可以用一个棘手的装饰器来简化：

LANGUAGE_MAP = {}
def language_handler(lang):
    def wrapper(fn):
        LANGUAGE_MAP[lang] = fn
        return fn
    return wrapper

@language_handler('en')
def handler():
    return LanguageData(1,2,3,4)

@language_handler('ru')
def handler():
    return LanguageData(5,6,7,8)

还要注意，如果数据是“恒定的”（即不依赖于上下文），那么您可以完全省略可调用对象以使一切变得更简单：

LANGUAGE_MAP = {
    'en': LanguageData(1,2,3,4),
    'ru': LanguageData(5,6,7,8),
}
def run_workflow(provider, language, workflow):
    data = LANGUAGE_MAP[language]
    ...

Answer 2

语言和提供者的组合可以组成方法名，调用将被动态调用。

例子：

import sys

def provider1_lang2():
    pass

def provider2_lang4():
    pass

 # get the provider / lang and call the method dynamically
 provider = 'provider2'
 lang = 'lang4' 
 method_name = '{}_{}'.format(provider,lang)
 method =  getattr(sys.modules[__name__], method_name)
 method()