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。表格中有2个条件需要检查：

语言
提供商

取决于这两个值，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种提供程序和7种不同的语言，我试图找出最有效（和Pythonic的方式）来处理这种情况，并提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 ...

...

也许有更好的方法来解决这个问题？

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}

为清楚起见，我已经编造了'en', 'ru'个值。看来0在您的情况下？还要注意，english和russian是独立的函数。最后，LanguageData类不是强制性的，您只需从这些函数返回一个字典即可。但是使用属性而不是字符串键进行维护似乎更容易。

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，如果存在两个以上的可能值，workflow可以用类似的方式包装。

与provider类似。除非操作同时依赖于provider和language，否则在这种情况下，您需要一个双重映射：

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

原始代码可以用棘手的装饰器简化：

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)

还请注意，如果数据是“恒定的”（即不依赖于上下文），则可以完全省略可调用对象以使所有内容更简单：

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

处理大量常数的最有效方法

1 个答案: