python - 根据不同的标准规则检查列表

Question

给定一个项目列表，我想创建一个函数来检查它是否符合特定条件并返回 true 或 false。

这是标准和预期输出的示例:

    #the to check against with criteria[1]
    langList = ['spanish','english','russian','persian']

    criteria = ['any'] #matches any language
    result: match

    #langList does not contain one or all
    criteria = ['any-but-or',[english,chinese]] 
    result: no match

    #langList does not contain all
    criteria = ['any-but-and',[english,chinese]] 
    result: match

    #langList contains one or all
    criteria = ['any-with-or',[english,chinese]] 
    result: match

    #langList contains all
    criteria = ['any-with-and',[english,chinese]] 
    result: no match

    #langList contains one or all and no others
    criteria = ['only-with-or',[english,chinese]] 
    result: no match

    #langList contains all and no others
    criteria = ['only-with-and',[english,chinese]] 
    result: no match

执行此操作的最佳方法是什么？ 我的想法如下所示，但我还没有掌握列表理解，我认为这是这里的关键。

    def check_criteria_match(criteria, languageList):

        rule = criteria[0]
        criteriaLanguages = criteria[1]
        match = True

        if rule != 'any':
            continue
        elif rule = 'any-but-or' #check languageList:
            match = False; break

        #[...]

        return match

编辑: 基于 Praveen Gollakota 的 answer , 这是最后一个函数，她是不是很漂亮:)

def check_rule(rule, o, p):
    # o = original, p = pattern
    if rule == 'any':
        return True
    elif rule == 'any-but-or':
        return not bool(set(p)-set(o))
    elif rule == 'any-but-and':
        return len(set(p)-set(o)) != 0
    elif rule == 'any-with-or':
        return len(set(p)-set(o)) <= 1
    elif rule == 'any-with-and':
        return len(set(p)-set(o)) == 0
    elif rule == 'only-with-or':
        return len(set(o)-set(p)) <= 1
    elif rule == 'only-with-and':
        return len(set(o)-set(p)) == 0

Answer 1

您想进行成员资格测试，sets 是合适的数据结构，set 操作运行良好。

您可以像这样使用集合和 lambda 函数来构建基于规则的调度程序。

>>> rule_checker = {'any': lambda o, p: True, 
...                 'any-but-or': lambda o, p: not bool(set(p) - set(o)),
...                 'any-but-and': lambda o, p: len(set(p) - set(o)) != 0,}

>>> lang_list = ['spanish','english','russian','persian']
>>> rule_checker['any-but-and'](lang_list, ['english','chinese'])
True
>>> rule_checker['any-but-or'](lang_list, ['english','chinese'])
False

当然，您需要在其中添加其余规则。

编辑:您也可以像这样在函数中编写规则。

>>> def check_rule(rule, o, p):
...     # o = original, p = pattern
...     if rule == 'any':
...         return True
...     elif rule == 'any-but-or':
...         return not bool(set(p)-set(o))
...     elif rule == 'any-but-and':
...         return len(set(p)-set(o)) != 0
... 
>>> check_rule('any-but-and', lang_list, ['english','chinese'])
True