编写干净的代码：在 Pyramid 中嵌套 For/If/Elif

Question

目标： 尝试打印 未决结果 （Assessment.name，例如 'Becoming a Leader'、'Something Else' 等...——打印用户在 all_assessments) and completed results(assessment.name from Assessment_Results.assessment.name, e.g. 'Becoming a Leader' --print what a user has completed as shown in user_results) 使用带有必要条件的干净快速的 for 循环。

问题： 当前代码未实现上述目标。

非常感谢任何建议！我还是个新手，所以真的欢迎任何指导！

Views.py

   def view_assessments(request):
        owner = authenticated_userid(request)
        print 'login owner', owner
        if owner is None:
            raise HTTPForbidden()
        all_assessments = api.retrieve_assessments()
        print 'these are all the assessments:', all_assessments
        print 'and type:', type(all_assessments)
        all_results = api.retrieve_assessment_results() # all the assessment results in a list
    
for x in all_assessments:
    alls = x.name
    if alls is not None: 
        for x in all_results: #found user based on all results
            assessment = x.assessment.name
            user = x.owner.username
            if user == owner:
                print 'completed', assessment
            elif assessment != alls: # DOES NOT WORK
                alls.index(assessment)

return {'assessments': all_assessments, 'assessment_results': all_results, 'loggedin': owner, 'user_results': user_results}

api 功能的细分：

目前 all_assessments 打印出所有现有评估名称和文本的列表。

all_assessments = [<Assessment(name='Becoming a Leader', text='better decisions')>, <Assessment(name='Good work', text='working on these skills')>, <Assessment(name='Teaching NTS', text='Series 1.1')>]

while all_results 打印出列表中每个用户的所有结果。此处显示：

all_results [<Assessment_Result(owner='<User(username ='baseball', password='...', firstname ='Jenny', lastname ='Jen', email='dance@aol.com')>', assessment='<Assessment(name='Becoming a Leader', text='better decisions')>')>, <Assessment_Result(owner='<User(username ='donald', password='...', firstname ='Drew', lastname ='James', email='cool@gmail.com')>', assessment='<Assessment(name='Good work', text='working on these skills')>')>]

最后，user_results 打印通过用户名找到的结果（基于登录的人）。

retrieved by username: [<Assessment_Result(owner='<User(username ='dorisday', password='..', firstname ='Doris', lastname ='Day', email='dorisday@gmail.com')>', assessment='<Assessment(name='Becoming a Leader', text='better decisions')>')>, <Assessment_Result(owner='<User(username ='dorisday', password='..', firstname ='Doris', lastname ='Day', email='dorisday@gmail.com')>', assessment='<Assessment(name='Good work', text='working on these skills')>')>]

Answer 1

我会从这样的事情开始：

def view_assessments(request):
     logged_in_userid = authenticated_userid(request)
     if logged_in_userid is None:
         raise HTTPForbidden()
     all_assessments = api.retrieve_assessments()
     all_results = api.retrieve_assessment_results()

     completed_assessments = []
     pending_assessments = []

     for assessment in all_assessments:
         if assessment.name is None: 
             continue

         found_assessment_result = False
         for result in all_results:
             if result.owner.username == logged_in_userid and result.assessment == assessment:
                 found_assessment_result = True
                 break  # no need to check further

         if found_assessment_result:
             compleded_assessments.append(assessment)
         else:
             pending_assessments.append(assessment)


      return {'completed_assessments': completed_assessments, 'pending_assessments': pending_assessments, 'loggedin': owner, 'user_results': user_results}

这里的诀窍是，当遍历两个嵌套列表时，有一个 "found" 布尔值，在进入内部循环之前将其设置为 False - 内部循环完成后，您可以检查变量，并且，根据其价值，将评估推入两个列表之一。

如您所料，此代码可能效率很低，因为它必须迭代所有评估和所有结果的乘积，因此如果您有 10 个评估和 10 个结果，则需要 100 次迭代，但是如果您有 100 次评估和 100 次结果，它将是 10.000 次迭代。但它会作为一个学习练习。

编写干净的代码：在 Pyramid 中嵌套 For/If/Elif

Writing clean code: Nested For/If/Elif in Pyramid

python

for-loop

if-statement

pyramid

Views.py