编写干净的代码:在金字塔中嵌套/If/Elif

Question

目标：试图打印未决结果(Assessment.name，例如'Becoming a Leader'、'Something Else'等.--打印用户在all_assessments中未完成的内容)和已完成的结果( Assessment_Results.assessment.name中的assessment.name，例如'Becoming a Leader' --如user_results中所示)，使用带必要条件的干净和快速的for循环打印用户已完成的结果。

问题：当前代码没有达到上述目标。

任何建议都是非常感谢的！我还是个新手，所以任何指导都是受欢迎的！

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所做工作的细目：

当前，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')>]

而all_results则输出列表中每个用户的所有结果。如下所示：

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打印由用户名(基于登录的人)找到的结果。

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，当迭代两个嵌套列表时，有一个“查找”布尔值，在进入内环之前将其设置为False --内环结束后，您可以检查变量，并根据其值将评估结果推入两个列表中的一个。

正如您所怀疑的，这段代码可能效率很低，因为它必须对所有评估和所有结果的乘积进行迭代，因此，如果您有10个评估和10个结果，它将需要100个迭代，但是如果您有100个评估和100个结果，则需要10.000次迭代。但它可以作为一种学习练习。