在python中将字典值追加到列表

我在将与键“id”一致的学生词典值添加到列表中时遇到问题。非常感谢您的帮助

students = list();
students.append( {'id':12345, 'first_name':'Alice', 
'last_name':'Anderson','assignments':[('assignment_1',0),('assignment_2',2), 
('assignment_3',3)]})
students.append({'id':22345, 'first_name':'John', 
'last_name':'Sparks','assignments':[('assignment_1',2),('assignment_2',3), 
('assignment_3',4)]})
students.append({'id':32345, 'first_name':'Taylor', 
'last_name':'Mason','assignments':[('assignment_1',3),('assignment_2',2), 
('assignment_3',3)]})

def return_passing(students):    
    grade_sum = 0
    counter = 0
    for s in students:  #loop thru students
        for assignment, grade in s['assignments']: 
            grade_sum += grade
            counter += 1
            average = grade_sum / counter
            lst = list()
        if average >= 2.0:
        lst.append((s['id']))
        return lst

return_passing(students) 

print(return_passing(students))   

---

您在错误的位置初始化东西时遇到了一些问题，因此它们会被重置。评论中对此作了解释：

def return_passing(students):    
    lst = [] # initialize lst here
    for s in students:  #loop thru students
        grade_sum = 0 # reset these for each student
        counter = 0
        for assignment, grade in s['assignments']: 
            grade_sum += grade
            counter += 1
        # now that we have gone throug all assignments
        # compute average
        average = float(grade_sum) / counter # convert to float for precision
        if average >= 2.0:
            lst.append(s['id'])
    return lst # return only after you've gone through all students

---

你有什么错误？我没有错误。我似乎无法获得附加到列表中的值。所以通过考试的学生是：2234532345对吗？很好的解释！非常感谢。