Question

我正在编写以下代码来计算学生的综合成绩以管理他们的成绩单，但没有获得所需的输出，有人可以帮助我吗，Thanx！

def transcript(coursedetails, studentdetails, grades):
    studentdetails.sort()
    coursedetails.sort()
    grades.sort()
    result=[(student,grade,coursedetail)
            for student in studentdetails 
            for grade in grades for coursedetail in coursedetails 
            if((student[0]==grade[0])and(grade[1]==coursedetail[0]))]
    print(result)

我给的输入是---

transcript([("MA101","Calculus"),("PH101","Mechanics"),("HU101","English")],[("UGM2018001","Rohit Grewal"),("UGP2018132","Neha Talwar")],[("UGM2018001","MA101","AB"),("UGP2018132","PH101","B"),("UGM2018001","PH101","B")])

实际输出---

[(('UGM2018001', 'Rohit Grewal'), ('UGM2018001', 'MA101', 'AB'), ('MA101', 'Calculus')), (('UGM2018001', 'Rohit Grewal'), ('UGM2018001', 'PH101', 'B'), ('PH101', 'Mechanics')), (('UGP2018132', 'Neha Talwar'), ('UGP2018132', 'PH101', 'B'), ('PH101', 'Mechanics'))]

期望输出---

[('UGM2018001', 'Rohit Grewal', [('MA101', 'Calculus', 'AB'), ('PH101', 'Mechanics', 'B')]), ('UGP2018132', 'Neha Talwar', [('PH101', 'Mechanics', 'B')])]

Answer 1

您可以考虑将结构从元组列表更改为字典，因为操作信息会更容易。

def transcript(coursedetails, studentdetails, grades):
    result = {}
    students = {}
    courses = {}
    for course, descript in coursedetails:
        courses[course] = descript
    for sid, student in studentdetails:
        students[sid] = {'name':student}
    for sid, course, grade in grades:
        details = (course, courses[course], grade)
        if not 'grades' in students[sid]:
            students[sid]['grades'] = []
        students[sid]['grades'].append(details)
    return  students

这将返回您的成绩单词典字典，其中学生ID是返回包含学生姓名和成绩的字典的键。

测试：

trans = transcript([("MA101","Calculus"),("PH101","Mechanics"),("HU101","English")],[("UGM2018001","Rohit Grewal"),("UGP2018132","Neha Talwar")],[("UGM2018001","MA101","AB"),("UGP2018132","PH101","B"),("UGM2018001","PH101","B")])

结果：

{'UGM2018001': {'name': 'Rohit Grewal', 'grades': [('MA101', 'Calculus', 'AB'), ('PH101', 'Mechanics', 'B')]}, 'UGP2018132': {'name': 'Neha Talwar', 'grades': [('PH101', 'Mechanics', 'B')]}}

如果您仍希望在上述所需输出中获取此信息，则以下内容将完全按照您的需要创建元组列表。

list((k,v['name'], v['grades']) for k,v in trans.items())

Answer 2

这可以通过collections.defaultdict：

实现

from collections import defaultdict

inputs = [("MA101","Calculus"),("PH101","Mechanics"),("HU101","English")],\
         [("UGM2018001","Rohit Grewal"),("UGP2018132","Neha Talwar")],\
         [("UGM2018001","MA101","AB"),("UGP2018132","PH101","B"),("UGM2018001","PH101","B")]

def transcript(coursedetails, studentdetails, grades):

    d = defaultdict(list)

    courses, students = dict(coursedetails), dict(studentdetails)

    for grade in grades:
        d[(grade[0], students[grade[0]])].append((grade[1], courses[grade[1]], grade[2]))

    return d

res = transcript(*inputs)

<强>结果

defaultdict(list,
            {('UGM2018001', 'Rohit Grewal'): [('MA101', 'Calculus', 'AB'),
                                              ('PH101', 'Mechanics', 'B')],
             ('UGP2018132', 'Neha Talwar'): [('PH101', 'Mechanics', 'B')]})

如果确实需要一个元组列表，这很容易转换：

res2 = [(k[0], k[1], v) for k, v in res.items()]

# [('UGM2018001', 'Rohit Grewal', [('MA101', 'Calculus', 'AB'),
#                                  ('PH101', 'Mechanics', 'B')]),
#  ('UGP2018132', 'Neha Talwar', [('PH101', 'Mechanics', 'B')])]

通过列表解析聚合来自多个列表的数据

2 个答案: