当 If 为 True 时编程诉诸 Else 语句

Question

我正在处理字典列表：

student_list = [{'id': 12341, 'first_name': 'Alice', 'last_name': 'Anderson',
     'assignments': [('assignment_1', 0), ('assignment_2', 2), ('assignment_3', 4)]},

 {'id': 12342, 'first_name': 'Boris', 'last_name': 'Bank',
   'assignments': [('assignment_1', 1), ('assignment_2', 3), ('assignment_3', 0)]},

 {'id': 12343, 'first_name': 'Carl', 'last_name': 'Cape',
   'assignments': [('assignment_1', 2), ('assignment_2', 4), ('assignment_3', 1)]},

 {'id': 12344, 'first_name': 'Didi', 'last_name': 'Dawson',
   'assignments': [('assignment_1', 3), ('assignment_2', 0), ('assignment_3', 2)]},

 {'id': 12345, 'first_name': 'Ed', 'last_name': 'Enders',
   'assignments': [('assignment_1', 4), ('assignment_2', 1), ('assignment_3', 3)]}]

我需要编写一个函数，通过学生姓名搜索字典，查看是否有匹配的作业名称和成绩。如果有则返回 True，如果没有则返回 False 并将其添加到列表中。我与我所拥有的很接近，但无论我输入一个 true 还是 false 集，如果参数它读取为 false 并将分配添加为元组。

您能帮我找到此功能所需的缺失指令吗？

def add_grade(student, assignment_name, grade):
  for s in student_list:
    if s['first_name'] == student:
        if any(s['assignments']) == (assignment_name, grade):
          print(True)
        else:
          s['assignments'].append((assignment_name, grade))
          print(False)

add_grade('Carl', 'assignment_2', 2)
print(student_list)

编辑： 第一个答案帮助我使代码正常工作（谢谢！）

现在我已将其扩大并使其更加草率，以便将一个作业添加两次并使用两个不同的等级，它会输出它已经存在的输出。我所做的既笨重又凌乱，看起来修复应该很容易，但我想不通。想法？

def add_grade2(student, assignment_name, grade):
  for s in student_list:
    if s['first_name'] == student:
        if (assignment_name, grade) in s['assignments']:
          print(True)
        elif (assignment_name, 0) in s['assignments']:
          print('That assignment has a different grade. ')
          break
        elif (assignment_name, 1) in s['assignments']:
          print('That assignment has a different grade. ')
          break
        elif (assignment_name, 2) in s['assignments']:
          print('That assignment has a different grade. ')
          break
        elif (assignment_name, 3) in s['assignments']:
          print('That assignment has a different grade. ')
          break
        elif (assignment_name, 4) in s['assignments']:
          print('That assignment has a different grade. ')
          break
        else:
          s['assignments'].append((assignment_name, grade))
          print(False)


add_grade2('Alice', 'assignment_1', 3)
print(student_list)

Answer 1

问题在于，对于单个学生，"assignments" 是一个元组列表，在第一种情况下，您将整个列表与单个元组进行比较，在第二种情况下，您只需检查 @ 987654322@（元组的第一个位置）在列表中。

您可以将分配存在检查更改为：

if (assignment_name, grade) in s['assignments']:
    print(True)
else:
    s['assignments'].append((assignment_name, grade))
    print(False)

此外，为了限制缩进并使内容更易于阅读，您可以使用next 查找正确的学生词典。然后你将拥有：

def add_grade(student, assignment_name, grade):
    s = next(s for s in student_list if s['first_name'] == student)
    if (assignment_name, grade) in s['assignments']:
        print(True)
    else:
        s['assignments'].append((assignment_name, grade))
        print(False)

编辑：每个作业只允许一个成绩（不能覆盖）：

for i in range(len(s['assignments'])):
    if s['assignments'][i][0] == assignment_name:
        if s['assignments'][i][1] == grade:
            print(True)
        else:
            print('That assignment has a different grade.')
        return
s['assignments'].append((assignment_name, grade))
print(False)

或（逻辑更简单，但计算量略多）：

if (assignment_name, grade) in s['assignments']:
    print(True)
elif any(a_name == assignment_name for a_name, _ in s['assignments']):
    print('That assignment has a different grade.')
else:
    s['assignments'].append((assignment_name, grade))
    print(False)

覆盖：

for i in range(len(s['assignments'])):
    if s['assignments'][i][0] == assignment_name:
        s['assignments'][i] = (assignment_name, grade)
        print(True)
        return
s['assignments'].append((assignment_name, grade))
print(False)

Answer 2

any 返回的结果将是True 或False。这将永远比较等于(assignment_name, grade)。

不清楚您是否只想匹配名称，或者名称​​和等级。无论哪种方式，您都需要一个循环来搜索可能性。