Question

我需要创建一个名为stats的函数来返回列表列表，其中每个内部列表中的第一项是教师的名字，而教师所拥有的课程数中的第二项。它应该返回：[[“Tom Smith”，6]，[“Emma Li”，3]]

Argument是一个字典，如下所示：

teacher_dict = {'Tom Smith': ['a', 'b', 'c', 'd', 'e', 'f'], 'Emma Li': ['x', 'y', 'z']}

这是我的尝试：

def stats(teacher_dict):
    big_list = []
    for teacher, courses in teacher_dict.items():
        number_of_courses = []
            for key in teacher_dict:
            teacher = ''
            num = 0
            for item in teacher_dict[key]:
                num += 1
            number_of_courses.append((key,num))
    return big_list.append([teacher, number_of_courses])

另一次尝试：

def stats(teacher_dict):
    big_list = []
    for teacher in teacher_dict.items():
        number_of_courses = len(teacher_dict[teacher])
    return big_list.append([teacher, number_of_courses])

非常感谢任何帮助！这两个脚本都有错误，我在Python中仍然非常初级，但我真的想弄清楚这一点。谢谢。

Answer 1

使用列表理解

[[k, len(v)] for k, v in teacher_dict.items()]

Answer 2

teacher_dict = {'Tom Smith': ['a', 'b', 'c', 'd', 'e', 'f'], 'Emma Li': ['x', 'y', 'z']}
stats = lambda teacher_dict: [[teacher, len(courses)] for teacher, courses in teacher_dict.items()]
stats(teacher_dict)

输出： [['Emma Li'，3]，['Tom Smith'，6]]

Answer 3

您的代码示例几乎是好的，问题是return语句过早地结束了该函数。 append调用也必须在循环内：

def stats(teacher_dict):
    big_list = []
    for teacher in teacher_dict.items():
        number_of_courses = len(teacher_dict[teacher])
        big_list.append([teacher, number_of_courses])
    return big_list

用于创建列表列表的功能

3 个答案: