Question

在使用Excel多年并学习VBA之后，我现在尝试学习Python。这是场景：

我问了7个夏令营辅导员，他们希望负责哪些活动。每个学生的回答数量都是随机的，选择的活动数量没有上限。但是，每项活动都是唯一的，一旦被学生“要求”，其他任何辅导员都无法要求。结果是：

Adam: archery, canoeing
Bob: frisbee, golf, painting, trampoline
Carol: tennis, dance, skating
Denise: cycling
Eddie: horseback, fencing, soccer
Fiona: painting
George: basketball, football

我对VB最熟悉（我是一个老家伙），过去我将上述信息存储在一个锯齿状的数组中。但是由于我是Python的新手，所以我对如何执行此操作感到困惑。我认为列表列表对我有用，这是我的代码。假设我有一个顾问列表，每个顾问活动的列表都是单独的。如何合并它们或将它们放在一个数据结构中？我在下面做什么错了？谢谢。

counselors = []
counselors = ['Adam','Bob','Carol','Denise','Eddie','Fiona','George']

#create a list of Carol's activities 
activities = []
activities = ['tennis','dance','skating']

counselors[2].append[(activities)]

Answer 1

Python中的锯齿状数组几乎就像您提到的列表列表一样。

我将使用字典来存储咨询员的活动信息，其中键是咨询员的姓名，值是咨询员将负责的活动列表。

counselors_activities = {"Adam": ["archery", "canoeing"],
                      "Bob": ["frisbee", "golf", "painting", "trampoline"],
                      "Carol": ["tennis", "dance", "skating"],
                      "Denise": ["cycling"],
                      "Eddie": ["horseback", "fencing", "soccer"],
                      "Fiona": ["painting"],
                      "George": ["basketball", "football"]}

并按以下方式访问字典中的每个顾问：

counselors_activites["Adam"] # when printed will display the result => ['archery', 'canoeing']

关于这个问题，我会将可用的活动列表存储在一个列表中，并且每当选择了一个活动时，就将其从列表中删除，并按如下方式将其添加到词典的咨询员中：

list_of_available_activities.remove("archery")
counselors_activities["Adam"].append("archery")

如果辅导员不再负责该活动，请将其从他们中删除，并将其重新添加到可用活动列表中。

更新：根据您的评论要求，我在下面提供了更为完善的解决方案。

文本文件activites.txt：

Adam: archery, canoeing
Bob: frisbee, golf, painting, trampoline
Carol: tennis, dance, skating
Denise: cycling
Eddie: horseback, fencing, soccer
Fiona: painting
George: basketball, football

代码：

#Set of activities available for counselors to choose from

set_of_activities = {"archery",
                  "canoeing",
                  "frisbee",
                  "golf",
                  "painting",
                  "trampoline",
                  "tennis",
                  "dance",
                  "skating",
                  "cycling",
                  "horseback",
                  "fencing",
                  "soccer",
                  "painting",
                  "basketball",
                  "football"}

with open('activities.txt', 'r') as f:
    for line in f:

        # Iterate over the file and pull out the counselor's names
        # and insert their activities into a list

        counselor_and_activities = line.split(':')
        counselor = counselor_and_activities[0]
        activities = counselor_and_activities[1].strip().split(', ')

    # Iterate over the list of activities chosen by the counselor and
    # see if that activity is free to choose from and if the activity
    # is free to choose, remove it from the set of available activities
    # and if it is not free remove it from the counselor's activity list

    for activity in activities:
        if activity in set_of_activities:
            set_of_activities.remove(activity)
        else:
            activities.remove(activity)

    # Insert the counselor and their chosen activities into the dictionary

    counselors_activities[counselor] = activities

# print(counselors_activities)

在这个新示例中，我做出了一个假设，那就是您已经有了一系列可以从已经可用的活动中选择的活动：

我使文本文件与问题中列出的咨询员及其活动的格式相同，但是该逻辑可以应用于其他存储方式。

作为旁注和之前第二个示例的更正，我使用了一个集合来表示活动列表，而不是本示例中的列表。该集合仅用于验证没有辅导员负责已经分配给其他人的活动；也就是说，在最坏的情况下，从集合中删除活动要比从列表中删除活动快。

辅导员可以从记事本文件插入字典中，而不必将其插入列表中。

当打印字典时，将产生结果：

{"Adam": ["archery", "canoeing"],
 "Bob": ["frisbee", "golf", "painting", "trampoline"],
 "Carol": ["tennis", "dance", "skating"],
 "Denise": ["cycling"],
 "Eddie": ["horseback", "fencing", "soccer"],
 "Fiona": [], # Empty activity list as the painting activity was already chosen by Bob
 "George": ["basketball", "football"]}

什么是锯齿状数组的Python等价物？

1 个答案: