Question

dictionary = {'Jason Seifer': ['Ruby Foundations', 'Ruby on Rails Forms', 'Technology Foundations'], 'Kenneth Love': ['Python Basics', 'Python Collections'], 'Daniel': ['Python Basics', 'Python Collections', 'test', 'test']}

def most_class(dictionary):
    current_count = 0
    max_count = 0
    best_teacher = "none"
    for key in dictionary:
        for values in dictionary.values():
            current_count += 1
            if current_count > max_count:
                max_count = current_count
                best_teacher = key
    return best_teacher

目标是确定字典中哪个键具有最多值。答案应该是Daniel，但每次运行代码时我都会得到不同的答案（大概是由于字典的蟒蛇散列造成的。

任何人都可以解释为什么我会得到这个结果，以及如何修复它？

Answer 1

你在词典上迭代的次数超过你需要的次数，并且使得这比你需要的要困难得多。您可以通过将Python内置max()与comprehensions结合使用来实现更多目的。试试这个：

dictionary = {
    'Jason Seifer': ['Ruby Foundations', 'Ruby on Rails Forms',
                     'Technology Foundations'], 
    'Kenneth Love': ['Python Basics', 'Python Collections'], 
    'Daniel': ['Python Basics', 'Python Collections', 'test', 'test']}

# get maximum number of items any one person has
maximum_len = len(max(dictionary.values(), key=len))

# build a list of everyone with that number of items
most = [name for name, classes in dictionary.items() 
            if len(classes) == maximum_len]

print(most)

此代码打印：

[ '丹尼尔']

这不仅可以用理解来代替循环，如果出现平局，你会获得一个名单列表，而不仅仅是找到的第一个名称。例如，如果我们将字典定义更改为：

dictionary = {
    'Jason Seifer': ['Ruby Foundations', 'Ruby on Rails Forms', 
                     'Technology Foundations', 'other_item'], 
    'Kenneth Love': ['Python Basics', 'Python Collections'], 
    'Daniel': ['Python Basics', 'Python Collections', 'test', 'test']}

然后将打印出来：

['Daniel'，'Jason Seifer']

Answer 2

你有两个问题。

您需要每次都重置current_count，而不仅仅是在开始时。
dictionary.values()为您提供了图书清单列表，而不是该作者的图书清单。

要解决第二个问题，我会使用for author, books in dictionary.items():而不是for key in dictionary:。然后，您不必每次都使用dictionary[key]。完成这些更改后，我们有：

def most_class(dictionary):
    max_count = 0
    best_teacher = "none"
    for author, books in dictionary.items():
        current_count = 0
        for book in books:
            current_count += 1
            if current_count > max_count:
                max_count = current_count
                best_teacher = author
    return best_teacher

此外，我可能会将if拉出循环：

def most_class(dictionary):
    max_count = 0
    best_teacher = "none"
    for author, books in dictionary.items():
        current_count = 0
        for book in books:
            current_count += 1
        if current_count > max_count:
            max_count = current_count
            best_teacher = author
    return best_teacher

此时，很明显我们根本不需要for循环，只能使用len：

def most_class(dictionary):
    max_count = 0
    best_teacher = "none"
    for author, books in dictionary.items():
        if len(books) > max_count:
            max_count = len(books)
            best_teacher = author
    return best_teacher

您的代码也可能像其他人建议的那样使用max函数受益，尽管我可能会以不同的方式使用它：

def most_class(dictionary):
    return max(dictionary.items(), default=('none', 0),
               key=lambda item: len(item[1]))[0]

Answer 3

你必须知道如何使用字典，然后我建议你这样做：

dictionary = {'Jason Seifer': ['Ruby Foundations', 'Ruby on Rails Forms', 'Technology Foundations'], 'Kenneth Love': ['Python Basics', 'Python Collections'], 'Daniel': ['Python Basics', 'Python Collections', 'test', 'test']}

def most_class(dictionary):
    best_teacher = None
    class_of_best_teacher = 0
    for teacher_name in dictionary.keys():
        class_of_teacher = len(dictionary[teacher_name])
        if class_of_teacher > class_of_best_teacher :
            best_teacher = teacher_name
            class_of_best_teacher = class_of_teacher
    return best_teacher
print most_class(dictionary)

Answer 4

您遍历字典中的键，但是您必须在每个键之间将 current_count 重置为零。因此， current_count 和 max_count 一起计数最多为9，而 best_teacher 将是for循环遍历的最后一个键。

修复你的代码。

...
for key in dictionary:
    current_count = 0
    ...

它有效，但我想这样做：

best_teacher = sorted(dictionary.keys(), key=lambda x: len(dictionary[x]))[-1]

Answer 5

为每个密钥重置current_count
应使用dictionary[key]代替dictionary.values()

以下代码返回Daniel。

dictionary = {'Jason Seifer': ['Ruby Foundations', 'Ruby on Rails Forms', 'Technology Foundations'], 'Kenneth Love': ['Python Basics', 'Python Collections'], 'Daniel': ['Python Basics', 'Python Collections', 'test', 'test']}

def most_class(dictionary):
    current_count = 0
    max_count = 0
    best_teacher = "none"
    for key in dictionary:
        current_count = 0
        for values in dictionary[key]:
            current_count += 1
            if current_count > max_count:
                max_count = current_count
                best_teacher = key
    return best_teacher

代码意外的Python结果

5 个答案: