Question

我被要求解决以下问题，但无法提供正确的代码：

此练习涉及构建非平凡的字典。主题是书籍。

每本书的关键是书名与该键关联的值是字典

在该词典中将有三个键：它们都是字符串，它们是： "Pages", "Author", "Publisher"

"Pages"与一个值关联-int "Author"与字典关联为值该"Author"字典有两个键："First"和"Last"每个都有一个字符串值 "Publisher"与字典关联为值该"Publisher"字典有一个键"Location"，其中有一个字符串作为值。

示例可能如下：

{"Harry Potter": {"Pages":200, "Author":{"First":"J.K", "Last":"Rowling"}, "Publisher":{"Location":"NYC"}},
"Fear and Lothing in Las Vegas": { ...}}

编写一个名为"build_book_dict"的函数接受五个输入，所有列表均为n长度标题，页面，第一，最后和位置的列表。

如上所述返回字典。键的拼写必须与上方显示的一样-正确且大写。

这里是一个例子：

titles = ["Harry Potter", "Fear and Lothing in Las Vegas"]
pages = [200, 350]
firsts = ["J.K.", "Hunter"]
lasts = ["Rowling", "Thompson"]
locations = ["NYC", "Aspen"]

book_dict = build_book_dict(titles, pages, firsts, lasts, locations)
print(book_dict)

{'Fear and Lothing in Las Vegas': {'Publisher': {'Location': 'Aspen'}, 'Author': {'Last': 'Thompson', 'First': 'Hunter'}, 'Pages': 350} 'Harry Potter': {'Publisher': {'Location': 'NYC'},'Author': {'Last': 'Rowling', 'First': 'J.K.'}, 'Pages': 200}}

我的代码当前看起来如下：

def build_book_dict(titles, pages, firsts, lasts, locations):
    inputs = zip(titles, pages, firsts, lasts, locations)
    for titles, pages, firsts, lasts, locations in inputs:
        dict = {
            titles : {
                "Pages" : pages,
                "Author" : {
                    "First" : first,
                    "Last" : last
                            },
                "Publisher" : {
                    "Location" : locations
                                },
                        },
                }

    return dict

但是它仅存储最后一本“书”的信息。

Answer 1

使用此功能，主要更改是我添加了d.update：

def build_book_dict(titles, pages, firsts, lasts, locations):
    inputs = zip(titles, pages, firsts, lasts, locations)
    d = {}
    for titles, pages, firsts, lasts, locations in inputs:
        d.update({
            titles : {
                "Pages" : pages,
                "Author" : {
                    "First" : firsts,
                    "Last" : lasts
                            },
                "Publisher" : {
                    "Location" : locations
                                },
                        },
                })

    return d

现在：

print(build_book_dict(titles, pages, firsts, lasts, locations))

成为：

{'Harry Potter': {'Pages': 200, 'Author': {'First': 'J.K.', 'Last': 'Rowling'}, 'Publisher': {'Location': 'NYC'}}, 'Fear and Lothing in Las Vegas': {'Pages': 350, 'Author': {'First': 'Hunter', 'Last': 'Thompson'}, 'Publisher': {'Location': 'Aspen'}}}

您的代码不起作用，因为您每次都在创建一个新词典，而不是将字典加在一起，但是d.update克服了这个问题。

此外，我将变量dict重命名为d，因为dict是默认关键字，而当您命名变量dict时，您将无法来访问实际的dict关键字。

Answer 2

在for循环之前创建一个空字典。然后在循环中像这样添加它：

def build_book_dict(titles, pages, firsts, lasts, locations):
    inputs = zip(titles, pages, firsts, lasts, locations)
    d = dict()
    for titles, pages, firsts, lasts, locations in inputs:
        d[titles] = {whatever}
    return d

Answer 3

您可以使用字典理解：

titles = ["Harry Potter", "Fear and Lothing in Las Vegas"]
pages = [200, 350]
firsts = ["J.K.", "Hunter"]
lasts = ["Rowling", "Thompson"]
locations = ["NYC", "Aspen"]
data = {a:{'Pages':b, 'Author':{'First':c, 'Last':d}, 'Publisher': {'Location': e}} for a, b, c, d, e in zip(titles, pages, firsts, lasts, locations)}

输出：

{'Harry Potter': {'Pages': 200, 'Author': {'First': 'J.K.', 'Last': 'Rowling'}, 'Publisher': {'Location': 'NYC'}}, 'Fear and Lothing in Las Vegas': {'Pages': 350, 'Author': {'First': 'Hunter', 'Last': 'Thompson'}, 'Publisher': {'Location': 'Aspen'}}}

从列表创建嵌套字典的功能

3 个答案: