Question

这是我应该编码的问题：

编写一个函数showCast的合同，文档字符串和实现，该函数获取电影标题，并按照字母顺序从给定的电影中打印出具有相应演员/女演员的角色。列必须对齐（在演员/演员的名字前20个字符（包括角色的名字）。）如果找不到电影，则会打印出错误信息。

它给出了一个应该在这里发生的事情的例子

>>> showCast("Harry Potter and the Sorcerer's Stone")

Character           Actor/Actress

----------------------------------------

Albus Dumbledore    Richard Harris

Harry Potter        Daniel Radcliffe

Hermione Granger    Emma Watson

Ron Weasley         Rupert Grint



>>> showCast('Hairy Potter')

No such movie found

以下是我为同一个项目编写的其他功能，可能有助于回答这个问题。到目前为止，我必须做的总结是，我正在创建一个名为myIMDb的字典，其中包含电影标题的键，以及另一个字典的值。在那个字典中，键是电影的一个字符，值是actor。而且我已经完成了它的工作。 myIMDb是记录的全局变量。

其他功能，他们所做的是docString

def addMovie (title, charList, actList):
    """The function addMovie takes a title of the movie, a list of characters,
    and a list of actors. (The order of characters and actors match one
    another.) The function addMovie adds a pair to myIMDb. The key is the title
    of the movie while the value is a dictionary that matches characters to
    actors"""

    dict2 = {}
    for i in range (0, len(charList)):
        dict2 [charList[i]] = actList[i]
    myIMDb[title] = dict2
    return myIMDb

我添加了三部电影，

addMovie("Shutter Island", ["Teddy Daniels", "Chuck Aule"],["Leonardo DiCaprio, ","Mark Ruffalo"])

addMovie("Zombieland", ["Columbus", "Wichita"],["Jesse Eisenberg, ","Emma Stone"])

addMovie("O Brother, Where Art Thou", ["Everett McGill", "Pete Hogwallop"],["George Clooney, ","John Turturro"])



def listMovies():
    """returns a list of titles of all the movies in the global variable myIMDb"""

    return (list(myIMDb.keys()))


def findActor(title, name):
    """ takes a movie title and a character's name and returns the
    actor/actress that played the given character in the given movie. If the
    given movie or the given character is notfound, it prints out an error
    message"""
    if title in myIMDb:
        if name in myIMDb[title]:
            return myIMDb[title][name]
        else:
            return "Error:  Character not in Movie"
    else:
        return "Error: No movie found"

现在我遇到了麻烦

我应该写showCast函数，但是我遇到了很多麻烦。我一直在修补它，但是当我调用myIMDb.values（）时，一切都会回来。我似乎无法遍历它来对它们进行排序以创建表格。

这是我到目前为止所提出的，但它并没有做我所希望的。我只是希望你们中的一个能引导我朝着正确的方向前进。（注释掉的区域就是我以前做过的事情，所以你可以看到我的思路。[print（alist）和print（alist [0]）只是为了确认它是列表中的一个大条目，而不是完全分开]）

def showCast(title):

    if title in myIMDb:
        actList=[]
        chList=[]
        aList = myIMDb[title]
        print (aList)

          """"for i in range (len(aList)):
              if i%2==0:
                  chList.append(aList[i])
              else:
                  actList.append(aList[i])
          print(chList)
          print(actList)""""

else:
    return "Movie not Found"

Answer 1

首先，我认为你不应该在addMovie函数中返回任何内容。只需将其添加到全局变量：

myIMDb = {}

def addMovie (title, charList, actList):
    global myIMDb

    """The function addMovie takes a title of the movie, a list of characters,
    and a list of actors. (The order of characters and actors match one
    another.) The function addMovie adds a pair to myIMDb. The key is the title
    of the movie while the value is a dictionary that matches characters to
    actors"""

    dict2 = {}
    for i in range (0, len(charList)):
        dict2 [charList[i]] = actList[i]
    myIMDb[title] = dict2

虽然我不建议经常使用全局变量，但我认为在这种情况下它是可以原谅的：D

之后，在showCast函数中，我会使用它：

def showCast(title):
    if title in myIMDb:
        actList=[]
        chList=[]
        movie = myIMDb[title]
        for character, cast in movie.keys(), movie.values(): #grab the character from
        #the keys, and cast from the values. 
              chList.append(character)
              actList.append(cast)

        print (chList, actList)
    else:
        return "Movie not Found"

这是我的输出：

['Columbus', 'Jesse Eisenberg, '] ['Wichita', 'Emma Stone']

它按预期工作，希望这有帮助！

Answer 2

也许这些片段可以帮助您：

打印表格行

def printRow(character, actor):
    print(character + (20 - len(character)) * ' ' + actor))

对字符进行排序

def getSortedTitleList(titleList):
    sortedList = []
    characters = sorted(titleList)
    for character in characters:
        sortedList.append((character, titleList[character]))
    return sortedList

注意：我将dict.keys().sort()更改为sorted(dict)以与Python 3兼容。

字典帮助！提取值并制作表格

2 个答案: