为什么我的函数仅返回列表中的一项？

Question

我正在学习python以提高自己的技能。我正在构建一个Web应用程序，以跟踪与我一起工作的自由职业者。我正在学习一个教程，但自己编写。这是代码：freelancers = ["juan", "andre"] #original listdef get_freelancer_titlecase(): # makes freelancers TitleCased    freelancers_in_titlecase = []#empty dict    for freelancer in freelancers:        freelancers_in_titlecase = freelancer.title()        return freelancers_in_titlecasedef print_freelancers_in_titlecase():#print freelancers in TitleCase    print_freelancers_in_titlecase = get_freelancer_titlecase()    print(print_freelancers_in_titlecase)freelancer_list = get_freelancer_titlecase() # Place the function inside a variableprint(freelancer_list)# here is my problem, this returns only the first entry "Juan", in title case but stops there.print(freelancers)# this is a check to see the items on the list and they are ["juan", "andre"]为什么print(freelancer_list)只退货一件？我需要能够调用列表，并让所有自由职业者拥有大写字母。后来，它应该是一本字典，其中也要保留一个ID，当然还要包含每个自由职业者的输入。这是我的第一个问题，所以先谢谢您。

Answer 1

您get_freelancer_titlecase(在获得第一个元素后立即从）返回 。

您需要使用进行存储以列出列表 freelancers_in_titlecase，append并在for循环外结束时返回。

freelancers = ["juan", "andre"] #original list

def get_freelancer_titlecase(): # makes freelancers TitleCased

    freelancers_in_titlecase = []

    for freelancer in freelancers:

        freelancers_in_titlecase.append(freelancer.title())

    return freelancers_in_titlecase

def print_freelancers_in_titlecase():#print freelancers in TitleCase

    print_freelancers_in_titlecase = get_freelancer_titlecase()

    print(print_freelancers_in_titlecase)

freelancer_list = get_freelancer_titlecase() # Place the function inside a variable

print(freelancer_list)  # ['Juan', 'Andre']

print(freelancers)  # ['juan', 'andre'] 

print_freelancers_in_titlecase()  # This function prints titlecased list.

Answer 2

您将freelancers_in_titlecase在第一个循环之后返回，此外，您将覆盖该变量。因此，您应该将和变量带入return外部，for loop而append不是替换它们：

def get_freelancer_titlecase(): # makes freelancers TitleCased

    freelancers_in_titlecase = []#it is actually a list

    for freelancer in freelancers:

        freelancers_in_titlecase.append(freelancer.title())

    return freelancers_in_titlecase

您可以考虑使用List Comprehensions重写此函数：

def get_freelancer_titlecase():

    freelancers_in_titlecase = [freelancer.title() for freelancer in freelancers]

    return freelancers_in_titlecase

Answer 3

问题出在您的get_freelancer_titlecase()职能上。

def get_freelancer_titlecase(): # makes freelancers TitleCased

    freelancers_in_titlecase = []#empty dict

    for freelancer in freelancers:

        freelancers_in_titlecase = freelancer.title()

        return freelancers_in_titlecase

此函数执行以下操作：

创建一个名为的空列表（不是字典） freelancers_in_titlecase

循环浏览freelancers列表中的项目。

将空列表替换freelancers_in_titlecase为第一个自由职业者的职务。

返回该值（循环完成之前）。

试试这个吧。

def get_freelancer_titlecase(): # makes freelancers TitleCased

    freelancers_in_titlecase = [] #empty list

    for freelancer in freelancers:

        freelancers_in_titlecase.append(freelancer.title())

    return freelancers_in_titlecase

这会将每个自由职业者的标题大小写添加到列表中，并且仅在循环完成后才返回列表。

如果您要更改它以返回一个dict以原始名称为键，标题大小写为值的方式，则可以这样操作：

def get_freelancer_titlecase(): # makes freelancers TitleCased

    freelancers_in_titlecase = {} # empty dict

    for freelancer in freelancers:

        freelancers_in_titlecase[freelancer] = freelancer.title()

    return freelancers_in_titlecase