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。我正在学习教程，但要自己制作。

代码如下：

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


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

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)# 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"]

print(freelancer_list)为什么只退回一件物品？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，当然还有每个自由职业者的输入。

这是我的第一个问题，因此，谢谢您。

Answer 1

获取第一个元素后，您将立即从函数get_freelancer_titlecase(返回。

您需要使用freelancers_in_titlecase存储到列表append，并在for循环外返回末尾。

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

在此功能中：

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

您应该附加到“ freelancers_in_titlecase”，因此代码应更改为：

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

Answer 3

问题出在您的get_freelancer_titlecase()函数中。

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

此功能执行以下操作：

创建一个名为freelancers_in_titlecase的空列表（而不是字典）
浏览freelancers列表中的项目。
将空列表freelancers_in_titlecase替换为第一个自由职业者的标题大小写。
返回该值（在循环结束之前）。

试试看。

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头衔加到列表中，并且仅在循环结束后才返回列表。

如果您要更改它以返回dict，且其原始名称为键，标题大小写为值，则可以这样做：

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

Answer 4

您将在第一个循环之后返回freelancers_in_titlecase，而且您将覆盖该变量。因此，您应该将return带到for loop和append变量之外，而不要替换它们：

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重写此功能：

def get_freelancer_titlecase():
    freelancers_in_titlecase = [freelancer.title() for freelancer in freelancers]
    return freelancers_in_titlecase

为什么我的函数仅返回列表中的一项？

4 个答案: