为什么我的函数只返回列表中的一个项？

Question

我在学习python来提高我的技能。我正在建立一个网络应用程序来跟踪与我一起工作的自由职业者。我遵循一个教程，但使我自己。你知道吗

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"]

为什么print(freelancer_list)只返回一个项目？ 我需要能够打电话给名单，让所有的自由职业者在标题的情况下。后来，它应该是一个字典，以容纳一个ID也，当然输入每个自由职业者。你知道吗

这是我的第一个问题，所以提前谢谢你。你知道吗

Answer 1

问题出在get_freelancer_titlecase()函数中。你知道吗

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：

1. 创建名为freelancers_in_titlecase的空列表（不是dict）
2. 循环遍历freelancers列表中的项。你知道吗
3. 将空列表freelancers_in_titlecase替换为第一个自由职业者的标题大小写。你知道吗
4. 返回该值（在循环完成之前）。你知道吗

试试这个。你知道吗

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

这将将每个自由职业者的头衔大小写添加到列表中，并且只在循环完成后返回列表。你知道吗

如果要将其更改为返回一个dict，原始名称作为键，标题大小写作为值，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 2

在此函数中：

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

您应该附加到“自由职业者在标题中”，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

在第一个循环之后返回freelancers_in_titlecase，并且重写变量。因此，应该将return置于变量for loop和append之外，而不是替换它们：

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

您可以考虑使用列表理解重写此函数：

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