Python中如何将类中字符串属性首字母大写输出？

你可以利用Python字符串的title()方法，它会把每个单词的首字母转为大写、其余字母转为小写，刚好匹配你的需求。下面提供两种实现方式：

方式一：初始化时统一处理名称（推荐） #

在类的__init__方法中直接格式化餐厅名称，后续所有用到名称的方法都无需重复处理：

class Restaurant:
    def __init__ (self, restaurant_name, cuisine_type):
        # 初始化时将名称转为首字母大写格式
        self.name = restaurant_name.title()
        self.type = cuisine_type
    def describe_restaurant(self):
        print(f"This restaurants name is: '{self.name}' and serves {self.type}")
    def restaurant_open(self):
        print(f"'{self.name}' is currently open")

res1 = Restaurant('pho hung','vietnamese pho')
res1.describe_restaurant()

res2 = Restaurant('macdonalds','american fast')
res2.describe_restaurant()
res2.restaurant_open()

方式二：仅在输出时临时处理 #

如果不想修改原始存储的名称，可在输出方法中单独处理：

class Restaurant:
    def __init__ (self, restaurant_name, cuisine_type):
        self.name = restaurant_name
        self.type = cuisine_type
    def describe_restaurant(self):
        # 输出时调用title()格式化名称
        print(f"This restaurants name is: '{self.name.title()}' and serves {self.type}")
    def restaurant_open(self):
        # 输出时调用title()格式化名称
        print(f"'{self.name.title()}' is currently open")

res1 = Restaurant('pho hung','vietnamese pho')
res1.describe_restaurant()

res2 = Restaurant('macdonalds','american fast')
res2.describe_restaurant()
res2.restaurant_open()

两种方式运行后都会输出符合要求的结果：

This restaurants name is: 'Pho Hung' and serves vietnamese pho
This restaurants name is: 'Macdonalds' and serves american fast
'Macdonalds' is currently open

内容的提问来源于stack exchange，提问作者Minh Tee