标签:span 简单 求和 标记 hello 没有 开头 括号 head
name = ‘李四‘ age = 18 a = "姓名:%s,年龄:%s"%(name,age) print(a) #姓名:李四,年龄:18 ? b = "%(name)s,%(age)s"%{‘name‘:‘张三‘,‘age‘:18} print(b) #张三,18
这种格式化并不是很好,因为它很冗长并且容易导致错误,比如没有正确显示元组或字典
name = ‘李四‘ age = 18 # 替换字段用大括号进行标记 a1 = "hello, {}. you are {}?".format(name,age) print(a1) #hello, 李四. you are 18? ? # 通过索引来以其他顺序引用变量 a2 = "hello, {1}. you are {0}?".format(age,name) print(a2) #hello, 李四. you are 18? ? # 通过参数来以其他顺序引用变量 a3 = "hello, {name}. you are {age1}?".format(age1=age,name=name) print(a3) #hello, 李四. you are 18? ? # 从字典中读取数据时还可以使用 ** data = {"name":"张三","age":18} a4 = "hello, {name}. you are {age}?".format(**data) print(a4) #hello, 李四. you are 18?
在处理多个参数和更长的字符串时仍然可能非常冗长
f-strings 是指以 f 或 F 开头的字符串,其中以 {} 包含的表达式会进行值替换。
name = ‘李四‘ age = 18 # F 和 f 的简单使用 b1 = f"hello, {name}. you are {age}?" b2 = F"hello, {name}. you are {age}?" print(b1) # hello, 李四. you are 18? print(b2) # hello, 李四. you are 18? # 字典也可以 teacher = {‘name‘: ‘meet‘, ‘age‘: 18} msg = f"The teacher is {teacher[‘name‘]}, aged {teacher[‘age‘]}" print(msg) # The comedian is meet, aged 18 # 列表也行 l1 = [‘meet‘, 18] msg = f‘姓名:{l1[0]},年龄:{l1[1]}.‘ print(msg) # 姓名:meet,年龄:18. #可以插入表达式 def sum_a_b(a,b): return a + b a = 1 b = 2 print(‘求和的结果为‘ + f‘{sum_a_b(a,b)}‘) #多行f 反斜杠 name = ‘barry‘ age = 18 ajd = ‘handsome‘ speaker = f‘Hi {name}.‘ f‘You are {age} years old.‘ f‘You are a {ajd} guy!‘ print(speaker) #Hi barry.You are 18 years old.You are a handsome guy! print(f"{You are very \"handsome\"}") #报错 #括号的处理 -->重点:两对为一组 print(f"{{73}}") # {73} print(f"{{{73}}}") # {73} print(f"{{{{73}}}}") # {{73}} m = 21 # ! , : { } ;这些标点不能出现在{} 这里面。 # print(f‘{;12}‘) # 报错 # 所以使用lambda 表达式会出现一些问题。 # 解决方式:可将lambda嵌套在圆括号里面解决此问题。 x = 5 print(f‘{(lambda x: x*2) (x)}‘) # 10
标签:span 简单 求和 标记 hello 没有 开头 括号 head
原文地址:https://www.cnblogs.com/lilinyuan5474/p/11700812.html