name = '李四'
age = 18
a = "姓名:%s,年齡:%s"%(name,age)
print(a)   #姓名:李四,年齡:18
​
b = "%(name)s,%(age)s"%{'name':'張三','age':18}
print(b)   #張三,18

name = '李四'
age = 18
# 替換字段用大括號(hào)進(jìn)行標(biāo)記
a1 = "hello, {}. you are {}?".format(name,age)
print(a1)   #hello, 李四. you are 18?
​
# 通過索引來以其他順序引用變量
a2 = "hello, {1}. you are {0}?".format(age,name)
print(a2)  #hello, 李四. you are 18?
​
# 通過參數(shù)來以其他順序引用變量
a3 = "hello, {name}. you are {age1}?".format(age1=age,name=name)
print(a3)  #hello, 李四. you are 18?
​
# 從字典中讀取數(shù)據(jù)時(shí)還可以使用 **
data = {"name":"張三","age":18}
a4 = "hello, {name}. you are {age}?".format(**data)
print(a4)  #hello, 李四. you are 18?

name = '李四'
age = 18
# F 和 f 的簡(jiǎn)單使用
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]},年齡：{l1[1]}.'
print(msg) # 姓名：meet,年齡：18.

#可以插入表達(dá)式
def sum_a_b(a,b):
  return a + b
a = 1
b = 2
print('求和的結(jié)果為' + 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\"}")   #報(bào)錯(cuò)


#括號(hào)的處理 -->重點(diǎn)：兩對(duì)為一組
print(f"{{73}}") # {73}
print(f"{{{73}}}") # {73}
print(f"{{{{73}}}}") # {{73}}


m = 21
# ! , : { } ;這些標(biāo)點(diǎn)不能出現(xiàn)在{} 這里面。
# print(f'{;12}') # 報(bào)錯(cuò)
# 所以使用lambda 表達(dá)式會(huì)出現(xiàn)一些問題。
# 解決方式：可將lambda嵌套在圓括號(hào)里面解決此問題。
x = 5
print(f'{(lambda x: x*2) (x)}') # 10