def show(*args): print(args,type(args)) #以元组的形式向列表传递参数 show(11,22,33,44,55,66)
def show(**kwargs): print(kwargs,type(kwargs)) #以字典的形式向函数传递参数 show(k1=80,k2="alex")
[11, 22] <class 'list'> (33, 44, 55, 88) <class 'tuple'>
def show(*args,**kwargs): print(args,type(args)) print(kwargs,type(kwargs)) show(123,"alex",666,alex="sb",nanyang="degnzhou")
(123, 'alex', 666) <class 'tuple'>
{'nanyang': 'degnzhou', 'alex': 'sb'} <class 'dict'>
def show(*args,**kwargs):
print(args,type(args))
print(kwargs,type(kwargs))
l = [11,22,33,44]
d = {"n1":88,"alex":"sb"}
#我们想把列表l传递给形参*args,把字典传递给形参**kwargs,看下面方式是否可以
show(l,d) (1)
([11, 22, 33, 44], {'alex': 'sb', 'n1': 88}) <class 'tuple'>
{} <class 'dict'>
(11, 22, 33, 44) <class 'tuple'>
{'alex': 'sb', 'n1': 88} <class 'dict'>
s1 = "{0} is {1}."
result = s1.format("alex","sb")(1)
l = ["alex","sb"] (2)
res = s1.format(*l)
print(result)
print(res)
s1 = "{name} is {acter}."
result = s1.format(name="alex",acter="sb")
d = {"name":"alex","acter":"sb"}
#向列表中传递字典形式的参数
res = s1.format(**d)
print(result)
print(res)
>>> func = lambda a:a+1 >>> ret = func(99) >>> print(ret) 100
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