a,b = 1,2
print(a,b)
a,b = b,a
print(a,b)

tu1 = (1)
print(tu1,type(tu1))    #1 <class 'int'>
tu2 = (1,)
print(tu2,type(tu2)) #(1,) <class 'tuple'>

tu3 = ([1]) 
print(tu3,type(tu3)) #[1] <class 'list'> 
tu4 = ([1],) 
print(tu4,type(tu4)) #([1],) <class 'tuple'>

li = [1,2,33,33,2,1,4,5,6,6]
set1 = set(li)
print(set1) #{1, 2, 33, 4, 5, 6}
li = list(set1)
print(li)   #[1, 2, 33, 4, 5, 6]

for i in range(0,10,-1):
    print(i)  

d =lambda  p:p*2
t =lambda  p:p*3
x = 2
x = d(x)    #4
x = t(x)    #12
x = d(x)    #24
print(x)    #24

print(list(map(lambda tup:{tup[0]:tup[1]},zip((('a'),('b')),(('c'),('d'))))))
# 上面一行代码等于下面三行，可以实现本题功能
ret = zip((('a'),('b')),(('c'),('d')))
res = map(lambda tup:{tup[0]:tup[1]},ret)
print(list(res))    #数据类型强转可以代替循环打印

def multipliers():
    return [lambda x:i*x for i in range(4)]
print([m(2) for m in multipliers()])    #[6, 6, 6, 6]


# 请修改multipliers的定义来产生期望的结果
def multipliers():
    return (lambda x:i*x for i in range(4))
print([m(2) for m in multipliers()])    #[0, 2, 4, 6]