30以内能够整除3的平方
ret = [i**2 for i in range(30) if i%3 ==0]
print(ret)
#找到嵌套列表中名字含有两个或两个以上e的所以名字
names =[['jeera''bxhjbdw''eeeee','ceshi'],['emoe','shezhou','eeewom']]
ret = [name for lst in names for name in lst if name.count('e') >= 2]
print(ret)
#字典推导式
mess = {'a':10,'b':20}
mess_ce ={mess[k]:k for k in mess}
print(mess_ce)
#集合推导式
#有自动去重功能
seq = {x **2 for x in [1,-1,2]}
print(seq)