今日作业

有两个列表，分别存放来老男孩报名学习linux和python课程的学生名字
linux=[‘钢弹‘,‘小壁虎‘,‘小虎比‘,‘alex‘,‘wupeiqi‘,‘yuanhao‘]
python=[‘dragon‘,‘钢弹‘,‘zhejiangF4‘,‘小虎比‘]

问题一：得出既报名linux又报名python的学生列表
问题二：得出只报名linux，而没有报名python的学生列表
问题三：得出只报名python，而没有报名linux的学生列表

-------------------------------------------------------------------------------------
#作业1，2，3
i=[i for i in linux if i in python]
print(i)

i=[ i for i in linux if not i in python]
print(i)

i=[i for i in python if not i in linux]
print(i)

-----------------------------------------------------------------------------------------------
shares={
‘IBM‘:36.6,
‘lenovo‘:27.3,
‘huawei‘:40.3,
‘oldboy‘:3.2,
‘ocean‘:20.1
}

问题一：得出股票价格大于30的股票名字列表
for i in shares:
if shares[i] > 30:
print (i)
------------------------------------------------------------
i=[i for i in shares if shares[i] > 30]
print(i)
----------------------------------------------------------------------

问题二：求出所有股票的总价格
print((i for i in shares.values()))
----------------------------------------------------------------

作业3

l=[10,2,3,4,5,6,7]
得到一个新列表l1，新列表中每个元素是l中对应每个元素值的平方
过滤出l1中大于40的值，然后求和
---------------------------------------------------------------------------------------
l = [10, 2, 3, 4, 5, 6, 7]
l1=[ i*i for i in l]
print( sum( i for i in l1 if i >40 ))

day24,python