列表生成式练习一

基于列表生成式

linux = [‘钢弹‘,‘小壁虎‘,‘小虎比‘,‘alex‘,‘wupeiqi‘,‘yuanhao‘]

python = [‘dragon‘,‘钢弹‘,‘zhejiangF4‘,‘小虎比‘]

一：得出既报名linux又报名python的学生列表

p_some_l  = []
p_some_l = [p_python for x_linux in linux for p_python in python if x_linux == p_python]
print(p_some_l) #[‘钢弹‘, ‘小虎比‘]

二：得出只报名linux，而没有报名python的学生列表

1 l_not_py = [x_linux for x_linux in linux if x_linux not in python]
2 print(l_not_py)#[‘小壁虎‘, ‘alex‘, ‘wupeiqi‘, ‘yuanhao‘]

三：得出只报名python，而没有报名linux的学生

py_no_l = [p_python for p_python in python if p_python not in linux]
print(py_no_l) # [‘dragon‘, ‘zhejiangF4‘]

列表 l=[10,2,3,4,5,6,7]

1、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

l = [10,2,3,4,5,6,7]
l1= [x * x for x in l ]  #把迭代对象迭代并返回自乘方的一个新列表
tmp = [x for x in l1 if x > 40] # [100, 49]  #筛选出大于40的值
print(tmp)
print (l1) #[100, 4, 9, 16, 25, 36, 49]

2、过滤出l1中大于40的值，然后求和

print(sum(tmp)) #149   #对刚才使用者产生的新表【100，49】求和

一：得出股票价格大于30的股票名字列表

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

dicg = ((i,v) for i ,v in shares.items()) #定制生成器
P = [i[0] for i in dicg if float(i[-1]) > 30] # [‘huawei‘, ‘lenovo‘, ‘oldboy‘, ‘IBM‘, ‘ocean‘]
print(P) # [‘IBM‘, ‘huawei‘]

二：求出所有股票的总价格

dicg = ((i,v) for i ,v in shares.items())
P_price = (i[-1] for i in dicg)
print(sum(P_price)) #127.5