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已经列表points中保存的是每个点的坐标(坐标是用元组表示的,第一个值是x坐标,第二个值是y坐标)
points = [ (10, 20), (0, 100), (20, 30), (-10, 20), (30, -100) ]
以下问题使用实参高阶函数来解决
1)获取列表中y坐标最大的点
y_max = max(points,key=lambda x:x[-1]) print(y_max)
2)获取列表中x坐标最小的点
x_min = min(points,key=lambda x:x[0]) print(x_min)
3)获取列表中距离原点最远的点
far = max(points,key=lambda x:x[0]**2 + x[1]**2) print(far)
4)将点按照点到x轴的距离大小从大到小排序
x_far = sorted(points,key=lambda x:x[1]**2,reverse=True) print(x_far)
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已经三个列表分别表示5个学生的姓名、学科和班号,使用map将这个三个列表拼成一个表示每个学生班级信息的的字典
names = ['小明', '小花', '小红', '老王'] nums = ['1906', '1807', '2001', '2004'] subjects = ['python', 'h5', 'java', 'python'] 结果:{'小明': 'python1906', '小花': 'h51807', '小红': 'java2001', '老王': 'python2004'}students_message = map(lambda x,y,z:(x,z+y),names,nums,subjects ) print(list(students_message)) -
已经一个列表message, 使用reduce计算列表中所有数字的和
message = ['你好', 20, '30', 5, 6.89, 'hello'] 结果:31.89
from functools import reduce message = ['你好', 20, '30', 5, 6.89, 'hello'] new_list = [] for x in message: if type(x) == float or type(x) == int: new_list.append(x) print(reduce(lambda z,y:z+y,new_list,0)) -
已知一个字典列表中保存的是每个学生各科的成绩,
1)计算并添加每个学生的平均分
def studentss(z): for x in studens: x['avg'] = (x['math']+x['English']+x['Chinese']) / 3 return studens print(studentss(studens))2)按照平均分从高到低排序
studens = [ {'name': 'stu1', 'math': 97, 'English': 67, 'Chinese': 80}, {'name': 'stu2', 'math': 56, 'English': 84, 'Chinese': 74}, {'name': 'stu3', 'math': 92, 'English': 83, 'Chinese': 78}, {'name': 'stu4', 'math': 62, 'English': 90, 'Chinese': 88} ] # 计算平均分 studens = [ {'name': 'stu1', 'math': 97, 'English': 67, 'Chinese': 80, 'avg':81}, {'name': 'stu2', 'math': 56, 'English': 84, 'Chinese': 74, 'avg':71}, {'name': 'stu3', 'math': 92, 'English': 83, 'Chinese': 78, 'avg':87}, {'name': 'stu4', 'math': 62, 'English': 90, 'Chinese': 88, 'avg':80} ] # 按照平均分从高到低排序 ...def studentss(z): for x in studens: x['avg'] = (x['math']+x['English']+x['Chinese']) / 3 return studens print(studentss(studens)) new_avg = sorted(studentss(studens),key=lambda a:a['avg'],reverse=True) print(new_avg) -
封装函数实现斗地主游戏发牌功能(函数返回四个数据,分别是:三个玩家手中的牌、底牌)
paiku = iter((['2','3','4','5','6','7','8','9','10','J','Q','K','A'])*4 +['大0','小王'])#转变为迭代器前打款顺序
player1 = []
player2 = []
player3 = []
dipai = []
def pai(y):
for x in paiku:
player1.append(x)
player2.append(x)
player3.append(x)
if len(player1) == len(player2) == len(player1) == 17:
dipai.append(x)
return player1,player2,player3,dipai
print(pai(paiku))



