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day2023-3-21-函数进阶作业

吴靖
2023-12-01
  1. 写一个匿名函数,判断指定的年是否是闰年 (先直接用普通函数)
  # 匿名函数
  is_leap_year = lambda year: (year % 4 == 0 and year % 100 != 0) or year % 400 == 0
  
  
  print(is_leap_year(2020))
  
  # 普通函数
  def is_leap_year(year):
      if (year % 4 == 0 and year % 100 != 0) or year % 400 == 0:
          result = True
      else:
          result = False
      return result
  
  
  print(is_leap_year(2020))
  1. 写一个函数将一个指定的列表中的元素逆序( 如[1, 2, 3] -> [3, 2, 1])(注意:不要使用列表自带的逆序函数)
# 方法1:添加新的列表
def reversed1(list1:list):
   new_list = []
   for index in range(len(list1)-1,-1,-1):
       new_list.append(list1[index])
   return new_list

# 方法2:切片
def list_reverse(list1:list):
   return list1[::-1]


# 方法3:交换前后元素的位置
def list_reverse2(list1:list):
   len1 = len(list1)
   for i in range(len1 // 2):
       j = len1 - 1 -i
       list1[i], list1[j] = list1[j], list1[i]
   return list1
  1. 编写一个函数,计算一个整数的各位数的平方和
例如: sum1(12) -> 5(1的平方加上2的平方)    sum1(123) -> 14
# 方法1
def quadratic_sum(num1):
    sum1 = 0
    for x in str(num1):
        sum1 += int(x) ** 2
    return sum1


# 方法2
def sum_square(num:int):
    return reduce(lambda x, item: x + int(item)**2, str(num), 0)


# 方法3
def sum_square2(num:int):
    return sum([int(x)**2 for x in str(num)])

4.求列表 nums 中绝对值最小的元素

例如:nums = [-23, 100, 89, -56, -234, 123], 最大值是:-23
nums = [-23, 100, 89, -56, -234, 123]
result = min(nums, key=lambda item: item ** 2)
print(result)

5.已经两个列表A和B,用map函数创建一个字典,A中的元素是key,B中的元素是value

A = ['name', 'age', 'sex']
B = ['张三', 18, '女']
新字典: {'name': '张三', 'age': 18, 'sex': '女'}
A = ['name', 'age', 'sex']
B = ['张三', 18, '女']
result = map(lambda i1, i2: (i1, i2), A, B)
print(dict(result))
  1. 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'}
    
    names = ['小明', '小花', '小红', '老王']
    nums = ['1906', '1807', '2001', '2004']
    subjects = ['python', 'h5', 'java', 'python']
    # {'小明': 'python1906', '小花': 'h51807', '小红': 'java2001', '老王': 'python2004'}
    result = map(lambda i1, i2, i3: (i1,i2+i3), names, nums, subjects)
    print(dict(result))
    
  2. 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
    
    # 方法1
    message = ['你好', 20, '30', 5, 6.89, 'hello']
    result = reduce(lambda x, item: x + item if type(item) == int or type(item) == float else x + 0, message, 0)
    print(result)
    
    # 方法2
    message = ['你好', 20, '30', 5, 6.89, 'hello']
    num = [x for x in message if type(x) == int or type(x) == float]
    result = reduce(lambda y, x: y + x, num, 0)
    print(result)
    
    # 方法3
    def is_number1(list1):
        sum1 = 0
        for item in list1:
            if type(item) == int or type(item) == float:
                sum1 += item
        return sum1
    
    
    message = ['你好', 20, '30', 5, 6.89, 'hello']
    print(is_number1(message))
    
    # 方法4
    result = reduce(lambda a, item: a + item, [x for x in message if type(x) in (int, float)], 0)
    print(result)
    
  3. 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)
    ]
    

    1)获取列表中y坐标最大的点

    result = max(points, key=lambda item:item[1])
    print(result)
    

    2)获取列表中x坐标最小的点

    result = min(points, key=lambda item:item[0])
    print(result)
    

    3)获取列表中距离原点最远的点

    result = max(points, key=lambda item: item[0]**2 + item[1]**2)
    print(result)
    

    4)将点按照点到x轴的距离大小从大到小排序

    result = sorted(points, key=lambda item:item[1]**2, reverse=True)
    print(result)
    
  4. 封装一个函数完成斗地主发牌的功能。

from random import shuffle

# 方法1
def landlord_deal_cards(cards):
    shuffle(cards)
    i1 = iter(cards)
    player1 = []
    player2 = []
    player3 = []
    bottom_cards = []
    for x in range(17):
        player1.append(next(i1))
        player2.append(next(i1))
        player3.append(next(i1))
    for y in i1:
        bottom_cards.append(y)
    return player1,player2,player3,bottom_cards


cards = ['大王', '小王', '黑桃A', '黑桃2', '黑桃3', 
         '黑桃4', '黑桃5', '黑桃6', '黑桃7', '黑桃8', 
         '黑桃9', '黑桃10', '黑桃J', '黑桃Q', '黑桃K', 
         '红桃A', '红桃2', '红桃3', '红桃4', '红桃5', 
         '红桃6', '红桃7', '红桃8', '红桃9', '红桃10', 
         '红桃J', '红桃Q', '红桃K', '方片A', '方片2', 
         '方片3', '方片4', '方片5', '方片6', '方片7', 
         '方片8', '方片9', '方片10', '方片J', '方片Q', 
         '方片K', '梅花A', '梅花2', '梅花3', '梅花4', 
         '梅花5', '梅花6', '梅花7', '梅花8', '梅花9', 
         '梅花10', '梅花J', '梅花Q', '梅花K']

print(landlord_deal_cards(cards))

# 方法2


def deal_cards():
    # 1)准备一副新的牌
    colors = ['♥', '♠', '♣', '♦']
    nums = [str(x) for x in range(2, 11)] + ['J', 'Q', 'K', 'A']
    pokers = ['joker', 'JOKER']
    for n in nums:
        for c in colors:
            pokers.append(c + n)

    # 2)洗牌
    shuffle(pokers)

    # 3)发牌
    pokers = iter(pokers)
    player1 = []
    player2 = []
    player3 = []
    for _ in range(17):
        player1.append(next(pokers))
        player2.append(next(pokers))
        player3.append(next(pokers))

    # 4)理牌
    values = {'J': 11, 'Q':12, 'K':13, 'A':14, '2':15, 'oker':16, 'OKER':17}
    values.update({str(x): x for x in range(3, 11)})

    player1.sort(key=lambda item: values[item[1:]],reverse=True)
    player2.sort(key=lambda item: values[item[1:]], reverse=True)
    player3.sort(key=lambda item: values[item[1:]], reverse=True)
    return player1, player2, player3,list(pokers)


p1, p2, p3, di = deal_cards()
print(p1)
print(p2)
print(p3)
print(di)
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