-
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)
# 方法1: def leap_year(year1): if (year1 % 4 == 0 and year1 % 100 != 0) or year1 % 400 == 0: return f'{year1}年是闰年' else: return f'{year1}年不是闰年' print(leap_year(2020)) # 方法2: year = lambda item : (item % 4 == 0 and item % 100 != 0) or item % 400 == 0 print(year(2300))
-
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)
def new_review(list1:list): newlist = [] for x in list1[-1:-len(list1)-1:-1]: newlist.append(x) return newlist list1 = [1,2,3,5,6] print(new_review(list1))
-
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
def sum1(num:int):
count = 0
for x in str(num):
count += int(x)**2
return count
print(sum1(1235))
- 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]
review = min(nums,key=lambda x : x ** 2)
print(review)
-
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'] new_dict = map(lambda x,y:[x,y],A,B) print(dict(new_dict))
-
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'] new_class = map(lambda x,z,y:[x,y+z],names,nums,subjects) print(dict(new_class))
-
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'] review = reduce(lambda i1,item :i1 + item if type(item) == int or type(item) == float else i1 + 0,message,0) print(review)
-
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
review = max(points,key=lambda index:index[-1]) print(review)
2)获取列表中x坐标最小的点
review = min(points,key=lambda index:index[0]) print(review)
3)获取列表中距离原点最远的点
review = max(points,key=lambda x:x[0]**2 and x[-1]**2) print(review)
4)将点按照点到x轴的距离大小从大到小排序
review = sorted(points,key=lambda x:x[0]**2,reverse=True) print(review)
-
封装一个函数完成斗地主发牌的功能。
from random import shuffle
card = ['红心3','红心4','红心5','红心6','红心7','红心8','红心9','红心10','红心J','红心Q','红心K','红心A','红心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','黑桃A','黑桃2',
'方块3','方块4','方块5','方块6','方块7','方块8','方块9','方块10','方块J','方块Q','方块K','方块A','方块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','梅花A','梅花2',
'小王','大王']
# 打乱
shuffle(card)
#发牌
new_card = iter(card)
player1 = []
player2 = []
player3 = []
count = 0
while count < 51:
player1.append(next(new_card))
count +=1
player2.append(next(new_card))
count +=1
player3.append(next(new_card))
count +=1