python中的关键字---4（匿名/递归函数）

Posted 2021-01-22 God is a girl

匿名函数

# 匿名函数 别称 : lambda表达式
# 函数,没有名字
# def wahaha(n):
#     return n**2
# print(wahaha.__name__)

# qqxing = lambda n : n**2
# print(qqxing.__name__)
# ret = qqxing(8)
# print(ret)

# f = lambda *args:sum(args)
# ret = f(1,2,3,4,5)
# print(ret)

# 写匿名函数 : 接收两个参数,返回较大值(不用max内置函数)
# f1 = lambda a,b : max(a,b)
# f2 = lambda a,b : a if a>b else b

# 匿名函数
# 定义
# 匿名函数的内存地址 = lambda 参数1,参数2 : 返回值/返回值的表达式
# 调用
# 接收返回值 = 匿名函数的内存地址(参数)

# dic={‘k1‘:10,‘k2‘:100,‘k3‘:30}
# def func(k):
#     return dic[k]
#
# m = max(dic,key = func)
# print(m)

# m = max(dic,key = lambda k:dic[k])
# print(m)

# 内置函数+匿名函数
l1 = [1,5,7,4,8]
# 每一个数的平方%2
# ret = map(lambda n : n**2%2 ,l1)
# print(list(ret))

l2 = [3,10,20,50,22,7]
# 找到大于20的所有数
# ret = filter(lambda n : n>20,l2)
# print(list(ret))

l3 = [{‘name‘:‘www‘,‘age‘:83},{‘name‘:‘asd‘,‘age‘:38}]
# 1.找到年龄大于40的值  filter
# 2.将列表中的项按照年龄从小到大排序 sorted
# 3.给列表中的每一个人的年龄+1
# 4.给列表中名叫www的人的年龄+1  三元运算符+字典拼接

# 1
# def func(dic):
#     return dic[‘age‘]>40
# ret = filter(lambda dic:dic[‘age‘]>40,l3)
# print(list(ret))

# 2
# def func(dic):
#     return dic[‘age‘]
# ret = sorted(l3,key=lambda dic:dic[‘age‘])
# print(ret)

# 3 lambda 并不能取代函数
# def func(dic):
#     dic[‘age‘] += 1
#     return dic
# ret = map(lambda dic : {‘name‘:dic[‘name‘],‘age‘:dic[‘age‘]+1},l3)
# print(list(ret))

# 4
# 现有两个元组((‘a‘),(‘b‘)),((‘c‘),(‘d‘))，
# 请使用python中匿名函数生成列表[{‘a‘:‘c‘},{‘b‘:‘d‘}]
# t1 = ((‘a‘),(‘b‘))
# t2 = ((‘c‘),(‘d‘))
# print(t1,t2)
# ret = zip(t1,t2)
# def func(tup):
#     return {tup[0]:tup[1]}
# map(func,ret)
# iterator1 = map(lambda tup:{tup[0]:tup[1]} ,ret)
# print(list(iterator1))
匿名函数lambda常配合内置函数组成表达式，进行简单的判断计算，节省代码，逻辑性强

---

递归函数

# 1.递归函数的定义 :一个函数在执行的过程中调用了自己
# 2.递归在python中的最大深度 : 1000/998
# 3.递归的最大深度是可以自己修改的,但是不建议你修改

# 停不下来的递归函数早晚会报错
# 了解递归/正确的递归函数



# n = 1,age = q(n+1) -2
# n = 2,age = q(n+1) -2
# n = 3,age = q(n+1) -2
# n = 4,age = q(n+1) -2
# n = 5,age = 18

# def q(n):
#     if n == 1:
#         age = q(n + 1) - 2
#         return age
#     elif n == 2:
#         age = q(n + 1) - 2
#         return age
#     elif n == 3:
#         age = q(n + 1) - 2
#         return age
#     elif n == 4:
#         age = q(n + 1) - 2
#         return age
#     elif n == 5:
#         return 18
# ret = q(1)
# print(ret)

# return 能够帮助你停止递归,并且把最内层的结果一层一层的返回到最外层
# 在最内层函数的返回值,返回的时候永远是离最内层最近的上一层才能接收到返回值

# def q(1):
#     if 1 == 1:
#         age = 12 - 2
#         return 10
#
# def q(2):
#     elif 2 == 2:
#         age = 14 - 2
#         return 12
#
# def q(3):
#     elif n == 3:
#         age = 16 - 2
#         return 14
#
# def q(4):
#     elif n == 4:
#         age = 16
#         return age
#
# def q(5):
#     elif n == 5:
#         return 18

# def q(n):
#     if n < 5:
#         age = q(n + 1) - 2
#         return age
#     elif n == 5:
#         return 18
# ret = q(1)
# print(ret)

---

二分查找法

l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]

# aim = 66
# # l中找到中间的值 和 66 对比
# def search(l):
#     mid = len(l)//2
#     mid_value = l[mid]
#     if aim > mid_value:
#         start = mid+1
#         new_l = l[start:]
#         search(new_l)
#     elif aim < mid_value:
#         end = mid
#         new_l = l[:end]
#         search(new_l)
#     else:
#         print(‘找到了‘,mid,l[mid])
#
# search(l)

# 递归过程
# def search(l):       # l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]
#     mid = len(l)//2  # 25//2 = 12
#     mid_value = l[mid] # l[12] = 41
#     if aim > mid_value: # 66 > 41
#         start = mid+1   # 12+1 = 13
#         new_l = l[start:]  # new_l = [42,43,55,56,66,67,69,72,76,82,83,88]
#         search(new_l)      # search(new_l)
#     elif aim < mid_value:
#         end = mid
#         new_l = l[:end]
#         search(new_l)
#     else:
#         print(‘找到了‘)