递归，匿名函数

Posted 2021-12-28 1naonao

递归

函数的嵌套调用：函数嵌套函数。函数的递归调用：它是一种特殊的嵌套调用，但是它在调用一个函数过程中，有直接间接调用了自身。

def foo():
    print('from foo')
    foo()

foo()  # 进入死循环

• 直接调用

  import sys
  
  # 修改递归层数
  sys.setrecursionlimit(10000)
  def foo(n):
      print('from foo',n)
      foo(n+1)
  foo(0)

• 间接调用

def bar():
    print('from bar')
    foo()
    
def foo():
    print('from foo')
    bar()
    
bar()

递归必须要有两个明确的阶段：

1. 递推：一层一层递归调用下去，进入下一层递归的问题规模都将会减小
2. 回溯：递归必须要有一个明确的结束条件，在满足该条件开始一层一层回溯。

递归的精髓在于通过不断地重复逼近一个最终的结果。

def age(n):
    if == 1:
        return 26
    res = age(n-1)+2
    return res
print(f"age(5):age(5)")

age(5):34

二分法的应用

from random import randint
nums = [randint(1, 100) for i in range(100)]
nums = sorted(nums)
print(nums)

[1, 2, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 10, 11, 11, 11, 11, 12, 13, 13, 15, 16, 16, 20, 21, 21, 23, 24, 26, 26, 27, 28, 28, 31, 33, 33, 34, 35, 38, 38, 39, 40, 42, 43, 45, 45, 46, 46, 47, 47, 51, 52, 52, 53, 53, 55, 55, 56, 56, 57, 57, 57, 58, 59, 61, 62, 64, 66, 66, 67, 68, 69, 69, 71, 72, 72, 74, 74, 75, 76, 78, 78, 79, 79, 79, 79, 80, 82, 85, 88, 89, 90, 90, 91, 91, 91, 94, 99, 99, 100]

def search(search_num,nums):
    mid_index=len(nums)/2
    print(nums)
    if not nums:
        print("not exists")
        return 
    if search_num>nums[mid_index]:
        nums=nums[mid_index+1:]
        search(search_num,nums)
    elif search_num<nums[mid_index]:
        nums=nums[:mid_index]
        search(search_num,nums)
    else:
        print('find it')
search(7,nums)
        

[1, 2, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 10, 11, 11, 11, 11, 12, 13, 13, 15, 16, 16, 20, 21, 21, 23, 24, 26, 26, 27, 28, 28, 31, 33, 33, 34, 35, 38, 38, 39, 40, 42, 43, 45, 45, 46, 46, 47, 47, 51, 52, 52, 53, 53, 55, 55, 56, 56, 57, 57, 57, 58, 59, 61, 62, 64, 66, 66, 67, 68, 69, 69, 71, 72, 72, 74, 74, 75, 76, 78, 78, 79, 79, 79, 79, 80, 82, 85, 88, 89, 90, 90, 91, 91, 91, 94, 99, 99, 100]
[1, 2, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 10, 11, 11, 11, 11, 12, 13, 13, 15, 16, 16, 20, 21, 21, 23, 24, 26, 26, 27, 28, 28, 31, 33, 33, 34, 35, 38, 38, 39, 40, 42, 43, 45, 45, 46, 46, 47, 47]
[1, 2, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 10, 11, 11, 11, 11, 12, 13, 13, 15, 16, 16, 20, 21]
[1, 2, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7]
[6, 6, 7, 7, 7]
find it

• 普通版本和递归版本比较

import time


def rec_find_num(num, lis):
    """递归版本"""
    lis_len = int(len(lis) / 2)  # 10.0

    binary_num = lis[lis_len]  # 10

    if len(lis) == 1:
        print('没找到')
        return

    if binary_num > num:
        lis = lis[:lis_len]
        rec_find_num(num, lis)
    elif binary_num < num:  # 10 < 18
        lis = lis[lis_len + 1:]
        rec_find_num(num, lis)
    else:
        print('找到了')


lis = [i for i in range(100000000)]  # [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
start = time.time()
rec_find_num(4567899900, lis)
end = time.time()
print(end - start)  # 1.1569085121154785

import time

lis = [i for i in range(100000000)]


def time_count(func):
    def wrapper(*args, **kwargs):
        start = time.time()
        res = func(*args, **kwargs)
        end = time.time()
        print(end - start)
        return res

    return wrapper


@time_count
def find_num(num):
    """普通版本"""
    for i in lis:
        if i == num:
            print('找到了')
            break
    else:
        print('没有被找到')


find_num(4567899900)  # 2.293410062789917

匿名函数

• 有名函数

  我们之前定的函数都是有名函数，它是基于函数名使用。

  def func():
      print('from func')
  
  
  func()
  func()
  func()
  print(func)

  from func
  from func
  from func
  <function func at 0x10518b268>

• 匿名函数

  匿名函数，他没有绑定名字，使用一次即被收回，加括号既可以运行。

  lambda x, y: x+y

  <function __main__.<lambda>(x, y)>

  res = (lambda x, y: x+y)(1, 2)
  print(res)

  3

  与内置函数联用

  1.如果我们想从上述字典中取出薪资最高的人，我们可以使用max()方法，但是max()默认比较的是字典的key。

  1. 首先将可迭代对象变成迭代器对象
  2. res=next(迭代器对象)，将res当做参数传给key指定的函数，然后将该函数的返回值当做判断依据

salary_dict = 
    'nick': 3000,
    'jason': 100000,
    'tank': 5000,
    'sean': 2000


print(f"max(salary_dict): max(salary_dict)")


def func(k):
    return salary_dict[k]


print(f"max(salary_dict, key=func()): max(salary_dict, key=func)")
print(
    f"max(salary_dict, key=lambda name: salary_dict[name]): max(salary_dict, key=lambda name: salary_dict[name])")

max(salary_dict): tank
max(salary_dict, key=func()): jason
max(salary_dict, key=lambda name: salary_dict[name]): jason

sorted()、filter()、sorted()方法联用。