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：
- 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
- 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。
- 首先将可迭代对象变成迭代器对象
- 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。