def func(n)return n * n
ret = func(9)
print(ret)

a = lambda n: n * n
ret = a(9)
print(ret)

print(func.__name__)  #查看函数名
print(a.__name__)       #__name__的值都是<lambda>

def func(a, b)return a + bx = lambda a, b : a+b
print(x(a , b))

def func(x, y)return x, ys = lambda x, y :(x, y)#如果不放(),s会被分为两部分 lambda x, y :x和y
print(s(250, 38))

fn = lambda *args :max(args) #匿名函数一定是一行函数
print(fn(1,3,8,56))#56

lst = ["聊斋", "西游记", "三国演义", "葫芦娃", "水浒传", "年轮", "亮剑"]
def func(s)return len(s)
li = sorted(lst, key = func)
print(li)
#key: 排序方案, sorted函数内部会把可迭代对象中的每一个元素拿出来交给#后面的key,后面的key计算出一个数字. 作为当前这个元素的权重, 整个函根
#据权重进行排序

lst = [{'name':"汪峰","age":48},{"name":"章子怡",'age':38},{"name":"alex","age":39},{"name":"wusir","age":32},{"name":"赵一宁","age":28}]ll = sorted(lst, key=lambda el: len(el['name']), reverse=True)
print(ll)

#普通函数
lst = ["张无忌", "张铁林", "赵一宁", "石可心","马大帅"]
def func(el)if el[0] == "张"return False  #不想要的else:return True  #想要的
f = filter(func, lst)
print("__iter__" in dir(f)) # 判断是否可以进行迭代
for e in f:print(e)#匿名函数
f = filter(lambda el :el[0] != "张", lst)
print("__iter__" in dir(f)) # 判断是否可以进行迭代
for e in f:print(e)

 lst = [{"name":"汪峰", "score":48},{"name":"章子怡", "score":39},{"name":"赵一宁","score":97},{"name":"石可心","score":90}]f = filter(lambda el: el['score'] < 60 , lst) # 去16期的人print(list(f))

lst = [1,4,7,2,5,8]
li = []
#第一种方法:算法
for el in lst:li.append(el**2)
#第二种方法:普通函数
def func(el)return el**2
m = map(func, lst)
print(list(m))
#第三种方法:匿名函数
m = map(lambda el: el**2, lst)
print(list(m))
# 把后面的可迭代对象中的每一个元素传递给function, 结果就是function返
#回值

map(func1, map(func2, map(func3 , lst)))

import sys
sys.setrecursionlimit()

# 遍历 D:/sylar文件夹, 打印出所有的文件和普通文件的文件名
import os
def func(filepath, n): # d:/sylar/# 1,打开这个文件夹files = os.listdir(filepath)# 2. 拿到每一个文件名for file in files:  # 文件名# 3. 获取到路径f_d = os.path.join(filepath, file) # d:/sylar/文件名/# 4. 判断是否是文件夹if os.path.isdir(f_d):# 5. 如果是文件夹. 继续再来一遍print("\t"*n, file,":") # 打印文件名func(f_d, n + 1)else:   #  不是文件夹. 普通文件print("\t"*n, file)func("d:/sylar",0)

def func(n,s):    #计算n的阶乘if n > 0:s = n * sn = n - 1return func(n,s)else:return s
ret = func(8,1)
print(ret)

def func(n,i,a,b):#a第一个数,b第二个数,i是第几次计算c = a + b # 第三个数if n > i:a = bb = ci+=1return func(n,i,a,b )else:return c
ret = func(400,3,1,1)
print(ret)

def func(n,i,a,b):#a第一个数,b第二个数,i是第几次计算c = a + b # 第三个数if n > c:a = bb = ci+=1return func(n,i,a,b )else:return i
ret = func(4000000,3,1,1)
print(ret)

 lst = [22, 33, 44, 55, 66, 77, 88, 99, 101 , 238 , 345 , 456 , 567 , 678 , 789 ]
n = 79for el in lst:if el == n: # O(1)print("找到了")break
else:print("没有")

lst = [22, 33, 44, 55, 66, 77, 88, 99, 101 , 238 , 345 , 456 , 567 , 678 , 789]
n = 88
left = 0
right = len(lst)-1
while left <= right: # 边界, 当右边比左边还小的时候退出循环mid = (left + right)//2 # 必须是整除. 因为索引没有小数if lst[mid] > n:right = mid - 1if lst[mid] < n:left = mid + 1if lst[mid] == n:print("找到了这个数")break
else:print("没有这个数")

lst = [22, 33, 44, 55, 66, 77, 88, 99, 101 , 238 , 345 , 456 , 567 , 678 , 789]
def func(n, left, right):if left <= right: # 边界print("哈哈")mid = (left + right)//2if n > lst[mid]:left = mid + 1return func(n, left, right) # 递归  递归的入口elif n < lst[mid]:right = mid - 1# 深坑. 函数的返回值返回给调用者return func(n, left, right)    # 递归elif n == lst[mid]:print("找到了")return mid# return  # 通过return返回. 终止递归else:print("没有这个数") # 递归的出口return -1 # 1, 索引+ 2, 什么都不返回, None
# 找66, 左边界:0,  右边界是:len(lst) - 1
ret = func(70, 0, len(lst) - 1)
print(ret) # 不是None

def binary_search(ls, target):left = 0right = len(ls) - 1if left > right:print("不在这里")middle = (left + right) // 2if target < ls[middle]:return binary_search(ls[:middle], target)elif target > ls[middle]:return binary_search(ls[middle+1:], target)else:print("在这里")
binary_search(lst, 567)