1 # 计算n的n次⽅
2 def func(n):
3   return n**n
4 print(func(10))
5 
6 
7 f = lambda n: n**n
8 print(f(10))

print(a(5)) # 函数的名字可以认为是a

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

匿名函数, 给函数传递2给参数. 返回最大值
fn = lambda *args: max(args) # 单行函数

print(fn(1,2,5,2,3,4,156,3,2,2,4,5,56,34,34,34,34,88))

1 lst = [1,5,3,4,6]
2 lst2 = sorted(lst)
3 print(lst) # 原列表不会改变
4 print(lst2) # 返回的新列表是经过排序的
5 dic = {1:'A', 3:'C', 2:'B'}
6 print(sorted(dic)) # 如果是字典. 则返回排序过后的ke

1 # 根据字符串⻓度进⾏排序
2 lst = ["麻花藤", "冈本次郎", "中央情报局", "狐仙"]
3 # 计算字符串⻓度
4 def func(s):
5   return len(s)
6 print(sorted(lst, key=func))

 1 # 根据字符串⻓度进⾏排序
 2 lst = ["麻花藤", "冈本次郎", "中央情报局", "狐仙"]
 3 # 计算字符串⻓度
 4 def func(s):
 5   return len(s)
 6 print(sorted(lst, key=lambda s: len(s)))
 7 lst = [{"id":1, "name":'alex', "age":18},
 8   {"id":2, "name":'wusir', "age":16},
 9   {"id":3, "name":'taibai', "age":17}]
10 # 按照年龄对学⽣信息进⾏排序
11 print(sorted(lst, key=lambda e: e['age']))

 1 lst = [
 2     {'name':"汪峰","age":48},
 3     {"name":"章子怡",'age':38},
 4     {"name":"alex","age":39},
 5     {"name":"wusir","age":32},
 6     {"name":"赵一宁","age":28}
 7     ]
 8 
 9 
10 
11 ll = sorted(lst, key=lambda el: len(el['name']), reverse=True)
12 print(ll)

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

筛选,
f = filter(lambda el: el[0]!="张", lst) # 将lst中的每一项传递给func, 所有返回True的都会保留,
所有返回False都会被过滤掉

print("__iter__" in dir(f)) # 判断是否可以进行迭代
for e in f:
    print(e)

 1 lst = [
 2     {"name":"汪峰", "score":48},
 3     {"name":"章子怡", "score":39},
 4     {"name":"赵一宁","score":97},
 5     {"name":"石可心","score":90}
 6 ]
 7 
 8 f = filter(lambda el: el['score'] < 60 , lst) # 去16期的人
 9 
10 print(list(f))

1 def func(e):
2   return e*e
3 mp = map(func, [1, 2, 3, 4, 5])
4 print(mp)
5 print(list(mp))

1 print(list(map(lambda x: x * x, [1, 2, 3, 4, 5])))

1 # 计算两个列表相同位置的数据的和
2 lst1 = [1, 2, 3, 4, 5]
3 lst2 = [2, 4, 6, 8, 10]
4 print(list(map(lambda x, y: x+y, lst1, lst2)))

1 lst1 = [1, 3, 5, 7]
2 lst2 = [2, 4, 6, 8, 10]
3 # 水桶效应, zip()
4 m = map(lambda x, y, z: x + y+ z, lst1, lst2, [5,1,2,3,6])
5 print(list(m))

1 def func():
2   print("我是谁")
3   func()
4 func()

1 def foo(n):
2   print(n)
3   n += 1
4   foo(n)
5 foo(1)

 1 import os
 2 def func(filepath, n): # d:/sylar/
 3     # 1,打开这个文件夹
 4     files = os.listdir(filepath)
 5     # 2. 拿到每一个文件名
 6     for file in files:  # 文件名
 7         # 3. 获取到路径
 8         f_d = os.path.join(filepath, file) # d:/sylar/文件名/
 9         # 4. 判断是否是文件夹
10         if os.path.isdir(f_d):
11             # 5. 如果是文件夹. 继续再来一遍
12             print("\t"*n, file,":") # 打印文件名
13             func(f_d, n + 1)
14         else:   #  不是文件夹. 普通文件
15             print("\t"*n, file)
16 
17 func("d:/sylar",0)

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

 1 lst = [22, 33, 44, 55, 66, 77, 88, 99, 101, 238, 345, 456, 567, 678, 789]
 2 n = 567
 3 left = 0
 4 right = len(lst) - 1
 5 count = 1
 6 while left <= right:
 7  middle = (left + right) // 2
 8  if n < lst[middle]:
 9  right = middle - 1
10  elif n > lst[middle]:
11  left = middle + 1
12  else:
13  print(count)
14  print(middle)
15  break
16  count = count + 1
17 else:
18  print("不存在")   

 1 def binary_search(ls, target):
 2  left = 0
 3  right = len(ls) - 1
 4  if left > right:
 5  print("不在这⾥")
 6  middle = (left + right) // 2
 7  if target < ls[middle]:
 8  return binary_search(ls[:middle], target)
 9  elif target > ls[middle]:
10  return binary_search(ls[middle+1:], target)
11  else:
12  print("在这⾥")
13 binary_search(lst, 567)

1  # 时间复杂度最低, 空间复杂度最低
2         lst1 = [5,6,7,8]
3         lst2 = [0,0,0,0,0,1,1,1,1]
4         for el in lst1:
5             lst2[el] = 1
6 
7         lst2[4] == 1   # o(1)