# ====================         内置函数            ===================    ================

#any 判断有一为真 才为真#all  判断 全部为真 才为真# print(any(["",1,0]))

#ord 输入字符，返回unicode 对应的编码位置# print(ord("a"))# print(ord("b"))# print(ord("中"))

#chr 输入编码位置，返回Unicode对应的字符# print(chr(97))# print(chr(20013))# print(ascii("a"))# print(ascii('国'))

#%s repr 原形毕露# name = "alex%r"%('sb')# print(name)# print(repr([1,2]))

#slice()切片# li = [11,22,33,44]# print(li[1::2])# sl = slice(1,4,2)# print(li[sl])

#reversed()  形成一个新的迭代器    反 转# li = [11,22,33,53,44,99,"alex"]# li.reverse()# print(li)# l2 = reversed(li)# for i in l2:#     print(i)

#重要的 内置函数

# dic = {"k1":20,"k2":30,"k3":100}# def func(x):#     return dic[x]# l1 = max(dic,key=func,)# print(l1)## dic = {3:30,2:30,1:100}# def func(x):#     return x# l2 = max(dic,key=func,)# print(l2)

# from collections import  Iterator#zip 可以防止多个可迭代对象，一i最少的可迭代对象的个数为基准，返回一个迭代器#拉链方法# li = [1,2,3,4,5]# l2 = ["a","b","v","s"]# l3 = [1,2,3]# print(zip(li,l2,l3))# print(isinstance(zip(li,l2,l3),Iterator))# for i in zip(li,l2,l3):#     print(i)

#map#[i*i for i in range(5)]# print(i*i for i in range(5))# print(map(abs,[1,2,3,-6,-7]))# for i in map(abs,[1,2,3,-4,-5,-7]):#     print(i)# def func(x):return x*x# for i in map(func,range(5)):#     print(i)# def func(x):return x*x# for i in map(func,[0,1,2,3,4]):#     print(i)

#filter 过滤#[i for i in [1,2,3,4,5] if i % 2 == 0]# def func1(x):return x % 2 == 0# for i in filter(func1,[1,2,3,4,5]):#     print(i)

# sorted  返回的是列表# 形成一个新列表，并返回#与list.sort() 改变原列表# li = [1,4,3,5,6]# print(sorted(li))# print(li)# l = ["sfdfrt",'dfdfrg','ggt','fdfd']# def func(x):#     return len(x)# for i in sorted(l,key=func):#     print(i)# print(sorted(l,key=len))

#都是带key的：zip max min map filter 返回的是迭代器 sorted 返回的是列表#bytes()#bytes# s = "中国".encode("utf-8")#编码# print(s)# s.decode("utf-8")#编码# #s gbk 的bytes -----> utf-8的 bytes# s = b"\xd6\xd0\xb9\xfa"# s1 = s.decode("gbk").encode("utf-8")# print(s1)

#方法二# s = '中国'.encode('utf-8')# 编码# s1 = '中国'# s2 = bytes(s1,encoding='utf-8').decode('utf-8')# print(s1)# import sys# sys.setrecursionlimit(100000)# count = 0# def func():#     global count#     count += 1#     print(count)#     func()# func()

# =========================================        匿名函数 =============# def func1(x):#     return x# func = lambda x:x# print(func1(3))# print(func(3))# def func(x,y):#     return x+y# func1 = lambda x,y:x+y# print(func1(1,3))# def func2(x):return x*x# for i in map(lambda x:x*x,[0,1,2,3,4]):#     print(i)

# l1 = ["234",'23','rtfggt','trtrtrt']# def func1(x):return  x % 2 == 0# for i in filter(func1,[1,2,3,4,5]):#     print(i)# for i in filter(lambda x:len(x) >= 3,l1):#     print(i)

#面试题# d = lambda x:x*2# t = lambda x:x*3# x = 2# x = d(x)# x = t(x)# x = d(x)# print(x)

# 现在有两个元组(('a'),('b')),(("c'),('d')), 请使用python 中的 匿名函数生成列标#[{'a':'c'},{"b":'d'}]# tu1 = (('a'),('b'))# tu2 = (('c'),('d'))# func = lambda  x,y:[{x[0]:y[0]},{x[1]:y[1]}]# print(func(tu1,tu2))## tu2 = (2)# print(tu2)# print(tu1)