# ### 高阶函数 :能够把函数当成参数传递的就是高阶函数 (map reduce sorted filter )
# map
'''
map(func,iterable)
功能:把iterable里面的数据一个一个的拿出来放到func函数中进行处理,最后把处理的结果返回到迭代器中
参数:
func : 自定义函数 或 内置函数
iterable: 可迭代性数据 (常用:容器类型数据,range对象,迭代器)
返回值:迭代器
'''
# 1. ["1","2","3","4"] => [1,2,3,4]
listvar = ["1","2","3","4"]
'''
# 普通写法
lst = []
for i in listvar:
res = int(i)
lst.append(res)
print(lst)
'''
from collections import Iterator , Iterable
it = map(int,listvar)
print(isinstance(it,Iterator))
print(isinstance(it,Iterable))
print(it)
# (1)使用for循环遍历迭代器
'''
for i in it:
print(i)
'''
# (2) 使用next获取迭代器中的数据
'''next调用迭代器中的数据,是单项不可逆,一条路走到黑'''
'''
res = next(it)
print(res)
res = next(it)
print(res)
res = next(it)
print(res)
res = next(it)
print(res)
'''
# (3) 使用list强转迭代器(瞬间得到迭代器中的所有数据)
res = list(it)
print(res)
# 2. [1,2,3,4,5] => [1,4,9,16,25]
# 普通写法:
listvar = [1,2,3,4,5]
lst = []
for i in listvar:
res = i ** 2
lst.append(res)
print(lst)
'''
map 如果是自定义函数,一定要有返回值
代码解析:
首先把listvar当中的第一个值1拿到func当中进行处理,返回1扔到迭代器里
然后把listvar当中的第二个值2拿到func当中进行处理,返回4扔到迭代器里
然后把listvar当中的第三个值3拿到func当中进行处理,返回9扔到迭代器里
..
依次类推,直到把列表里面的数据全部拿完为止.
'''
def func(n):
return n ** 2
it = map(func,listvar)
print(isinstance(it,Iterator))
listvar = list(it)
print(listvar)
# 3. {97:"a",98:"b",99:'c',100:'d',101:"e"} {"c","a","b"} => [99,97,98]
'''
# 普通写法:
dic = {97:"a",98:"b",99:'c',100:'d',101:"e"}
dic2 = {}
# 反转字典
for a,b in dic.items():
dic2[b] = a
print(dic2)
lst = ["c","a","b"]
lst2 = []
#{'a': 97, 'b': 98, 'c': 99, 'd': 100, 'e': 101}
for i in lst:
res = dic2[i]
lst2.append(res)
print(lst2)
'''
def func(n):
dic = {97:"a",98:"b",99:'c',100:'d',101:"e"}
dic2 = {}
# 反转字典
for a,b in dic.items():
dic2[b] = a
#{'a': 97, 'b': 98, 'c': 99, 'd': 100, 'e': 101}
return dic2[n]
it = map(func,["c","a","b"])
print(isinstance(it,Iterator))
lst = list(it)
print(lst)
