迭代器和高阶函数(map, filter, reduce, sorted)

from collections import Iterable,Iterator
res = isinstance(range(8),Iterable)
print(res) # True  判断这个数据类型是不是一个可迭代对象
res1 = isinstance(range(8),Iterator)
print(res1) # False  判断这个数据类型是不是一个迭代器

list1 = [1,2,"字符",9.9]
it = iter(list1)
# 方式一:
res = "__iter__" in dir(it) and "__next__" in dir(it)
print(res) # True
# 方式二:
print(it) # <list_iterator object at 0x000001F41C4DF1D0>
print(type(it)) # <class ‘list_iterator‘>
# 方式三:
from collections import Iterator,Iterable
res = isinstance(it,Iterator)
print(res) # True

list1 = [1,2,"字符",9.9]
it = iter(list1)
# print(it) # <list_iterator object at 0x000001976367F1D0>
# 调用迭代器里面的元素:
# 方式一:  用next()方法
res = next(it)
print(res) # 1
res = next(it)
print(res) # 2
res = next(it)
print(res) # 字符
res = next(it)
print(res) # 9.9
# res = next(it)
# print(res) # StopIteration  此时迭代器里的元素已经调用完,再调用的话就会报错,因为迭代器是一直向下执行的,不会再回过头去取值

it2 = iter(list1)
res2 = next(it2)
print(res2) # 1  如果要重新取迭代器里面的值的话,就要重置这个迭代器,也就是将容器类型重新转化成迭代器

# 方式二: 用for和next配合使用来取值
list1 = [1,2,"字符",9.9]
it = iter(list1)
for i in range(len(list1)):
    res = next(it)
    print(res)  # 这个是将值一次性全部取出来,如果数据量比较大的话,就不推荐这种方式

list1 = [1, 2, "字符", 9.9]
it = iter(list1)
for i in range(2):
    res = next(it)
    print(res)  #  这个是只取出迭代器里面的前面两个值,如果数据量比较大,而我们只需要前面某些数据的话,推荐这种方式

# 方式三: 直接使用for循环遍历迭代器里面的值
list1 = [1, 2, "字符", 9.9]
it = iter(list1)
for i in it:
    print(i)  #  遍历出迭代器里的所有值

# 方式四: 将迭代器强转为列表类型
list1 = [1, 2, "字符", 9.9]
it = iter(list1)
list1 = list(it)
print(list1) # [1, 2, ‘字符‘, 9.9]

# 案例一: 将列表里面的字符串元素变成整型
#方法一:
list1 = ["1","2","3","4"]  # ==> [1,2,3,4]
from collections import Iterator,Iterable
it = map(int,list1)
res = isinstance(it,Iterator)
print(res)  # True 所以map函数取值应该遵循迭代器取值的方法
list2 = list(it)
print(list2) # [1, 2, 3, 4]
# 方法二:
it = map(lambda n:int(n),list1)
list2 = list(it)
print(list2) # [1, 2, 3, 4]

# 案例二:
# 方法一:
list1 = [1,2,3,4]  # ==> [1,4,9,16]
def func1(n):
    return n**2
it = map(func1,list1)
list2 = list(it)
print(list2) # [1, 4, 9, 16]
# 方法二:
it = map(lambda n:n**2,list1)
list2 = list(it)
print(list2) # [1, 4, 9, 16]


# 案例三:
# dic = {97:"a",98:"b",99:"c"} 给["a","b","c"] => 返回 [97,98,99]
# 方法一:
dict1 = {97:"a",98:"b",99:"c"}
list1 = ["a","b","c"]
def func1(n):
    for k,v in dict1.items():
        if str(n) in v:
            return k
it = map(func1,list1)
res = list(it)
print(res) # [97, 98, 99]
# 方法二:
def func1(n):
    dict2 = {}
    for k,v in dict1.items():
        dict2[v]=k
    return dict2[n]
it = map(func1,list1)
res = list(it)
print(res) # [97, 98, 99]

list1 = [1,2,3,4,5,6,7]  # 需要保留奇数
# 方法一:
def func1(n):
    if n %2 == 1:
        return True
    else:
        return False
it = filter(func1,list1)
list2 = list(it)
print(list2) # [1, 3, 5, 7]
# 方法二:
it = filter(lambda n:True if n%2 == 1 else False,list2)
list2 = list(it)
print(list2) # [1, 3, 5, 7]

# 案例一:
list1 = [5,4,8,8]
# 方法一:
def func1(x,y):
    return x*10+y
from functools import reduce
res = reduce(func1,list1)
print(res,type(res)) # 5488 <class ‘int‘>
# 方法二:
from functools import reduce
res= reduce(lambda x,y:x*10+y,list1)
print(res,type(res)) # 5488 <class ‘int‘>

# 案例二:将字符串"123"变成123,不使用int方法
str1 = "123"
def dictvar(n):
    dict1 ={"0":0,"1":1,"2":2,"3":3,"4":4,"5":5,"6":6,"7":7,"8":8,"9":9}
    return dict1[n]
it = map(dictvar,str1)
list1 = list(it)
print(list1) # [1, 2, 3]  首先将字符串类型的数字转为整型的数字
def func1(x,y):
    return x*10+y
from functools import reduce
res = reduce(func1,list1)
print(res,type(res)) # 123 <class ‘int‘>

# 按照升序
tuple1 = (2,83,6,78)
list1 = sorted(tuple1)
print(list1) # [2, 6, 78, 83]

# 按照倒序
set1 = {2,84,52,58,23}
list1 = sorted(set1,reverse=True)
print(list1) # [84, 58, 52, 23, 2]

# 按照函数定义的功能排序
list1 = [26,84,52,1,32]
def func1(n):
    return n%10
res = sorted(list1,key=func1)
print(res) # [1, 52, 32, 84, 26]