python1.递归访问目录 2.定义一个嵌套函数 3.定义一个递归函数：打印斐波那契数列 4.对列表进行排序: 5.利用map函数： 计算三个列表，相同位置元素之和 6.利用filter函

前导知识：

1.切换目录: os.chdir(path)
2.列出当前目录中所有的文件和文件夹 os.listdir(path), path: 绝对路径
3.判断是否是文件: os.path.isfile(path)
4.判断是否是目录: os.path.isdir(path)
5.拼接路径： os.path.join(path1, path2, path3…)
且目录中嵌套目录，有层次的列出给定目录中所有的文件和文件夹

import os


def listdir(path, count=0):
    list_dir = os.listdir(path)
    for i in list_dir:
        path1 = os.path.join(path + "/", i)
        if os.path.isdir(path1):
            print(count * "    ", i)
            count1 = count + 1
            listdir(path1, count1)
        else:
            print(count * '    ', i)


listdir('D://text')

def outer():
    print('this is outing function')

    def inner():
        print('this is inner function')

    return inner


outer()()

def fibonacci_seq(arg):
    if arg == 0:
        return 0
    if arg == 1:
        return 1
    if arg >= 2:
        return fibonacci_seq(arg - 1) + fibonacci_seq(arg - 2)


print(fibonacci_seq(9))  # 34

list_data = [“grape”, “peach”, “berry”, “pineapple”, “apple”, “strayberry”, “watermelon”]
排序规则：按照最后一个字符进行排序，如果最后一个字符相等，按照第一个字符排序

list_data = ["grape", "peach", "berry", "pineapple", "apple", "strayberry", "watermelon"]
 
 
def sort_role(x):
    return x[-1], x[0]
 
 
# list_data.sort(key=sort_role)
list_data.sort(key=lambda x: (x[-1], x[0]))
print(list_data)

list1 = [1, 2, 3]
list2 = [4, 5, 6]
list3 = [7, 8, 9]
sum_list = map(lambda x, y, z: (x+y+z), list1, list2, list3)
# print(sum_list)  # 返回的是一个map对象
print(list(sum_list))
# 结果：[12, 15, 18]

list_data = ["grape", "what", "which", "you", "friend", "am"]
result = filter(lambda x: 'a' not in x, list_data)
# print(result)  # 返回一个filter类型的对象
print(list(result))
# 结果：['which', 'you', 'friend']

from functools import reduce
list_data = list(range(1, 101))
# print(list_data)
result = reduce(lambda x, y: x+y, list_data)
print(result)  # 结果：5050