递归访问目录,定义嵌套函数，打印斐波那契数列，对列表进行排序，map函数计算列表，filter函数过滤，reduce计算1~100的和

递归访问目录

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 list_dir_content(dir_path, count=0):
    for file_name in os.listdir(dir_path):
        print(file_name)
        sub_path = os.path.join(dir_path, file_name)

        if os.path.isdir(sub_path):
            print(count * "\t" + file_name)
            sub_count = count + 1
            list_dir_content(sub_path, sub_count)

        if os.path.isfile(sub_path):
            print(count * "\t" + file_name)


list_dir_content("D:\\test")
print(os.getcwd())

定义一个嵌套函数

外层函数打印this is outing function
内层函数功能:打印This is inner function

def outer():
    print("This is outing function")
    def inner():
        print("this is inner function")
    inner()
outer()

定义一个递归函数：打印斐波那契数列

F[n]=F[n-1]+F[n-2](n>=2,F[0]=0,F[1]=1)

def fibonacci_func(n):
    if n == 1:
        return 0
    if n == 2:
        return 1
    if n > 2:
        return fibonacci_func(n - 1) + fibonacci_func(n - 2)

list_data = []
for i in range(1, 20):
    data = fibonacci_func(i)
    list_data.append(data)
print(list_data)

对列表进行排序

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"]
list_data.sort(key=lambda x: (x[-1], x[0]))
print(list_data)

利用map函数： 计算三个列表，相同位置元素之和

list1 = [1, 2, 3]
list2 = [4, 5, 6]
list3 = [7, 8, 9]

list_data1 = [1, 2, 3]
list_data2 = [4, 5, 6]
list_data3 = [7, 8, 9]
map_obj = map(lambda x, y, z: x+y+z, list_data1, list_data2, list_data3)
print(map_obj, type(map_obj))
print(list(map_obj))

利用filter函数过滤列表中所有带a的字符串

list_data = [“grape”, “what”, “which”, “you”, “friend”, “am”]

list_data = ["grape", "what", "which", "you", "friend", "am"]
def func(list_data):
    for i in list_data:
        if i.find("a") >= 0:
            return i

print(list(filter(func, list_data)))

利用reduce计算1 + 2 + 3…+ 100之和

from functools import reduce


def coun_num(n1, n2):
    coun1 = n1 + n2
    return coun1


print(reduce(coun_num, [i for i in range(1, 101)]))