字符串格式化，函数，组，递归，练习

# -*- coding: UTF-8 -*-
msg = 'i am {} my hobby is {}'.format('lhf',18)
print(msg)

msg1 = 'i am %s my hobby is %s' % ('lhf',[1,2])
print(msg1)

name='lhf'
age=18
msg2 = 'i am %s my hobby is %s' % (name,age)
print(msg2)

tpl = 'percent %.2f %%' % 99.9762333333333
print(tpl)

tpl1 = 'i am %(name)s age %(age)d' % {'name':'alex','age':18}
print(tpl1)

msg3 = 'i am %(name)+60s my hobby is alex' % {'name':'lhf'}
print(msg3)

msg4 = 'i am 33[43;1m%(name)+60s33[0m my hobby is alex' %{'name':'lhf'}
print(msg4)

#print('root','x','0','0',sep=':') #结果(root:x:0:0)

print('root'+':'+'x'+':'+'0','0') #结果('root:x:0', '0')



result:
i am lhf my hobby is 18
i am lhf my hobby is [1, 2]
i am lhf my hobby is 18
percent 99.98 %
i am alex age 18
i am                                                          lhf my hobby is alex
i am                                                          lhf my hobby is alex
('root:x:0', '0')

# -*- coding: UTF-8 -*-
name = '海风'
def huangwei():
    name = '黄伟'
    print(name)
    def liuyang():
        name = '刘洋'
        print(name)
        def nulige():
            name = '炉指花'
            print(name)
        print(name)
        nulige()
    liuyang()
    print(name)
print(name)
huangwei()
print(name)

result:
海风
黄伟
刘洋
刘洋
炉指花
黄伟
海风

# -*- coding: UTF-8 -*-
def calc(n):
    print(n)
    if int(n/2) == 0:
        return n
    res = calc(int(n/2))
    return res
calc(20)

result:
20
10
5
2
1

集合特点：
1.由不同元素组成
2.集合无序
3.集合元素是不可变类型，包括（数字，字符串，元祖

#交叉补集
python_l = ['lcg','szw','zjw','lcg']
linux_l = ['lcg','szw','sb']
p_s = set(python_l)
l_s = set(linux_l)
print(p_s.symmetric_difference(l_s))
print(p_s^l_s)
p_s.difference_update(l_s) #等同于p_s = p_s - l_s
print(p_s)
s1 = {1,2}
s2 = {2,3,5}
print(s1.isdisjoint(s2)) #判断两个集合是否包含相同的元素

#子集，父集
s1 = {1,2}
s2 = {1,2,3}
print(s1.issubset(s2))#s1是s2的子集
print(s2.issubset(s1))
print(s2.issuperset(s1))#s1是s2的父集
s1.update(s2)#更新多个值
# s1.add(1,2,3,4)#更新一个值
# s1.union(s2)#不更新
print(s1)
s = frozenset('hello')#两种字典类型的区别
s5 = set('hello')
print(s,s5)
names = ['alex','alex','wupeiqi']
names = list(set(names))
print(names)
result:
{'zjw', 'sb'}
{'zjw', 'sb'}
{'zjw'}
False
True
False
True
{1, 2, 3}
frozenset({'e', 'o', 'l', 'h'}) {'e', 'o', 'l', 'h'}
['alex', 'wupeiqi']

可迭代类型：for循环可以遍历
函数形参与实参知识点：
　　位置参数（形如test(1,2,3)）：实参和形参一一对应，缺一不行，多一也不行
　　关键字参数（形如test(x=1,z=3,y=2)）：实参和形参无需一一对应，缺一不行，多一也不行
　　位置函数作为形参必须在关键字函数左边。
def test(x,*args) #*args用于函数扩展

模拟多级节点
方法一：
dic = {
    '植物':
        {'草本植物':
            ['牵牛花','瓜叶菊','葫芦','翠菊','冬小麦','甜菜'],
         '水木植物':
            ['乔木','灌木','丰灌木','如松','杉'],
         '水生植物':
            ['荷花','千屈菜','呂菊','黄奕菊','水葱','再力花','梭鱼草']},
     '动物':
        {'两栖动物':
            ['山龟','山鳖','石蛙','娃娃鱼','蟾蜍','龟','鳄鱼','蜥蜴','蛇'],
         '禽类':
            ['知鸡','原鸡','长鸣鸡','昌国鸡','斗鸡','长尾鸡','乌骨鸡'],
         '哺乳类动物':
            ['虎','狼','蛇','鹿','貂','猴','树懒','斑马','狗']}}
li = []
bool = True
while bool:
    bool1 = True
    for i,v in enumerate(dic,1):
        print(i,v)
        li.append(v)
    print('要添加新种类请输入i,否则输入对应下一级数字查看菜单')
    u_c = input('>>>')
    if u_c.isdigit():   #数字用来选择
        li1 = []
        while bool:
            u_c = int(u_c)
            for j,k in enumerate(dic[li[u_c-1]],1):
                print(j,k)
                li1.append(k)
            u_c1 = input('>>>>')
            if u_c1 == 'b':
                li.clear()
                break
            while bool:
                for x,y in enumerate(dic[li[int(u_c-1)]][li1[int(u_c1)-1]],1):
                    print(x,y)
                u_c2 = str(input('>>>>>'))
                u_c2 = u_c2.lower()
                if u_c2 == 'b':
                    # bool = False
                    li1.clear()
                    break
                elif u_c2 == 'q':
                    li.clear()
                    bool = False
                    break
                else:
                    continue
    elif u_c.isalpha(): #字母i用来添加新种类
        if u_c == 'i'or u_c == 'I':
            while bool1:
                lx = str(input('类型:'))
                if dic.get(lx,1) == 1: #得到lx中的字符串，如果该字符串在dic第一层中没有，则返回后面值‘1’
                    dic[lx] = {}
                zl = str(input('种类:'))
                if dic[lx].get(zl,1) == 1: #得到lx中的字符串，如果该字符串在dic第一层中没有，则返回后面值‘1’
                    dic[lx][zl] = []
                while bool1:
                    mc = input('名称:')
                    if mc == 'b':
                        li.clear()
                        bool1 = False
                        break
                    elif mc == 'q':
                        bool = False
                        break
                    elif mc in dic[lx][zl]:
                        print('已经存在')
                    else:
                        dic[lx][zl].append(mc)

方法二:（编程思路来自老男孩郑建文老师）
db = {
    '北京':{},
    '上海':{
        '虹口':{
            '外国语':'小明',
            '西藏北路':'布丁酒店',
            '上海南站':'高铁'},
        '宝山':{
            '大场':'蹦迪',
            '上海大学':'钱伟长'},
        '闵行':{
            '上海交大':'闵行校区'
        }
    }
}
path = []
while True:
    temp = db
    for item in path:
        temp = temp[item]
    print('当前可选的所有子节点：',list(temp.keys()))
    choice = input('1:添加节点; 2:查看节点(Q/B)；
 >>>')
    if choice == '1':
        name = input('请输入要添加的节点名称')
        temp[name] = {}
    elif choice == '2':
        name = input('请输入要查看的节点名称:')
        path.append(name)
    elif choice.lower() == 'b':
        if path:
            path.pop()
    elif choice.lower() == 'q':
        break
    else:
        print('输入错误，请重新输入。')