第一周代码整理
字典练习题
1 有如下值集合 [11,22,33,44,55,66,77,88,99,90...],
将所有大于 66 的值保存至字典的第一个key中,将小于 66 的值保存至第二个key的值中
即: {'k1': 大于66的所有值, 'k2': 小于66的所有值}
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l = [11,22,33,44,55,66,77,88,99,90] d = {'k1':[],'k2':[]} for i in l: if i >= 66: d['k1'].append(i) else: d['k2'].append(i) print(d)
2 统计s='hello alex alex say hello sb sb'中每个单词的个数
结果如:{'hello': 2, 'alex': 2, 'say': 1, 'sb': 2}
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s='hello alex alex say hello sb sb' l = s.split() #1 切成列表遍历单词 count统计重复次数 d = {} for i in l: d.setdefault(i,l.count(i)) #2 字典的setdefault功能:没有key就加 有key就不变 print(d)
集合练习题
一.关系运算
有如下两个集合,pythons是报名python课程的学员名字集合,linuxs是报名linux课程的学员名字集合
pythons={'alex','egon','yuanhao','wupeiqi','gangdan','biubiu'}
linuxs={'wupeiqi','oldboy','gangdan'}
1. 求出即报名python又报名linux课程的学员名字集合
2. 求出所有报名的学生名字集合
3. 求出只报名python课程的学员名字
4. 求出没有同时这两门课程的学员名字集合
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pythons={'alex','egon','yuanhao','wupeiqi','gangdan','biubiu'} linuxs={'wupeiqi','oldboy','gangdan'} print(pythons & linuxs) print(pythons | linuxs) print(pythons - linuxs) print(pythons ^ linuxs)
二.去重
1. 有列表l=['a.txt','b',1,'a.txt','a.txt'],列表元素均为可hash类型,去重,得到新列表,且新列表无需保持列表原来的顺序
2.在上题的基础上,保存列表原来的顺序
3.去除文件中重复的行,肯定要保持文件内容的顺序不变
4.有如下列表,列表元素为不可hash类型,去重,得到新列表,且新列表一定要保持列表原来的顺序
l = [
{'name': 'egon', 'age': 18, 'sex': 'male'},
{'name': 'alex', 'age': 73, 'sex': 'male'},
{'name': 'egon', 'age': 20, 'sex': 'female'},
{'name': 'egon', 'age': 18, 'sex': 'male'},
{'name': 'egon', 'age': 18, 'sex': 'male'},
]
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l=['a.txt','b',1,'a.txt','a.txt'] new_l = list(set(l)) print(new_l) l=['a.txt','b',1,'a.txt','a.txt'] new_l = [] for i in l: if i not in new_l: new_l.append(i) print(new_l) import os with open('db.txt','r',encoding='utf-8') as read_f,\ open('.db.txt.txt.swap','w',encoding='utf-8') as write_f: s=set() for line in read_f: if line not in s: s.add(line) write_f.write(line) os.remove('db.txt.txt') os.rename('.db.txt.txt.swap','db.txt.txt') l = [ {'name': 'egon', 'age': 18, 'sex': 'male'}, {'name': 'alex', 'age': 73, 'sex': 'male'}, {'name': 'egon', 'age': 20, 'sex': 'female'}, {'name': 'egon', 'age': 18, 'sex': 'male'}, {'name': 'egon', 'age': 18, 'sex': 'male'}, ] new_l = [] for i in l: if i not in new_l: new_l.append(i) print(new_l)
登陆接口
基础需求:
让用户输入用户名密码
认证成功后显示欢迎信息
输错三次后退出程序
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name_db = 'xjj' pwd_db = '123' count = 1 while count <= 3: name_inp = input('username>>: ').strip() pwd_inp = input('password>>: ') if name_db == name_inp and pwd_db == pwd_inp: print('login success') break else: print('login failure') count += 1
升级需求:
可以支持多个用户登录 (提示,通过列表存多个账户信息)
用户3次认证失败后,退出程序,再次启动程序尝试登录时,还是锁定状态(提示:需把用户锁定的状态存到文件里)
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#db.txt内容:egon1|egon2| dic={ 'egon1':{'password':'123','count':0}, 'egon2':{'password':'123','count':0}, 'egon3':{'password':'123','count':0}, } count=0 while True: name=input('u>>: ') if name not in dic: print('用户不存在') continue with open('a.txt','r') as f: lock_users=f.read().split('|') if name in lock_users: print('用户%s已经被锁定' %name) break if dic[name]['count'] > 2: print('尝试次数过多,锁定') with open('db.txt','a.txt') as f: f.write('%s|' %name) break password=input('p>>: ') if password == dic[name]['password']: print('登录成功') break else: print('用户名或密码错误') dic[name]['count']+=1
简单购物车
简单购物车,要求如下:
实现打印商品详细信息,用户输入商品名和购买个数,则将商品名,价格,购买个数加入购物列表,
如果输入为空或其他非法输入则要求用户重新输入
msg_dic = {
'apple': 10,
'tesla': 100000,
'mac': 3000,
'lenovo': 30000,
'chicken': 10,
}
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shoppint_cart = [] while True: for k,v in msg_dic.items(): info = '商品名:{} 价格:{}'.format(k,v) print(info.center(30,' ')) name = input('请输入商品名').strip() if name not in msg_dic: print('商品名不对,从新输入') continue while True: count = input('请输入购买个数').strip() if not count.isdigit(): print('购买个数必须为整数') else: count = int(count) break for item in shoppint_cart: if item['name'] == name: item['count'] += count break else: price = msg_dic[name] info = {'name':name,'count':count,'price':price} shoppint_cart.append(info) print(shoppint_cart)