标签：二进制 文件删除 匿名函数 需要 address coding iter binary over
# 1.看代码分析结果
# func_list = []
# for i in range(10):
# func_list.append(lambda: i)
# v1 = func_list[0]()
# v2 = func_list[5]()
# print(v1, v2)
# result:
# 9 9
# 2.看代码分析结果
# func_list = []
# for i in range(10):
# func_list.append(lambda x:x+i)
# v1 = func_list[0](2)
# v2 = func_list[5](1)
# print(v1,v2)
# result:
# 11 10
# 3.看代码分析结果
# func_list = []
# for i in range(10):
# func_list.append(lambda x:x+i)
# for i in range(0,len(func_list)):
# result = func_list[i](i)
# print(result)
# result:
# 0 2 4 6 8 10 12 14 16 18
# 4.看代码写结果（面试题）：
# def func(name):
# v = lambda x:x+name
# return v
#
# v1 = func(‘太白‘) # closure: v1: v name = ‘太白‘
# v2 = func(‘alex‘) # closure: v2: v name = ‘alex‘
# v3 = v1(‘银角‘) # v(‘银角‘)
# v4 = v2(‘金角‘) # v(‘金角‘)
# print(v1,v2,v3,v4)
# result:
# <function func.<locals>.<lambda> at 0x0000019B736BBD90> <function func.<locals>.<lambda> at 0x0000019B7387EEA0> 银角太白 金角alex
# 5.看代码写结果【面试题】
result = []
for i in range(10):
func = lambda: i # 注意：函数不执行，内部代码不会执行。
result.append(func)
print(i)
print(result)
v1 = result[0]()
v2 = result[9]()
print(v1, v2)
# result:
# 9
# [<function <lambda> at 0x0000016314AB1F28>, <function <lambda> at 0x0000016314C8BD90>, <function <lambda> at 0x0000016314E4EEA0>, <function <lambda> at 0x0000016314E4F510>, <function <lambda> at 0x0000016314E4F598>, <function <lambda> at 0x0000016314E4F620>, <function <lambda> at 0x0000016314E4F6A8>, <function <lambda> at 0x0000016314E4F730>, <function <lambda> at 0x0000016314E4F7B8>, <function <lambda> at 0x0000016314E4F840>]
# 9 9
# 6.看代码分析结果【面试题】
def func(num):
def inner(): # closure
print(num)
return inner
result = []
for i in range(10):
f = func(i)
result.append(f)
print(i)
print(result)
v1 = result[0]()
v2 = result[9]()
print(v1, v2)
# result:
# 9
# [<function func.<locals>.inner at 0x0000029DFF491F28>, <function func.<locals>.inner at 0x0000029DFF66BD90>, <function func.<locals>.inner at 0x0000029DFF82EEA0>, <function func.<locals>.inner at 0x0000029DFF82F510>, <function func.<locals>.inner at 0x0000029DFF82F598>, <function func.<locals>.inner at 0x0000029DFF82F620>, <function func.<locals>.inner at 0x0000029DFF82F6A8>, <function func.<locals>.inner at 0x0000029DFF82F730>, <function func.<locals>.inner at 0x0000029DFF82F7B8>, <function func.<locals>.inner at 0x0000029DFF82F840>]
# 0
# 9
# None None
# 7.看代码写结果【新浪微博面试题】
def func():
for num in range(10):
pass
v4 = [lambda: num + 10, lambda: num + 100, lambda: num + 100, ]
result1 = v4[1]()
result2 = v4[2]()
print(result1, result2)
func()
# result:
# 109 109
# 8.请编写一个函数实现将IP地址转换成一个整数。【面试题，较难,可以先做其他题】
# 如 10.3.9.12 转换规则为二进制：
# 10 00001010
# 3 00000011
# 9 00001001
# 12 00001100
# 再将以上二进制拼接起来计算十进制结果：00001010 00000011 00001001 00001100 = ？
def ip_to_decimal(ip_address):
ip_sp_list = ip_address.split(‘.‘)
ip_binary = ‘‘
for i in ip_sp_list:
ip_binary += bin(int(i))[2:].rjust(8, ‘0‘)
# print(ip_binary) # 00001010 00000011 00001001 00001100
return int(ip_binary, 2)
print(ip_to_decimal(‘10.3.9.12‘))
# 9.都完成的做一下作业（下面题都是用内置函数或者和匿名函数结合做出）：
# 9.1 用map来处理字符串列表,把列表中所有人都变成sb,比方alex_sb
# ? name=[‘oldboy’,‘alex‘,‘wusir‘]
name = [‘oldboy‘, ‘alex‘, ‘wusir‘]
name_sb = map(lambda str: str + ‘_sb‘, name)
print(list(name_sb))
# 9.2 用map来处理下述l，然后用list得到一个新的列表，列表中每个人的名字都是sb结尾
# l = [{‘name‘:‘alex‘},{‘name‘:‘y‘}]
# def func(dic):
# dic[‘name‘] += ‘_sb‘
# return dic
# l = [{‘name‘:‘alex‘},{‘name‘:‘y‘}]
# name_sb = map(func, l)
# print(list(name_sb))
# 9.3 用filter来处理,得到股票价格大于20的股票名字
# shares={
# ‘IBM‘:36.6,
# ‘Lenovo‘:23.2,
# ‘oldboy‘:21.2,
# ‘ocean‘:10.2,
# }
# shares = {
# ‘IBM‘: 36.6,
# ‘Lenovo‘: 23.2,
# ‘oldboy‘: 21.2,
# ‘ocean‘: 10.2,
# }
# a = filter(lambda name: shares[name] > 20, shares)
# print(list(a))
# 9.4 有下面字典，得到购买每只股票的总价格，并放在一个迭代器中结果：list一下[9110.0, 27161.0,......]
# portfolio = [
# {‘name‘: ‘IBM‘, ‘shares‘: 100, ‘price‘: 91.1},
# {‘name‘: ‘AAPL‘, ‘shares‘: 50, ‘price‘: 543.22},
# {‘name‘: ‘FB‘, ‘shares‘: 200, ‘price‘: 21.09},
# {‘name‘: ‘HPQ‘, ‘shares‘: 35, ‘price‘: 31.75},
# {‘name‘: ‘YHOO‘, ‘shares‘: 45, ‘price‘: 16.35},
# {‘name‘: ‘ACME‘, ‘shares‘: 75, ‘price‘: 115.65}]
portfolio = [
{‘name‘: ‘IBM‘, ‘shares‘: 100, ‘price‘: 91.1},
{‘name‘: ‘AAPL‘, ‘shares‘: 50, ‘price‘: 543.22},
{‘name‘: ‘FB‘, ‘shares‘: 200, ‘price‘: 21.09},
{‘name‘: ‘HPQ‘, ‘shares‘: 35, ‘price‘: 31.75},
{‘name‘: ‘YHOO‘, ‘shares‘: 45, ‘price‘: 16.35},
{‘name‘: ‘ACME‘, ‘shares‘: 75, ‘price‘: 115.65}]
# aggregate = map(lambda dic: round(dic[‘shares‘] * dic[‘price‘], 1), portfolio)
# print(list(aggregate))
# 9.5 还是上面的字典，用filter过滤出单价大于100的股票。
# price_over_100 = filter(lambda dic: dic[‘price‘] > 100, portfolio)
# print(list(price_over_100))
# 9.6 有下列三种数据类型，
# l1 = [1,2,3,4,5,6]
# l2 = [‘oldboy‘,‘alex‘,‘wusir‘,‘太白‘,‘日天‘]
# tu = (‘**‘,‘***‘,‘****‘,‘*******‘)
# 写代码，最终得到的是（每个元祖第一个元素>2,第三个*至少是4个。）
# [(3, ‘wusir‘, ‘****‘), (4, ‘太白‘, ‘*******‘)]这样的数据。
# l1 = [1,2,3,4,5,6]
# l2 = [‘oldboy‘,‘alex‘,‘wusir‘,‘太白‘,‘日天‘]
# tu = (‘**‘,‘***‘,‘****‘,‘*******‘)
# z1 = zip(l1,l2,tu)
# z1 = filter(lambda tu: tu[0] > 2, z1)
# print(list(z1))
# 9.7 有如下数据类型(实战题)：
# l1 = [ {‘sales_volumn‘: 0},
# {‘sales_volumn‘: 108},
# {‘sales_volumn‘: 337},
# {‘sales_volumn‘: 475},
# {‘sales_volumn‘: 396},
# {‘sales_volumn‘: 172},
# {‘sales_volumn‘: 9},
# {‘sales_volumn‘: 58},
# {‘sales_volumn‘: 272},
# {‘sales_volumn‘: 456},
# {‘sales_volumn‘: 440},
# {‘sales_volumn‘: 239}]
# ? 将l1按照列表中的每个字典的values大小进行排序，形成一个新的列表。
# l1 = [{‘sales_volumn‘: 0},
# {‘sales_volumn‘: 108},
# {‘sales_volumn‘: 337},
# {‘sales_volumn‘: 475},
# {‘sales_volumn‘: 396},
# {‘sales_volumn‘: 172},
# {‘sales_volumn‘: 9},
# {‘sales_volumn‘: 58},
# {‘sales_volumn‘: 272},
# {‘sales_volumn‘: 456},
# {‘sales_volumn‘: 440},
# {‘sales_volumn‘: 239}]
# sorted_l1 = sorted(l1, key=lambda dic: dic[‘sales_volumn‘])
# print(sorted_l1)
# 10.求结果(面试题)
# v = [lambda :x for x in range(10)]
# print(v)
# print(v[0])
# print(v[0]())
# result:
# [<function <listcomp>.<lambda> at 0x0000018196A9FAE8>, <function <listcomp>.<lambda> at 0x0000018196A9FB70>, <function <listcomp>.<lambda> at 0x0000018196A9FBF8>, <function <listcomp>.<lambda> at 0x0000018196A9FC80>, <function <listcomp>.<lambda> at 0x0000018196A9FD08>, <function <listcomp>.<lambda> at 0x0000018196A9FD90>, <function <listcomp>.<lambda> at 0x0000018196A9FE18>, <function <listcomp>.<lambda> at 0x0000018196A9FEA0>, <function <listcomp>.<lambda> at 0x0000018196A9FF28>, <function <listcomp>.<lambda> at 0x0000018196AA5048>]
# <function <listcomp>.<lambda> at 0x0000018196A9FAE8>
# 9
# 11.求结果(面试题)
# v = (lambda :x for x in range(10))
# print(v) # 生成器
# # print(v[0]) # 生成器不可用索引方式读取
# # print(v[0]()) # 生成器不可用索引方式读取
# print(next(v)) # 函数
# print(next(v)()) # 1
# 12.map(str,[1,2,3,4,5,6,7,8,9])输出是什么? (面试题)
# 一个迭代器，转换为list内容为[‘1‘, ‘2‘, ‘3‘, ‘4‘, ‘5‘, ‘6‘, ‘7‘, ‘8‘, ‘9‘]
# 13.求结果：（面试题，比较难，先做其他题）
# def num():
# return [lambda x:i*x for i in range(4)]
# print([m(2) for m in num()])
# [6, 6, 6, 6]
# 14.有一个数组[3,4，1,2,5,6,6,5,4，3,3]请写一个函数，找出该数组中没有重复的数
# 的总和（上面数据的么有重复的总和为1+2=3)(面试题)
def no_repeat_sum(array):
obj = filter(lambda x: array.count(x) == 1, array)
return sum(obj)
print(no_repeat_sum([3,4,1,2,5,6,6,5,4,3,3]))
# 15.写一个函数完成三次登陆功能：
# 用户的用户名密码从一个文件register中取出。
# register文件包含多个用户名，密码，用户名密码通过|隔开，每个人的用户名密码占用文件中一行。
# 完成三次验证，三次验证不成功则登录失败，登录失败返回False。
# 登陆成功返回True。
def get_userinfo():
users_dic = {}
with open(r‘04 作业\register‘, encoding=‘utf-8‘, mode=‘r‘) as file_handler:
for line in file_handler:
temp_list = line.split(‘|‘)
users_dic.setdefault(temp_list[0].strip(), temp_list[1].strip())
return users_dic
def login():
for i in range(3):
users_dic = get_userinfo()
user = input(‘USER:\n‘)
password = input(‘PASSWORD:\n‘)
if user in users_dic and password == users_dic[user]:
print(‘u success!‘)
return True
return False
# login()
# 16.再写一个函数完成注册功能：
# 用户输入用户名密码注册。
# 注册时要验证（文件regsiter中）用户名是否存在，如果存在则让其重新输入用户名，如果不存在，则注册成功。
# 注册成功后，将注册成功的用户名，密码写入regsiter文件，并以 | 隔开。
# 注册成功后，返回True,否则返回False。
def get_userinfo():
users_dic = {}
with open(r‘04 作业\register‘, encoding=‘utf-8‘, mode=‘r‘) as file_handler:
for line in file_handler:
temp_list = line.split(‘|‘)
users_dic.setdefault(temp_list[0].strip(), temp_list[1].strip())
return users_dic
def new_user(new_user_list):
users_dic = get_userinfo()
if new_user_list[0] in users_dic:
return False
with open(r‘04 作业\register‘, encoding=‘utf-8‘, mode=‘a‘) as file_handler:
file_handler.write(‘\n‘ + new_user_list[0] + ‘|‘ + new_user_list[1])
return True
def create_account():
for i in range(3):
new_user_list =[]
user = input(‘USER:\n‘)
password = input(‘PASSWORD:\n‘)
new_user_list.append(user)
new_user_list.append(password)
if new_user(new_user_list):
print(‘U create a new account!‘)
return True
return False
# 17.完成一个员工信息表的增删功能（选做题，有时间做，没时间周末做）。
# 文件存储格式如下：
# id，name，age，phone，job
# 1,Alex,22,13651054608,IT
# 2,太白,23,13304320533,Tearcher
# 3,nezha,25,1333235322,IT
# 现在要让你实现两个功能：
# 第一个功能是实现给文件增加数据，用户通过输入姓名，年龄，电话，工作，
# 给原文件增加数据（增加的数据默认追加到原数据最后一行的下一行），但id要实现自增（id自增有些难度，id是不需要用户输入的但是必须按照顺序增加）。
# 第二个功能是实现给原文件删除数据，用户只需输入id，则将原文件对应的这一条数据删除
# （删除后下面的id不变，比如此时你输入1，则将第一条数据删除，但是下面所有数据的id值不变及太白，nezha的 id不变）。
def del_employee(id):
flag = 0
import os
with open(r‘04 作业\employees‘, encoding=‘utf-8‘, mode=‘r‘) as file_handler_1, open(r‘04 作业\employees.bak‘, encoding=‘utf-8‘, mode=‘w‘) as file_handler_2:
for line in file_handler_1:
temp_line = line
temp_line.split(‘,‘)
if temp_line[0] != id:
flag = 1
file_handler_2.write(line)
os.remove(r‘04 作业\employees‘)
os.rename(r‘04 作业\employees.bak‘, r‘04 作业\employees‘)
if flag == 1:
return True
print(‘No Found!‘)
return False
def add_employee(employee):
with open(r‘04 作业\employees‘, encoding=‘utf-8‘, mode=‘r+‘) as file_handler:
for line in file_handler:
pass
line.split(‘,‘)
id = int(line[0]) + 1
file_handler.write(‘\n‘ + str(id))
for item in employee:
file_handler.write(‘,‘ + employee[item])
def employee_management():
for i in range(3):
mode = input(‘select ur mode!\ninput 1 to add a new employee\ninput 2 to del an employee\n‘)
if mode == ‘1‘:
employee = {}
employee.setdefault(‘name‘, input(‘name?\n‘))
employee.setdefault(‘age‘, input(‘age?\n‘))
employee.setdefault(‘phone‘, input(‘phone?\n‘))
employee.setdefault(‘job‘, input(‘job?\n‘))
add_employee(employee)
return True
elif mode == ‘2‘:
id = input(‘del id?\n‘)
if del_employee(id):
print(‘Del succeed.\n‘)
else:
print(‘Del failed.\n‘)
return True
else:
print(‘PLS INPUT CORRECT MODE NUM.\n‘)
return False
employee_management()
标签：二进制 文件删除 匿名函数 需要 address coding iter binary over
原文地址：https://www.cnblogs.com/raygor/p/13285660.html
