python 基础习题

1、8<<2等于？

8  ---> 1000
32 ---> 100000
-----------结果---
32

2、通过内置函数计算5除以2的余数

print(dir())    #不带参数时，返回当前范围内的变量、方法和定义的类型列表，#找到__builtins__
print(dir(__builtins__))    #找内置函数

print(divmod(5,2)[1])
----------------------结果-----------
1　

3、s=[1,"h",2,"e",[1,2,3],"l",(4,5),"l",{1:"111"},"o"],将s中的5个字符提取出来并拼接成字符串。

方法一：

s1 = []
for n in s:
    if type(n) == str:
        s1.append(n)
print("".join(s1))

方法二：

print("".join([i for i in s if type(i) == str]))

4、判断"yuan"是否在[123,(1,"yuan"),{"yuan":"handsome"},"yuanhao"],如何判断以及对应结果？

s1 = [123,(1,"yuan"),{"yuan":"handsome"},"yuanhao"]

def foo(name):
    if "yuan" in name:
        print(name)
    for i in name:
        if type(i) == list or type(i) == tuple:
            foo(i)
        if type(i) == dict:
            foo(i.keys())
            foo(i.values())

foo(s1)

-----------结果为---------
(1, 'yuan')
dict_keys(['yuan'])

6、 a=[1,2,[3,"hello"],{"egon":"aigan"}]
b=a[:]

a[0]=5
a[2][0]=666

print(a)
print(b)
#计算结果以及为什么？

[5, 2, [666, 'hello'], {'egon': 'aigan'}]
[1, 2, [666, 'hello'], {'egon': 'aigan'}]

b相当于a的浅拷贝，当拷贝a中[3,"hello"]相当于只拷贝了一个内存地址，当劣列表里的元素改变时，b指向的内存地址并未发生改变，所以列表元素跟着一起改变

7 使用文件读取，找出文件中最长的行的长度（用一行代码解决）？

print(max([len(line) for line in open('file')]))

10 .通过函数化编程实现5的阶乘

方式一：

def func(n):
    if n == 1:
        return 1
    else:
        return n * func(n-1)

obj = func(3)
print(obj)

方式二：

from functools import reduce

def func(number):
    obj = reduce(lambda x,y:x*y,range(1,number + 1))
    return obj

print(func(4))

11 打印如下图案：

            ＊
           *＊*
          **＊**
         ***＊***
          **＊**
           *＊*
            *

def func(number):
    for i in range(1,number,2):
        print(("*" * i).center(number))
    for i in range(number,0,-2):
        print(("*" * i).center(number))

func(7)

12.

def outer():
    count = 10
    def inner():
        nonlocal count     #nonlocal  作用于外部作用域
        count = 20
        print(count)
    inner()
    print(count)
outer()

1.分析运行结果？
    20
    10
2.如何让两个打印都是20

    def outer():
        count = 10
        def inner():
            nonlocal count     #nonlocal  作用于外部作用域
            count = 20
            print(count)
        inner()
        print(count)
    outer()

　　

13 输入一个年份，判断是否是闰年？

def func(year):
    if (year%4 == 0 and year%100 != 0) or year%400 == 0:
        return True
    else:
        return False

print(func(2005))

judge  = lambda year: True if (year%4 == 0 and year%100 != 0) or (year%400 == 0) else False
print(judge(2004))　　

14 任意输入三个数，判断大小？

def func(a,b,c):
    if a >b:
        if a >c:
            print("%s最大"% a)
        else:
            print("%s最大" % c)
    else:
        if b >c:
            print("%s最大" % b)
        else:
            print("%s最大" % c)

func(1,2,3)

15 求s=a+aa+aaa+aaaa+aa...a的值，其中a是一个数字。例如2+22+222+2222+22222，几个数相加以及a的值由键盘控制。

def func(number,count):
    ret = 0
    if int(count) > 1:
        for i in range(1,int(count) + 1 ):
            ret += int((str(number) * i))
        return ret
    else:
        return number

obj = func(2,3)
print(obj)

16.

f=open("a")

while 1:
    choice=input("是否显示:[Y/N]:")
    if choice.upper()=="Y":
        for i in f:
            print(i)
    else:
        break

请问程序有无bug，怎么解决？

--------------------------------结果---------------
f=open("a")

while 1:
    choice=input("是否显示:[Y/N]:")
    if choice.upper()=="Y":
        for i in f:
            print(i)
        f.seek(0)
    else:
        break

f.close("a")

17
def foo():
　　print('hello foo')
　　return()
def bar():
　　print('hello bar')

（1）为这些基础函数加一个装饰器，执行对应函数内容后，将当前时间写入一个文件做一个日志记录。
（2）改成参数装饰器，即可以根据调用时传的参数决定是否记录时间，比如@logger(True)

import time

def init(func):
    def wrapper(*args,**kwargs):
        a= str(time.time()) + "执行%s\n" % func
        with open("record.txt","a+") as f:
            f.write(a)
        func(*args,**kwargs)
    return wrapper

@init
def foo():
    print('hello foo')
    return()

@init
def bar():
    print('hello bar')

foo()
bar()

import time

def auth(flag):
    def init(func):
        def wrapper(*args,**kwargs):
            if flag == True:
                a= str(time.strftime("%Y-%m-%d %H:%M:%S", time.localtime())) + "执行%s\n" % func
                with open("record.txt","a+") as f:
                    f.write(a)
            func(*args,**kwargs)
        return wrapper
    return init

@auth(True)
def foo():
    print('hello foo')
    return()

@auth(True)
def bar():
    print('hello bar')

foo()
bar()

18.

list3 = [
    {"name": "alex", "hobby": "抽烟"},
    {"name": "alex", "hobby": "喝酒"},
    {"name": "alex", "hobby": "烫头"},
    {"name": "alex", "hobby": "Massage"},
    {"name": "wusir", "hobby": "喊麦"},
    {"name": "wusir", "hobby": "街舞"},]

a  = [
    {"name":"alex","hobby":[1,2,3,4]},
    {"name":"wusir","hobby":[1,2,3,4]}
]

　　

19 三次登陆锁定：要求一个用户名密码输入密码错误次数超过三次锁定？

accounts = {}

def lock_user(name):
    with open("lock_user", mode="r+", encoding="utf8") as f_lock:
        for line in f_lock:
            if line.strip().split()[0] == name:
                print("Lock user")
                exit()

def lockd_user(**kwargs):
    with open("lock_user",mode="a+",encoding="utf8") as f_lockd_user:
        for key in kwargs:
            if kwargs[key] >2:
                 f_lockd_user.write(key + "\n")

def check_user(name,passwd):
    with open("user",mode="r",encoding="utf8") as f_check:
        for line in f_check:
            if name == line.strip().split()[0]:
                if passwd == line.strip().split()[1]:
                    print("Success")
                    exit()
                else:
                    add_error(name)
                    return name
        return name

def add_error(name):
    if accounts:
        if name in accounts:
            accounts[name] += 1
        else:
            accounts[name] = 1
    else:
        accounts[name] = 1

def main():
    count = 0
    while True:
        name = input("input name: ")
        passwd = input("input passwd: ")
        lock_user(name)    #判断用户是否锁定
        name = check_user(name,passwd)       #判断用户
        count += 1
        if count > 2:
            lockd_user(**accounts)
            print("exit than three")
            break

if __name__ == '__main__':
        main()