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
" % 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
" % 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 + "
")
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()