python3.9笔记1

str1 = 'Hello World!'
str2 = 'Hello Nowcoder!'
print(str1,str2,sep='\n')
输出： 
Hello World!
Hello Nowcoder!

print(r'\nmenkim is me') #\nmenkim is me
name = r'\nmenkim is me'
print(name) #\nmenkim is me
print('\nmenkim is me') #会先换行然后输出menkim is me

print('0.343'.isnumeric()) #False
print('1'.isnumeric()) #True

#实现输入整数，输出内容和并转化为int类型
num = input()
if num.isdigit():
    a = int(num)
    print(a,type(a),sep="\n")
else:
    print("非数字")

name = input()
print("I am %s and I am studying Python in Nowcoder!" %name)
# print(f"I am {name} and I am studying Python in Nowcoder!")
# print("I am {} and I am studying Python in Nowcoder!".format(name))

int() #输出0
int(23.6) #浮点数向下取整，23，默认10进制
int(23) #输入整数时，默认10进制
int('0xa',16) 
#如果有base输入，且x以字符串形式输入，
#代表x为base进制，则按照输入的base将x转换为10进制输出
输出：10
int('10',8) #将8进制转化为10进制
输出：8

#将十六进制整数ABCDEF，转换为十进制整数
dic = {'A':10,'B':11,'C':12,'D':13,'E':14,'F':15}
n = 0
m = 'ABCDEF'
l = len(m)
for i in m:
    n = n + dic[i]*16**(l-1)
    l = l - 1
print("%15d" % n)

n = int(input()) #输入120
m = ''
while n != 0:
    m += str(n%6) #将余数拼接
    n = n//6
print(m[::-1]) #将得到的数转制

str1 = input()
if str1.encode('utf8').isalpha(): #判断字符串是否均为字母
    print(len(str1))
else:
    print("请重新输入")

str1 = "hello world"
str1.title()
输出：'Hello World'

import re
def titlecase(s):
    return re.sub(r"[A-Za-z]+('[A-Za-z]+)?",
                  lambda mo: mo.group(0).capitalize(),
                  s)


titlecase("they're bill's friends.")
输出："They're Bill's Friends."

" hEllo world".capitalize() #输出 ' hello world'
"hEllo world".capitalize() #输出 'Hello world'
"hello world".capitalize() #输出 'Hello world'

'www.example.com'.strip('cmowz.')
输出：'example'

a = "Hello"
print(a[:4]) 
输出：'Hell'

c = '123'
print(c[2::-1]) #等同于c[::-1]
print(c[::-1])
输出：
321
321

sorted([5, 2, 3, 1, 4])
输出：[1, 2, 3, 4, 5]

a = [5, 2, 3, 1, 4]
a.sort()
a   #输出：[1, 2, 3, 4, 5]

a = [5, 2, 3, 1, 4]
a.reverse()
a   #输出：[4, 1, 3, 2, 5]

a=''.join(list(reversed('1234')))
print(a) 
输出：
4321

'1234'[::-1]  #输出'4321'

a = "{1,2,3,4}"
b = eval(a)
print(b,type(b))
输出：{1, 2, 3, 4} <class 'set'> #其他同理

a = 12
print(eval('a+1')) 
输出：13

a = 12
g = {'a':4}
print(eval('a+1',g))
输出：5

a = 12
b = 11
c = 10
g = {'a':4,'b':3}
l = {'b':2,'c':1}
print(eval('a+b+c',g,l))
输出：7

x,y = map(int,('2','3'))
print(x,y,type(x),type(y))
输出：2 3 <class 'int'> <class 'int'>

r = map(lambda x:x*2,(2,3))
list(r)
输出：[4, 6]

0 and 2 #0
1 and 2 #2
1 and 0 #0
0 or 2 #2
1 or 2 #1
not 1 #False
not 0 #True

dt = {"A":4,"B":3,"C":2,"D":1,"F":0}
list1 = []
list2 = []


while True:
    s = input()
    if s == "False":
        break
    else:
        s1 = int(input())
        list1.append(dt[s] * s1)
        list2.append(s1)


print('%.2f'%(sum(list1)/sum(list2)))

round(5.5) #6
round(4.5) #4
round(3.2,2) #3.2
round(6.675,2) #6.67
round(6.575,2) #6.58
#两者舍入不同的原因是由于大多数十进制小数实际上都不能以浮点数精确地表示。

from decimal import Decimal
a = Decimal('23.3456').quantize(Decimal('0.01'),rounding='ROUND_HALF_UP')
print(a)  #23.35
b = Decimal('7.325').quantize(Decimal('1.'), rounding='ROUND_HALF_UP')
print(b) #7

df = [[1,2,3],[4,5,6],[7,8,9]]
for x,y,z in df[0],df[1],df[2]:
    print(x,y,z)
输出：
1 2 3
4 5 6
7 8 9
#此处会把df[0],df[1],df[2]当作一个元组，
#然后将每一项的三个值分别赋给x,y,z

a = [1,2,3]
b = zip(a)
list(b)
输出：[(1,), (2,), (3,)]

a = [1,2,3]
b = [4,5,6,7]
c = zip(a,b)
# dict(c) #{1: 4, 2: 5, 3: 6}
# list(c)  #[(1, 4), (2, 5), (3, 6)]
# tuple(c) #((1, 4), (2, 5), (3, 6))
x,y = zip(*zip(a,b)) 
# list(x)  #[1, 2, 3]
list(y) #[4, 5, 6]

var = input()
var_dict = {}
for i in var:
    var_dict[i] = var.count(i)
print(var_dict)

num_list = list(map(int,input().split()))
print(max(num_list),min(num_list),sep='\n')

'%x'%234 ,'%#x'%234 ,'%#X'%234 ,'%X'%234 
#('ea', '0xea', '0XEA', 'EA')

print((12.23).hex()) #0x1.875c28f5c28f6p+3

 '%#o' % 10, '%o' % 10 #('0o12', '12')


format(10, '#o'), format(10, 'o') #('0o12', '12')


 f'{10:#o}', f'{10:o}' #('0o12', '12')

bin(3) #输出'0b11'

var = 'NiuNiu won the prize!'
print(var.find('the',9)) #11

s = input()
n = []
# while True:  #一种方法
#     if s == '0':
#         break
#     else:
#         n.append(s)
#         s = input()
# print(' '.join(n))
while s != '0': #第二种方法
    n.append(s)
    s = input()
print(' '.join(n))

class Employee():
    def __init__(self,name,salary):
        self.name = name
        self.salary = salary


    def printclass(self):
        print(f"{self.name}'salary is {self.salary}, and his age is {self.age}")


name = input()
salary = int(input())
age = int(input())
e = Employee(name,salary)
print(hasattr(e,'age')) 
#判断e对象是否有为age的属性，有返回True,否则False
setattr(e,'age',age) 
#为e对象添加名为age的属性及其对应的值
e.printclass()

seasons = ['Spring', 'Summer', 'Fall', 'Winter']
print(list(enumerate(seasons)))
print(list(enumerate(seasons,1)))
输出：
[(0, 'Spring'), (1, 'Summer'), (2, 'Fall'), (3, 'Winter')]
[(1, 'Spring'), (2, 'Summer'), (3, 'Fall'), (4, 'Winter')]


seasons = ['Spring', 'Summer', 'Fall', 'Winter']
for index,value in enumerate(seasons,1):#索引从1开始计数
    print(index,value)
输出：
1 Spring
2 Summer
3 Fall
4 Winter

if 条件语句1:
    语句块1 #条件为真时执行
elif 条件语句2:
    语句块2 #条件为真时执行
else：
    语句块3 #条件为假时执行

n = 0
while n < 5:
    n += 1
    if n == 3:
        continue
    print('I am wrong %d'%n)
else:
    print(n)
输出：
I am wrong 1
I am wrong 2
I am wrong 4
I am wrong 5
5

n = 0
while n < 5:
    n += 1
    if n == 2:
        continue
    if n == 4:
        break
    print('I am wrong %d'%n)
else:
    print(n)
输出：
I am wrong 1
I am wrong 3

for iter in iters:#接收可迭代对象作为参数，每次迭代其中的一个元素
    循环语句块
else:
    语句块


while 条件语句: 
    循环语句块   #条件为真时执行
else:
    语句块  #条件为假时执行

def rabbit_num(n):
    num_list = [2,3]
    if n <= 2:
        return num_list[n-1]
    else:
        return rabbit_num(n-1) + rabbit_num(n-2)


n = int(input())
print(rabbit_num(n))

class Student():
    def __init__(self,name,number,score,grade):
        self.name = name
        self.number = number
        self.score = int(score)
        self.grade = grade.split()


    def info(self):
        print(f"{self.name}'s student number is {self.number}, and his grade is {self.score}. He submitted {len(self.grade)} assignments, each with a grade of {' '.join(self.grade)}")


a = Student('menkim','12345','90','A B C')
a.info()

def func(foo, bar=None): ...

def func(posonly1, posonly2, , positional_or_keyword): ...

def func(arg, *, kw_only1, kw_only2): ...

def func(*args, **kwargs): ...

def func(foo,bar=None,*args,**kwargs): ...

def func(name,age,score=90): ...
func(3, 5, 80)
func(*(3, 5), 80) #*后面是可迭代对象，例如列表、元组、集合等
#实际上 * 操作符的作用是将可迭代对象解包出来传递参数
#下面的 ** 操作符也是同样的作用

def func(name,age): ...
func(name=3, age=5)
func(**{'name': 3, 'age': 5}) 
#**后面要为mapping
#例如dict,collections.defaultdict,collections.OrderedDict等

def f(a, b):
     print(a, b)


# f(b=1, *(2,)) # 2 1
# f(a=1, *(2,)) 
# TypeError: f() got multiple values for argument 'a'
f(1, *(2,)) # 1 2

def foo(name, **kwargs):
    return 'name' in kwargs


foo(2,**{'name':3}) 
#报错：TypeError: foo() got multiple values for argument 'name'

def foo(name,/ , **kwargs):
    print(name,kwargs['name'])
    return 'name' in kwargs


foo(2,**{'name':3}) #可以使用，不报错
输出：
2 3
True

def cheeseshop(kind, *arguments, **keywords):
    print("-- Do you have any", kind, "?")
    print("-- I'm sorry, we're all out of", kind)
    for arg in arguments:
        print(arg)
    print("-" * 40)
    for kw in keywords:
        print(kw, ":", keywords[kw])


cheeseshop("Limburger", "It's very runny, sir.",
           "It's really very, VERY runny, sir.",
           shopkeeper="Michael Palin",
           client="John Cleese",
           sketch="Cheese Shop Sketch")
print("\n") #两行 ，上下两种调用方法输出一致
cheeseshop(*("Limburger", "It's very runny, sir.",
           "It's really very, VERY runny, sir."),
           **{"shopkeeper":"Michael Palin",
           "client":"John Cleese",
           "sketch":"Cheese Shop Sketch"})

def foo(*args, **kwargs):
    a = args
    b = kwargs
    print(args,kwargs)
    print(a,b)
    for i in args:
        print(i,end=' ')
    for j in kwargs:
        print(j,end=' ')


#foo(*[2,3],**{'hah':4,'www':6}) #实际传入了四个参数，可以被使用
foo(2,3,'hah'=4,'www'=6) #还可以这样写
输出：
#args 被初始化为一个元组接收额外的位置参数
(2, 3) {'hah': 4, 'www': 6} 
(2, 3) {'hah': 4, 'www': 6}
2 3 hah www

def f(a, L=[]):
    L.append(a)
    return L


print(f(1))
print(f(2))
print(f(3))
输出：
[1]
[1, 2]
[1, 2, 3]

def f(a, L=None):
    if L is None:
        L = []
    L.append(a)
    return L


print(f(3))
print(f(4))
输出：
[3]
[4]

from datetime import datetime 
dt_o = datetime.fromtimestamp(1233456) 
print(dt_o)
print(dt_o.year,dt_o.month,dt_o.day,dt_o.hour, dt_o.minute, dt_o.second)
输出：
1970-01-15 14:37:36
1970 1 15 14 37 36

from datetime import datetime,time
b = datetime(2022,4,9,0,0,0)
a = timedelta(seconds=3661.34,minutes=10,weeks=1)
c = b + a 
print(c.day,c.hour,c.minute,c.second)
print(datetime.today())
print(time(microsecond=5))
a
输出：
16 1 11 1
2022-09-07 22:28:48.454689
00:00:00.000005
datetime.timedelta(days=7, seconds=4261, microseconds=340000)

from datetime import datetime,timezone
a = datetime.fromtimestamp(0) #本地时间
b = datetime.fromtimestamp(0,tz=timezone.utc) #UTC标准时间
print(a,b,sep='\n')
输出：
1970-01-01 08:00:00
1970-01-01 00:00:00+00:00

import math
a,b,c = map(int,input().split())
p = (a + b + c)/2
s = math.sqrt(p*(p-a)*(p-b)*(p-c)) #面积值
print('circumference=%.2f'%(p*2),'area=%.2f'%s,sep=' ')

#欧几里得的辗转相除法：两正整数a,b相除的余数r（a%b）如果为零，则除数b为
#两数的最大公约数，如果余数r不为零，则将b作为被除数,余数r作为除数，继续b%r得到
#余数，一直如此迭代下去，知道余数为0，则将此次的余数作为a,b的最大公约数
#a,b的最小公倍数为(a*b)/最大公约数
n,m = map(int,input().split())
a,b = n,m
r = a % b 
while r != 0:
    a,b = b,r
    r = a % b 
d = (n*m)//b #b为最大公约数，d为最小公倍数
print(b+d) #时间复杂度约为O(logN)

http_status = {'200':'OK','202':'Accepted','400':'Bad Request',
            '403':'Forbidden','404':'Not Found',
            '500':'Internal Server Error','502':'Bad Gateway'}
while True:
    try:
        n = input()
        print('%s'%http_status[n])
    except:
        break

while True:
    try:
        a,b,c = map(float,input().split())
        if a + b > c and a + c > b and b + c > a:
            if a == b == c:
                print('Equilateral triangle!')
            elif a == b or a == c or b == c:
                print('Isosceles triangle!')
            else:
                print('Ordinary triangle!')
        else:
            print('Not a triangle!')
    except:
        break

import calendar
print(calendar.monthrange(2022,8))
print(calendar._monthlen(2022,8))
输出：
(0, 31) #0指的是星期一，0-6分别指星期一到星期日
31

msg = input()
try:
    for i in msg:
        if i.isnumeric() or i == '.':
            pass
        else:
            a,b = msg.split(i)
            print('%.4f%s%.4f=%.4f'%(eval(a),i,eval(b),eval(msg)))
except SyntaxError as e:
    print('Invalid operation!')
except ZeroDivisionError as e:
    print('Wrong!Division by zero!')

n = int(input())
total = (1+n)*n//2
print(total)

n = 0
for i in range(100,1000):
    flag = 1 #做个标记
    for j in range(2,int(i**0.5)+1):
        if i%j == 0:
            flag = 0
            break
    if flag == 1:
        n += 1
print(n)

def prime_Number(n):
    ls = [i for i in range(2, n + 1)]
    for i in range(2, n + 1):
        if i not in ls:
            continue
        else:
            l = ls.index(i)
            # 按照堆栈的思想循环,不破坏原来的列表的索引
            for j in range(len(ls) - 1, l, -1):
                if ls[j] % i == 0:
                    # ls.remove(ls[j])
                    ls.pop(j)
    print(*ls)
    print(n - 1 - len(ls))




while True:
    try:
        n = int(input())
        prime_Number(n)
    except:
        break

'menkim'.rjust(8) #'  menkim'
'menkim'.ljust(8) #'menkim  '

'menkim'.center(8) #' menkim '

#靠左
'{:<6}'.format(12) #等效于'{:<{}}'.format(12,6)和f'{12:<6}'
输出：
'12    '
#靠右
'{:>6}'.format(12)
输出：
'    12'
#居中
'{:^6}'.format(12)
输出：
'  12  '

# 按照对称的思想来解决这个问题
# 用列表存储每一行，再循环居中打印出每一行即可得到超级圣诞树的树枝部分
# 枝干部分按照输入的值循环居中打印*即可
# 关键点在于找到空格的规律和最大行的宽度
n = int(input())
# 构造基本单元
ls = ["*", "* *", "* * *"]
# 定义一个函数用于往列表中插入数据
def Write(ls):
    length = len(ls) #因为列表是可变的，所以要将长度定下来
    for i in range(length):
        ls.append(ls[i] + " " * (2 * length - 1 - 2 * i) + ls[i])
    return ls




# 将树枝存入列表中
for j in range(1, n):
    Write(ls)
# 求出最大宽度
width = 6 * (2 ** (n - 1)) - 1
# 输出树枝
for i in range(len(ls)):
    print(ls[i].center(width))
# 输出树干
for _ in range(n):
    print("*".center(width))

#创建n行m列的矩阵
n,m = map(int,input().split())
ls = [[0 for _ in range(m)] for _ in range(n)]
for i in range(n):
    ls[i] = list(map(int,input().split()))
#找到x行y列的元素
x,y = map(int,input().split())
print(ls[x-1][y-1])