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python 小项目练习

python 小程序

标签 ：python

1 Interweaving strings and removing digits

Your friend Rick is trying to send you a message, but he is concerned that it would get intercepted by his partner. He came up with a solution:

  1. Add digits in random places within the message.
  2. Split the resulting message in two. He wrote down every second character on one page, and the remaining ones on another. He then dispatched the two messages separately. Write a function interweave(s1, s2) that reverses this operation to decode his message! Example 1: interweave("hlo", "el") -> "hello" Example 2: interweave("h3lo", "el4") -> "hello" Rick's a bit peculiar about his formats. He would feel ashamed if he found out his message led to extra white spaces hanging around the edges of his message...
  • code
def interweave(s1,s2):
    s_1=""
    s_2=""
    out=""
    n=0
    for i in s1:
        if i not in '1234567890':
           t=''.join(i)
           s_1=s_1+t
    n1=len(s_1)
    for j in s2:
        if j not in '1234567890':
           p=''.join(j)
           s_2=s_2+p
    n2=len(s_2)
    #判断长度 但是 有待改进 ： 如果位数差的比较多 while
    if n1==n2:
       pass
    elif n1<n2:
       s_1=s_1+" "
    else:
       s_2=s_2+" "
    
    for i in s_1:
        out = out +i +s_2[n]
        n+=1
    out=out.strip()
    return out

2 count words

from collections import Counter

words = [
    'look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes',
    'the', 'eyes', 'the', 'eyes', 'the', 'eyes', 'not', 'around', 'the',
    'eyes', "don't", 'look', 'around', 'the', 'eyes', 'look', 'into',
    'my', 'eyes', "you're", 'under'
]


print (Counter(words))
Counter({'eyes': 8, 'the': 5, 'look': 4, 'into': 3, 'my': 3, 'around': 2, 'not': 1, "don't": 1, "you're": 1, 'under': 1})
print (Counter(words).most_common(4))
[('eyes', 8), ('the', 5), ('look', 4), ('into', 3)]

3 判断是否完美整数（记录几个大神的）

import math

def is_square(n):
    if n < 0:
        return False
    return (int(math.sqrt(n)) - math.sqrt(n)) == 0
import math
def is_square(n):
    return n > -1 and math.sqrt(n) % 1 == 0

def is_square(n): 
    i=1
    while n>0:
         n=n-i
         i=i+2
  #平方拆成等比数列 
    return n==0

4 translate 字符转换 删除

intab = "aeiou"
outtab = "12345"
trantab = str.maketrans(intab, outtab)   # 制作翻译表
 
str = "this is string example....wow!!!"
print (str.translate(trantab))
def disemvowel(string):
    return string.translate(None, 'aeiouAEIOU')

5 循环询问票价

"""
#循环性 咨询票价  直到输入错误数据   可以选择继续 或者 quit
无线循环
输入quit 直接退出程序
###如何控制再开始一段循环 solved
"""


import time

c=0
while c< 10:
    b=True
    while b:
        d=input("Enter your age :  ")#判断输入类型
        if d.isdigit():         #是不是全部是整数
            a=int(d)
            
            if a <=3 :
                print("It's free for you!")
                continue
            elif a>3 and a<12 :
                print("It costs 3$,please!")
                continue
            elif a>=12 and a< 100:
                print("It costs 6$, please!")
                continue
            else:
                b=False  # 超出售票年龄范围
                c+=1
                print('\nError!!!(out of range). Wait for 5s')
                time.sleep(5) # 暂停5秒
        
        elif d!='quit':    #输入有字符 不是quit  报错 继续运行
            print('\nError!!(Enter integer).   Wait for 3s')
            time.sleep(3)
            break
        else:
            break
    
    if d=='quit': #输入quit 退出程序
        print("It is quiting")
        c=1002
    else:
        prompt="\nif you wnat to quit please enter quit:" 
        prompt+="\nif you want to  continue please enter any  key:    "
        e=input(prompt)
        if e =='quit':   #输入有 quit  即时退出程序
            print("It is quiting")
            c=1001

6 Homework: A byte of python

Checking whether a text is a palindrome should also ignore punctuation, spaces and case. For example, "Rise to vote, sir." is also a palindrome but our current program doesn't say it is. Can you improve the above program to recognize this palindrome? If you need a hint, the idea is that...1

Use a tuple (you can find a list of all punctuation marks here) to hold all the forbidden characters, then use the membership test to determine whether a character should be removed or not, i.e. forbidden = (!, ?, ., ...).

7 心形图

grid=[['.','.','.','.','.','.','.'],
      ['.','*','*','.','.','.','.'],
      ['*','*','*','*','.','.','.'],
      ['*','*','*','*','*','.','.'],
      ['.','*','*','*','*','*','.'],
      ['*','*','*','*','*','.','.'],
      ['*','*','*','*','.','.','.'],
      ['.','*','*','.','.','.','.'],
      ['.','.','.','.','.','.','.']]
i=0
while i<7:
    for j in range(0,9):
        print(grid[j][i],end='')
    print('\n')##换行
    i+=1

8 tic tac toe

AI_tictactoe

###建立棋盘空格位置---棋子 现所有位置都是空格
the_board={'1-1':' ','1-2':' ','1-3':' ',
           '2-1':' ','2-2':' ','2-3': '',
           '3-1':' ','3-2':' ','3-3':' ',
           }

###打印棋盘
def print_board(board):
    ###参数为字典
    print(board['1-1']+'|'+board['1-2']+'|'+board['1-3'])
    print('-+-+-')
    print(board['2-1']+'|'+board['2-2']+'|'+board['2-3'])
    print('-+-+-')
    print(board['3-1']+'|'+board['3-2']+'|'+board['3-3'])


### 允许玩家修改棋子
turn='X'
for i in range(9):
    print_board(the_board)
    print('Turn for '+turn)
    move=input(" Move on which space?: ")
    the_board[move]=turn
    if turn == "X":
        turn='O'
    else:
        turn='X'

print_board(the_board)

9 列表物品按数量添加到字典

##展示 物品数目和分类
def display_stuff(stu):
    for k , v in stu.items():
        print(k + ': '+ str(v))
        total_number=0
        total_number+=v
    print("\ntotal_number: " +str(total_number))
    
### 添加物品----添加字典
def add_stuff(st,add_stu):
    for i in add_stu:
        st.setdefault(i,0) ###防止字典里面不含该物品
        st[i]=st[i]+1
    return st      ###注意函数需要返回值
stuff={'rope':1,'arrow':3,'torch':4,'coin':32,'gold':12,'cloth':12}
addstuff=['rope','rope','rope','gold','fish']
stuff_total=add_stuff(stuff,addstuff)
display_stuff(stuff_total)

10 寻找文本中的 电话号码 email

# -*- coding: utf-8 -*-
"""
Created on Tue May 22 16:11:33 2018

@author: 刘闯

提取 电话号码和 Email地址


1.从剪切板获取文本：
  1-1pyperclip
2.找出其中的 电话号码 和Email 地址：
   2-1 两个正则表达式
   2-2 找出所有的匹配
   2-3 放入字符串
   2-4 如果没有匹配 ；；；
   
3. 粘贴到剪切板

"""

import  pyperclip, re
text=str(pyperclip.paste())

# phone 010-8888-8888
phonenumber_regex=re.compile(r'''(
        (\d{4}|\(\d{4}\))?  ##区号  0101  or (0101) 且不一定必须
        (\s|-|\.)?    ##连接符 0101-1234 or 0101.1234 o1o1 1234 且不一定必须
        (\d{4})
        (\s|-|\.)?  
        (\d{4})
        )''',re.VERBOSE)


## email  ex: 1659608216@qq.com
email_regex=re.compile(r'''(
        [a-zA-Z0-9._%+-]+    # username
        @         
        [a-zA-z0-9-]+        #domin name
        (\.[a-zA-Z]{2,})+  #.cn or.com  (.edu.cn)
        )''',re.VERBOSE)

###findall 抓取信息 tuple--list
matches=[]
for groups in phonenumber_regex.findall(text):
    phonenumber='-'.join([groups[1],groups[3],groups[5]])
    matches.append(phonenumber)
for groups in email_regex.findall(text):
    matches.append(groups[0])


####copy to clipboard
if len(matches)>0:
    pyperclip.copy('\n'.join(matches))
    print("Copied to clipboard:")
    print('\n'.join(matches))
else:
    print('No number or email found')

###问题  电话号码和 手机号码 冲突显示 需要个判断（开头）
#### 校园邮箱 识别 缺失   solved
#### 随机8位数字 都添加识别   错误

#### 识别 3位  是数字还是电话号 

###类似程序：
 1. 寻找网站的URL https://开头
 
 2. 寻找整理日期格式
 
 3. 删除敏感信息 sub()
 
 4. 寻找常见的英文打字错误  如：单词间多个空格 重复单词 多个标点符号

11 记录时间秒表

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sun Jul 15 00:53:43 2018
@author: liuchuang
超级秒表 项目
1.记录从按下回车键开始，每次按键的时间，每次按键都是一个新的“单圈”。 
2.打印圈数、总时间和单圈时间。
这意味着代码将需要完成以下任务:
1.在程序开始时，通过调用 time.time()得到当前时间，将它保存为一个时间戳。 在每个单圈开始时也一样。
2.记录圈数，每次用户按下回车键时加 1。 用时间戳相减，得到计算流逝的时间。
3.处理 KeyboardInterrupt 异常，这样用户可以按 Ctrl-C 退出。
4.打开一个新的文件编辑器窗口，并保存为 stopwatch.py。
创建一个简单的工时表应用程序，当输入一个人的名字时，用当前的时间记录 下他们进入或离开的时间
"""

import time 


# Display the program's instructions.
print('Press ENTER to begin. Afterwards, press ENTER to "click" the stopwatch. Press Ctrl-C to quit.')

input()
print("Start!")

startTime = time.time()
lastTime = startTime
lapNum =1 


try:
    while True:
        input()
        lapTime =round(time.time()- lastTime,4)
        totalTime = round(time.time()-startTime,4)
        print("Lap %s : total time : %s  lap time: %s" %(lapNum,totalTime,lapTime),end="")
        lapNum+=1
        lastTime = time.time()
        
except KeyboardInterrupt:
    print("\n done")