python数据类型之字符串补充

拼搏现实的明天。 2022-05-16 09:48 304阅读 0赞

## 1.如何随机生成验证码，快速生成内推码 ##

import random   
    import string   ###导入字符串模块
    code_str = string.ascii_letters + string.digits  ###大小写所有字母和所有数字
    print(code_str)
    def gen_code(len=4):      
        return "".join(random.sample(code_str, len))   ###返回的是一个从code_str随机选len个字符组成的字符串
    print([gen_code(len=6) for i in range(100)])   ###列表生成式，结果
    print({gen_code(len=6) for j in range(100)})    ###但当数据太大时，必定会重复，所以这里用集和可以避免

![这里写图片描述][70]

## 2.凯撒加密—加密算法 ##

在密码学中，恺撒密码是一种最简单且最广为人知的加密技术。它是一种替换加密的技术，  
明文中的所有字母都在字母表上向后（或向前）按照一个固定数目进行偏移后被替换成密文。

![这里写图片描述][70 1]

## 3.暴力破解 ##

思路:  
测试文本中是否存在至少两个最常见的英文单词， 如果有， 则代表破解成功.

import string
    def kaisacrypt(text='hello', k=3):
        lower = string.ascii_lowercase[k:] + string.ascii_lowercase[:k]  # 对原有小写字母向右移动k位
        upper = string.ascii_uppercase[k:] + string.ascii_uppercase[:k]
        table = str.maketrans(string.ascii_letters, lower+upper)    # 用于创建字符串映射的转换表'hello'
        return text.translate(table)      # 根据转换表去转换对应的字符
    print(kaisacrypt())
    
    def check(text):
    
        mostCommonWords = ('the', 'is', 'to', 'not', 'have', 'than', 'for', 'ok', 'and' )
        return  len([1 for word in mostCommonWords  if word in text])>2
    
    def bruteForce(text):
        for i in range(26):
            t = kaisacrypt(text, -i)
            if check(t):
                print(i)
                print(t)
                break
        text = "If not to the sun for smiling, warm is still" \
           " in the sun there, but wewill laugh more confident calm; " \
           "if turned to found his own shadow, appropriate escape, " \
           "the sun will be through the heart,warm each place behind the corner; " \
           "if an outstretched palm cannot fall butterfly, then clenched waving arms, " \
           "given power; if I can't have bright smile, it will face to the sunshine, " \
           "and sunshine smile together, in full bloom."
    cryptStr = kaisacrypt(text=text, k=18)
    print(cryptStr)
    
    bruteForce(cryptStr)

![这里写图片描述][70 2]  
4.基于缓冲区的生产-消费者模型

import random
    import time
    
    cacheList = []   # 缓冲区列表， 存放生产的所有包子，工作方式是队列的工作方式(FIFO-first in firest out).
    cacheListLen = 5   # 如果缓冲区列表的长度为5， 缓冲区就满了， 生产者不能再生产了.
    
    def isfull():
        return  len(cacheList) == 5   # 缓冲区满了
    
    def consumer(name):    #消费者
        print("%s准备买包子......." %(name))
        while True:
            kind  = yield
            print("%s购买%s包子成功..." %(name, kind))
    
    def producer(name):
        print("%s厨师正在生产包子......." %(name))
        kinds = ['A', 'B', 'C', 'D']
        while True:
            if  not isfull():
                time.sleep(random.random())   # 模拟生产数据耗费的时间
                kind = random.choice(kinds)
                print("%s已经生产好%s包子了..." %(name, kind))
                cacheList.append(kind)
            else:
                print("已经有足够的包子")
                yield
    
    p = producer("hello")
    (next(p))
    consumers = [consumer('user'+str(i)) for i in range(10)]
    for i in consumers:
        if not cacheList:
            print("目前没有包子")
        else:
            if not isfull():   # 如果缓冲区没有满， 就让生产者继续生产包子， 丛上次停止的地方继续执行.
                next(p)
            kind = cacheList.pop()  # 从缓冲区拿出包子给消费者.
            next(i)
            i.send(kind)

![这里写图片描述][70 3]

[70]: /images/20220516/42f6fcd80c1a47c5b5b9c0c34179ae23.png
[70 1]: /images/20220516/bf15ad3adc0848e7b970babe64643310.png
[70 2]: /images/20220516/fdbc0e37cec94b97b8c96f8596544d0f.png
[70 3]: /images/20220516/eab90df601da47bbb373f46a9a798412.png