Carven 发表于 2021-1-18 20:25:36

python 20课练习,动动手2(请教大家怎么精简)

本帖最后由 Carven 于 2021-1-18 20:29 编辑

请教大家,第二个for 语句怎么精简?
还有想问下目前思路会不会有很大问题,里面有没有逻辑在一些别的应用上造成很大的问题?

谢谢解答,不胜感激。


xiaoxie = 'abcdefghijklmnopqrstuvwxyz'
daxie = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'

def find1(x):
    #temp为临时值,xx为转译为'1'和'0'后的字符
    temp = x.replace('\n','')
    xx = ''
    for each in list(temp):
      if each in xiaoxie:
            each = 0
      else:
            each = 1
      xx += str(each)
    # n为当前索引数,mima为密码,nember1/2为临时索引结果的和
    n = 0
    mima = ''
    for i in list(xx):
      if i == '0' and (n >= 4) and (n <= (len(xx)-3)):
            nember1 = int(xx) + int(xx) + int(xx) + int(xx) + int(xx) + int(xx)
            nember2 = int(xx) + int(xx)
            if (nember1 == 6) and (nember2 == 0):
                mima += temp
      n += 1      

    print('密码:',mima)
      

find1('''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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EthanHsiung 发表于 2021-1-20 21:21:13

第二个for循环里,i只用了一次,可以用索引n来代替
    mima = ''
    xx = list(xx)
    for n in range(len(xx)):
      ifxx == '0' and (n >= 4) and (n <= (len(xx)-3)):
            nember1 = int(xx) + int(xx) + int(xx) + int(xx) + int(xx) + int(xx)
            nember2 = int(xx) + int(xx)
            if (nember1 == 6) and (nember2 == 0):
                mima += temp   

    print('密码:',mima)

基本没什么问题

qq1151985918 发表于 2021-1-21 09:43:12

老板还是好好学习吧,当你学会正则表达式这个没那么烦了import re
getcode = re.findall(r"[^A-Z]{3}(){3}[^A-Z]",text)
code = "".join(x for x in getcode)
print(getcode,code)

Carven 发表于 2021-1-21 18:28:42

qq1151985918 发表于 2021-1-21 09:43
老板还是好好学习吧,当你学会正则表达式这个没那么烦了

明白,谢谢你!

Carven 发表于 2021-1-21 18:40:15

EthanHsiung 发表于 2021-1-20 21:21
第二个for循环里,i只用了一次,可以用索引n来代替




谢谢!懂了懂了,节省了一倍的运算(n += 1)
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