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[已解决]python 20课练习,动动手2(请教大家怎么精简)

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发表于 2021-1-18 20:25:36 | 显示全部楼层 |阅读模式

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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。


  1. xiaoxie = 'abcdefghijklmnopqrstuvwxyz'
  2. daxie = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'

  3. def find1(x):
  4.     #temp为临时值,xx为转译为'1'和'0'后的字符
  5.     temp = x.replace('\n','')
  6.     xx = ''
  7.     for each in list(temp):
  8.         if each in xiaoxie:
  9.             each = 0
  10.         else:
  11.             each = 1
  12.         xx += str(each)
  13.     # n为当前索引数,mima为密码,nember1/2为临时索引结果的和
  14.     n = 0
  15.     mima = ''
  16.     for i in list(xx):
  17.         if i == '0' and (n >= 4) and (n <= (len(xx)-3)):
  18.             nember1 = int(xx[n-3]) + int(xx[n-2]) + int(xx[n-1]) + int(xx[n+1]) + int(xx[n+2]) + int(xx[n+3])
  19.             nember2 = int(xx[n-4]) + int(xx[n+4])
  20.             if (nember1 == 6) and (nember2 == 0):
  21.                 mima += temp[n]
  22.         n += 1      

  23.     print('密码:',mima)
  24.         

  25. find1('''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
  26. lqNsYCyaQXBzrFUbkAUAWAKrDgDtAlGMBqWQhpEwquZqWZJpslUfMllCwWptqINjrOBTLuPzwvXNbLCx
  27. oFRritKRpJgBOaGPZdkUzvYnvYmAlEsVmKRXqyQUOdCBqLYyboOYeAQNLnkuiDXCiNiksSSRpDMVQQgs
  28. TmYThnppfKSmkpjjceGLaMOvYgsvNkGENKgGtUSzVPlLMeephDKrWGNpAxBqRiCnQIuKwDZurIRsznUp
  29. xhstMWpHdZzqeEnttlAHiWbhbIJpwKBAGsFWthBiwBDKTFzIUamELZJFLbmqOmwBIYBJIofAUyxTLUEm
  30. LcOXzjIHFcfUfEWccbWzhVuIZMdTVsOaNsLTpHjrvFaYTnJUrqaYnUmbobfOXXtkjKjFUEhVNlPWeaIY
  31. uABNuEDKFWiUaqQiBHGsgEmDSKiJKLkkzFWlobXIyUlDzGcwDnEOwwWxgFpebsslmKItazIYctPROSJp
  32. ImDHusQIkoFQSmuqwdcPNERvflfHtkGTrtbQXqZUsazoxmbVTPIuuxKvfALAaGnIuIZwczCcoIRMncBs
  33. timFGHPJPaHMPTLjsBjIwbbzKyghaDTUStzJceaYAneywwpvdiUJifPHtXmAwZPChWsDmwonzeeQCEiB
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  128. qKWGc''')

  129.         


复制代码
最佳答案
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
  1. import re
  2. getcode = re.findall(r"[^A-Z][A-Z]{3}([a-z])[A-Z]{3}[^A-Z]",text)
  3. code = "".join(x for x in getcode)
  4. print(getcode,code)
复制代码
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发表于 2021-1-20 21:21:13 | 显示全部楼层
第二个for循环里,i只用了一次,可以用索引n来代替
  1.     mima = ''
  2.     xx = list(xx)
  3.     for n in range(len(xx)):
  4.         if  xx[n] == '0' and (n >= 4) and (n <= (len(xx)-3)):
  5.             nember1 = int(xx[n-3]) + int(xx[n-2]) + int(xx[n-1]) + int(xx[n+1]) + int(xx[n+2]) + int(xx[n+3])
  6.             nember2 = int(xx[n-4]) + int(xx[n+4])
  7.             if (nember1 == 6) and (nember2 == 0):
  8.                 mima += temp[n]     

  9.     print('密码:',mima)
复制代码


基本没什么问题
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发表于 2021-1-21 09:43:12 | 显示全部楼层    本楼为最佳答案   
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
  1. import re
  2. getcode = re.findall(r"[^A-Z][A-Z]{3}([a-z])[A-Z]{3}[^A-Z]",text)
  3. code = "".join(x for x in getcode)
  4. print(getcode,code)
复制代码
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 楼主| 发表于 2021-1-21 18:28:42 | 显示全部楼层
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!
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 楼主| 发表于 2021-1-21 18:40:15 | 显示全部楼层
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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