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[已解决]代码不报错不给结果是什么情况?

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发表于 2020-5-12 23:56:01 | 显示全部楼层 |阅读模式
10鱼币
Python课后作业代码,写好后运行不给结果,也不报错。请问是什么情况?
运行结果如图,代码如下
  1. def find1(str1):
  2.     a = [] #建立空列表储存转化为大小写判断的flag
  3.     b = '' #建议空字符串以便拼接密码
  4.     for each in str1:
  5.         if each.islower() == 1: #判断是否为小写
  6.             flag = 1 #将小写标记为1
  7.             a.append(flag) #将flag储存到列表a中
  8.         else:
  9.             flag = 0 #将大写标记为0
  10.             a.append(flag) #将flag储存到列表a中
  11.     i = 3 #按照题目要求,从第4个元素开始判断
  12.     while i <= len(a)-3: #在倒数第4个元素处停止
  13.         if i == 3 or i == (len(a)-3): #在第4个和倒数第4个处判断[0,0,0,1,0,0,0]的组合
  14.             if a[i] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0:
  15.                 b = b + str1[i]
  16.         else: #在其他位置上的元素判断[1,0,0,0,1,0,0,0,1]的组合
  17.             if a[i] == 1 and a[i-4] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] == 1:
  18.                 b = b + str1[i]
  19.     print('密码是' + c)
  20.    
  21. x = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
  22. lqNsYCyaQXBzrFUbkAUAWAKrDgDtAlGMBqWQhpEwquZqWZJpslUfMllCwWptqINjrOBTLuPzwvXNbLCx
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  124. qKWGc'''
  125. find1(x)
复制代码
最佳答案
2020-5-12 23:56:02
比较新奇的算法了,就是效率有点低,不过对于现在的电脑来说不算什么,先说两个显性错误:
1.while循环中i值没有变化,死循环了,加一条i+=1或者换成for循环;
2.变量c没有,是变量b。
然后是算法bug:
1.没有考虑非字母,这里主要是'\n';
2.判断第四个字符和倒数第四个字符不是简单看前三后三,而是第四个是前三后四,倒数第四个是前四后三。
修正了代码,可以跑起来了,还有可以优化的地方再慢慢想吧:
  1. def find1(str1):
  2.     a = [] #建立空列表储存转化为大小写判断的flag
  3.     b = '' #建议空字符串以便拼接密码
  4.     for each in str1:
  5.         if each.islower() == 1: #判断是否为小写
  6.             flag = 1 #将小写标记为1
  7.             a.append(flag) #将flag储存到列表a中
  8.         elif each.isupper() == 1:
  9.             flag = 0 #将大写标记为0
  10.             a.append(flag) #将flag储存到列表a中
  11.         else:
  12.             flag = 2 #非字母标记为2
  13.             a.append(flag) #将flag储存到列表a中
  14.         
  15.     i = 3 #按照题目要求,从第4个元素开始判断
  16.     while i <= len(a)-3: #在倒数第4个元素处停止
  17.         if i == 3 : #在第4个处判断[0,0,0,1,0,0,0,非0]的组合
  18.             if a[i] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] != 0:
  19.                 b = b + str1[i]
  20.         if i == (len(a)-3): #在倒数第4个处判断[非0,0,0,0,1,0,0,0]的组合
  21.             if a[i] == 1 and a[i-4] != 0 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0:
  22.                 b = b + str1[i]
  23.         else: #在其他位置上的元素判断[非0,0,0,0,1,0,0,0,非0]的组合
  24.             if a[i] == 1 and a[i-4] != 0 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] != 0:
  25.                 b = b + str1[i]
  26.         i+=1
  27.     print('密码是' + b)
复制代码
1.png

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比较新奇的算法了,就是效率有点低,不过对于现在的电脑来说不算什么,先说两个显性错误: 1.while循环中i值没有变化,死循环了,加一条i+=1或者换成for循环; 2.变量c没有,是变量b。 然后是算法bug: 1.没有考虑非字母,这里主要是'\n'; 2.判断第四个字符和倒数第四个字符不是简单看前三后三,而是第四个是前三后四,倒数第四个是前四后三。 修正了代码,可以跑起来了,还有可以优化的地方再慢慢想吧:
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发表于 2020-5-12 23:56:02 | 显示全部楼层    本楼为最佳答案   
比较新奇的算法了,就是效率有点低,不过对于现在的电脑来说不算什么,先说两个显性错误:
1.while循环中i值没有变化,死循环了,加一条i+=1或者换成for循环;
2.变量c没有,是变量b。
然后是算法bug:
1.没有考虑非字母,这里主要是'\n';
2.判断第四个字符和倒数第四个字符不是简单看前三后三,而是第四个是前三后四,倒数第四个是前四后三。
修正了代码,可以跑起来了,还有可以优化的地方再慢慢想吧:
  1. def find1(str1):
  2.     a = [] #建立空列表储存转化为大小写判断的flag
  3.     b = '' #建议空字符串以便拼接密码
  4.     for each in str1:
  5.         if each.islower() == 1: #判断是否为小写
  6.             flag = 1 #将小写标记为1
  7.             a.append(flag) #将flag储存到列表a中
  8.         elif each.isupper() == 1:
  9.             flag = 0 #将大写标记为0
  10.             a.append(flag) #将flag储存到列表a中
  11.         else:
  12.             flag = 2 #非字母标记为2
  13.             a.append(flag) #将flag储存到列表a中
  14.         
  15.     i = 3 #按照题目要求,从第4个元素开始判断
  16.     while i <= len(a)-3: #在倒数第4个元素处停止
  17.         if i == 3 : #在第4个处判断[0,0,0,1,0,0,0,非0]的组合
  18.             if a[i] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] != 0:
  19.                 b = b + str1[i]
  20.         if i == (len(a)-3): #在倒数第4个处判断[非0,0,0,0,1,0,0,0]的组合
  21.             if a[i] == 1 and a[i-4] != 0 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0:
  22.                 b = b + str1[i]
  23.         else: #在其他位置上的元素判断[非0,0,0,0,1,0,0,0,非0]的组合
  24.             if a[i] == 1 and a[i-4] != 0 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] != 0:
  25.                 b = b + str1[i]
  26.         i+=1
  27.     print('密码是' + b)
复制代码
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发表于 2020-5-13 07:21:21 | 显示全部楼层
还在算吧
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发表于 2020-5-13 07:37:47 | 显示全部楼层
本帖最后由 heidern0612 于 2020-5-13 07:39 编辑

大致看了下,最后while的时候,i的变化呢?

i什么时候不满足while呢?

没退出的话,应该一直就循环,卡在循环里了吧?

你可以在if分支下面建个print,打印一句话。然后运行下,你会发现一直在打印这句话。

也就是说,你i没有小于你while条件的时候,自然就一直循环了。

你可以运行我这段代码试试,我就加了俩print,别的代码没动你的。

  1. def find1(str1):
  2.     a = [] #建立空列表储存转化为大小写判断的flag
  3.     b = '' #建议空字符串以便拼接密码
  4.     for each in str1:
  5.         if each.islower() == 1: #判断是否为小写
  6.             flag = 1 #将小写标记为1
  7.             a.append(flag) #将flag储存到列表a中
  8.         else:
  9.             flag = 0 #将大写标记为0
  10.             a.append(flag) #将flag储存到列表a中
  11.     i = 3 #按照题目要求,从第4个元素开始判断
  12.     while i <= len(a)-3: #在倒数第4个元素处停止
  13.         if i == 3 or i == (len(a)-3): #在第4个和倒数第4个处判断[0,0,0,1,0,0,0]的组合
  14.             if a[i] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0:
  15.                 b = b + str1[i]
  16.                 print("卡在if")
  17.         else: #在其他位置上的元素判断[1,0,0,0,1,0,0,0,1]的组合
  18.             if a[i] == 1 and a[i-4] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] == 1:
  19.                 b = b + str1[i]
  20.                 print("卡在else")
  21.     print('密码是' + c)
  22.    
  23. x = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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  120. xXhattDrfBGbZRveKbgjzxJLdYREQlMeLtcIEUoyJocdAfUbymxuFLVjGkOQniiPParqoyQYfDYAQTkM
  121. WLicLxpEFkBbwlKrTyYilKTtYkpVGxtOjYmcBDOrwFhFiGutmpTyTarUbVUeSevBTdaPDpjRkaEmLJMg
  122. WsMhSGfIcBChcqrRKgKpjvGnFipjswgjetRtniMagakbCXAjpzWTtMlgZGCJwGyglpcLebrKWhgwJfWV
  123. qGifWNEpCtjuejHoyVCdIxzMYGnfoslgTNAJdtVBWDVoGLzHSAVBTnhNIvAOExQNiJOIPPiHkdaRbfaP
  124. ixDDoCDOOeAqvQJFxLWDICfGmufyxmaMshbvcrtjqqVtffZTnbtCOQfzRMGwOQEKaAmSWjnYdNgvdkmd
  125. dQmaKZSdqKNrnvJlcyVMKuNWmuoOeyKecgjXbmSqnpjwJEaDYoehEklEgJyiksGxdEKgfYRXQecRZgfe
  126. qKWGc'''
  127. find1(x)
复制代码


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发表于 2020-5-13 07:40:02 | 显示全部楼层
对的,没有出现下一个“>>>”应该是还在算
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发表于 2020-5-13 09:32:05 | 显示全部楼层
刷个存在感
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 楼主| 发表于 2020-5-13 10:57:14 | 显示全部楼层
heidern0612 发表于 2020-5-13 07:37
大致看了下,最后while的时候,i的变化呢?

i什么时候不满足while呢?

感谢回复,确实卡在了if的位置。我都不知道啥时候不小心把i++给删掉了。我还一直没注意到。
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 楼主| 发表于 2020-5-13 11:00:04 | 显示全部楼层
txxcat 发表于 2020-5-12 23:56
比较新奇的算法了,就是效率有点低,不过对于现在的电脑来说不算什么,先说两个显性错误:
1.while循环中i ...

感谢老哥,前面写了i += 1后面不小心删掉没注意到。最后的b和c也是后面整理代码出现的错误。算法bug也指出来了。太棒了!
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