[已解决]020讲动动手第一题

fengchui

能不能帮忙看下我写出来的结果是：
密码是： ['u', 'i', 'l', 'o', 'v', 'e', 'f', 'i', 's', 'd', 'h', 'c']
多了447位的U和4578位的d，代码如下

1. string = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
   
2. lqNsYCyaQXBzrFUbkAUAWAKrDgDtAlGMBqWQhpEwquZqWZJpslUfMllCwWptqINjrOBTLuPzwvXNbLCx
   
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104. qKWGc'''
     
105. liststr= []
     
106. length = len(string)
     
107. for num0 in range(3,length-4):#从第三个开始索引
     
108.     numl = num0 - 3#密码左边三个数字
     
109.     numr = num0 + 4#密码右边三个数字
     
110.     if numr <= length -3:
     
111.         if (string[num0].islower()) and\
     
112.            (string[numl:num0].isupper()) and\
     
113.            (string[numl-1].islower()) and\
     
114.            (string[num0+1:numr].isupper()) and\
     
115.            (string[numr].islower()):
     
116.                #判断逻辑：密码本身是小写+密码左边三个数是大写+密码左边的左边是小写+密码右边三个数是大写+密码右边的右边是小写
     
117.             liststr.append(string[num0])
     
118. print('密码是：',''.join(liststr))
     
119.             
     

复制代码

最佳答案

月排行榜 / 总排行榜

yhhpf

2020-7-8 17:03:59

举个例子:

1. >>> s='''aaaa
   
2. bbbb
   
3. cccc'''
   
4. >>> s
   
5. 'aaaa\nbbbb\ncccc'
   
6. >>> len(s)
   
7. 14
   
8. >>> s[4]
   
9. '\n'
   
10. >>> s[4].isupper()
    
11. False
    
12. >>> s[4].islower()
    
13. False
    
14. >>> s[3:4].islower()
    
15. True

复制代码

你代码中包含了"\n"换行符，而且你使用到切片的方式去判断是否大小写，所以你说的那两个位置返回了True，导致你将其加入了结果；修改方法如下：

不知道为什么要审核，只发部分代码看下：

1. for num0 in range(3, length - 4):  # 从第三个开始索引
   
2.     count += 1
   
3.     numl = num0 - 3  # 密码左边三个数字
   
4.     numr = num0 + 4  # 密码右边三个数字
   
5.     if numr <= length - 3:
   
6.         if (string[num0].islower()) and \
   
7.                 (string[numl:num0].isupper()) and \
   
8.                 (string[numl - 1].islower()) and \
   
9.                 (string[num0 + 1:numr].isupper()) and \
   
10.                 (string[numr].islower()) and \
    
11.                 '\n' not in string[numl:numr]:     #加上这句判断是否存在\n换行符
    
12.             # 判断逻辑：密码本身是小写+密码左边三个数是大写+密码左边的左边是小写+密码右边三个数是大写+密码右边的右边是小写
    
13.             liststr.append(string[num0])
    
14. print('密码是：', ''.join(liststr))

复制代码

跳转到最佳答案楼层

yhhpf

举个例子:

1. >>> s='''aaaa
   
2. bbbb
   
3. cccc'''
   
4. >>> s
   
5. 'aaaa\nbbbb\ncccc'
   
6. >>> len(s)
   
7. 14
   
8. >>> s[4]
   
9. '\n'
   
10. >>> s[4].isupper()
    
11. False
    
12. >>> s[4].islower()
    
13. False
    
14. >>> s[3:4].islower()
    
15. True

复制代码

你代码中包含了"\n"换行符，而且你使用到切片的方式去判断是否大小写，所以你说的那两个位置返回了True，导致你将其加入了结果；修改方法如下：

不知道为什么要审核，只发部分代码看下：

1. for num0 in range(3, length - 4):  # 从第三个开始索引
   
2.     count += 1
   
3.     numl = num0 - 3  # 密码左边三个数字
   
4.     numr = num0 + 4  # 密码右边三个数字
   
5.     if numr <= length - 3:
   
6.         if (string[num0].islower()) and \
   
7.                 (string[numl:num0].isupper()) and \
   
8.                 (string[numl - 1].islower()) and \
   
9.                 (string[num0 + 1:numr].isupper()) and \
   
10.                 (string[numr].islower()) and \
    
11.                 '\n' not in string[numl:numr]:     #加上这句判断是否存在\n换行符
    
12.             # 判断逻辑：密码本身是小写+密码左边三个数是大写+密码左边的左边是小写+密码右边三个数是大写+密码右边的右边是小写
    
13.             liststr.append(string[num0])
    
14. print('密码是：', ''.join(liststr))

复制代码

fengchui

yhhpf 发表于 2020-7-8 17:03
举个例子:

你代码中包含了"\n"换行符，而且你使用到切片的方式去判断是否大小写，所以你说的那两个位置 ...

理解了，所以当“/n”在代码里的时候，占了一个字符位，但是实际上对整个字符的大小写判定都没有影响？

yhhpf

fengchui 发表于 2020-7-8 17:47
理解了，所以当“/n”在代码里的时候，占了一个字符位，但是实际上对整个字符的大小写判定都没有影响？

你看我的例子里面，如果单个去判断‘\n’是否大小写，返回的会是False，但是isupper和islower只会判断字符串中的【英文字母】是否为大小写，所以当你用字符串切片查找到类似“\nua”等字符串时，islower会返回True，isupper返回False

今天又起雾了

yhhpf 发表于 2020-7-8 17:03
举个例子:

你代码中包含了"\n"换行符，而且你使用到切片的方式去判断是否大小写，所以你说的那两个位置 ...

hi  鱼友   if numr <= length - 3  这句干嘛的呀
然后我看了一下代码  这句不对吧string[numl-1].islower()   假如numl = 0   0- 1就是string[-1]== 'c'.
小甲鱼老师给的最后一个字符是c，，如果是大写的，，，不就不对吗

yhhpf

今天又起雾了 发表于 2020-7-30 15:03
hi  鱼友   if numr

1， if numr <= length - 3 主要是判断有没有计算到最后3位，如果到了就没必要继续了，因为楼主的思路是一次判断前后的几位，已经包括后面的了。

2，嗯，这代码是在楼主代码基础上修改的，你说的BUG确实存在。

今天又起雾了

yhhpf 发表于 2020-7-30 15:29
1， if numr

谢谢鱼友