[已解决]大佬们，帮忙修改下代码

Newbie_XZW

代码在字符串下面，麻烦向下拖一下

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qKWGc'''

list1 = list(str1)
ler1 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
answer = []
def subgroup(sample,list):
    x = []
    for i in range(sample, len(list)):
        x.append(list[i - sample:i])
    return x

soup = subgroup(7,list1)

for each in soup:
    if each[0] in ler1:
        if each[1] in ler1:
            if each[2] in ler1:
                if each[4] in ler1:
                    if each[5] in ler1:
                        if each[6] in ler1:
                            if each[3] not in ler1:
                                if each[3] != '\n':
                                    print(each[3],end='')

大佬们，帮忙看看代码哪有问题

正确答案：ilovefishc

代码答案：llpuiallouovvhezfbilbjiirmsqqxthtgcnkdljoasncexw

最佳答案

月排行榜 / 总排行榜

太阳总会升起

2020-11-29 12:28:32

错误的地方：
判断不全

题目要求小写字母前三位和后三位有且仅有三个大写字母，也就是说小写字母的前面第四位和后面第四位是小写，所以要每9位分开组成大列表来判断前后第四位是否为小写

你的源代码应该是的字符串每7位分开组成一个大列表，然后判断第1， 2， 3， 5， 6， 7位是大写，第4位不是大写也不是换行就打印第三位

下面是你的源代码修改的

1. list1 = list(str1)
   
2. ler1 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
   
3. 
   
4. def subgroup(sample, list):
   
5.     x = []
   
6.     for i in range(sample, len(list)):
   
7.         x.append(list[i - sample:i])
   
8.     return x
   
9. 
   
10. 
    
11. soup = subgroup(9, list1)
    
12. 
    
13. for each in soup:
    
14.     if len(each) == 9:
    
15.         if each[0] not in ler1:
    
16.             if each[1] in ler1:
    
17.                 if each[2] in ler1:
    
18.                     if each[3] in ler1:
    
19.                         if each[5] in ler1:
    
20.                             if each[6] in ler1:
    
21.                                 if each[7] in ler1:
    
22.                                     if each[8] not in ler1:
    
23.                                         if each[4] not in ler1:
    
24.                                             if each[4] != '\n':
    
25.                                                 print(each[4], end='')

复制代码

因为我没看懂你的for循环那里，所以下面是我按照你的思路写的函数和调用

1. def choice(list):
   
2.     x = []
   
3.     length = len(list1)
   
4.     for i in range(length):
   
5.         x.append(list[i:i+9])
   
6.     return x
   
7. 
   
8. soup = choice(list1)

复制代码

跳转到最佳答案楼层

Newbie_XZW

帅哥美女们，快来帮忙看一下

suchocolate

题目是啥？

Newbie_XZW

suchocolate 发表于 2020-11-28 18:18
题目是啥？

第020讲：函数：内嵌函数和闭包 | 课后测试题及答案

动动手 最后一题

题目：请用已学过的知识编写程序，找出小甲鱼藏在下边这个长字符串中的密码，密码的埋藏点符合以下规律：

                    a) 每位密码为单个小写字母
                    b) 每位密码的左右两边均有且只有三个大写字母

suchocolate

1. import re
   
2. result = re.findall(r'[a-z][A-Z]{3}([a-z])[A-Z]{3}[a-z]', str1)
   
3. print(result)

复制代码

昨非

suchocolate 发表于 2020-11-28 21:05

啊这，人家问你拧螺丝，你这直接造火箭啊

太阳总会升起

你可以说一下你的思路吗，for循环里面看不太懂
enmmmm

太阳总会升起

错误的地方：
判断不全

题目要求小写字母前三位和后三位有且仅有三个大写字母，也就是说小写字母的前面第四位和后面第四位是小写，所以要每9位分开组成大列表来判断前后第四位是否为小写

你的源代码应该是的字符串每7位分开组成一个大列表，然后判断第1， 2， 3， 5， 6， 7位是大写，第4位不是大写也不是换行就打印第三位

下面是你的源代码修改的

1. list1 = list(str1)
   
2. ler1 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
   
3. 
   
4. def subgroup(sample, list):
   
5.     x = []
   
6.     for i in range(sample, len(list)):
   
7.         x.append(list[i - sample:i])
   
8.     return x
   
9. 
   
10. 
    
11. soup = subgroup(9, list1)
    
12. 
    
13. for each in soup:
    
14.     if len(each) == 9:
    
15.         if each[0] not in ler1:
    
16.             if each[1] in ler1:
    
17.                 if each[2] in ler1:
    
18.                     if each[3] in ler1:
    
19.                         if each[5] in ler1:
    
20.                             if each[6] in ler1:
    
21.                                 if each[7] in ler1:
    
22.                                     if each[8] not in ler1:
    
23.                                         if each[4] not in ler1:
    
24.                                             if each[4] != '\n':
    
25.                                                 print(each[4], end='')

复制代码

因为我没看懂你的for循环那里，所以下面是我按照你的思路写的函数和调用

1. def choice(list):
   
2.     x = []
   
3.     length = len(list1)
   
4.     for i in range(length):
   
5.         x.append(list[i:i+9])
   
6.     return x
   
7. 
   
8. soup = choice(list1)

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lijiachen

昨非 发表于 2020-11-29 00:44
啊这，人家问你拧螺丝，你这直接造火箭啊

正则有点秀

python_zbl