[已解决]大佬们，帮忙修改下代码

Newbie_XZW · 发表于 2020-11-27 17:03:23

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代码在字符串下面，麻烦向下拖一下

str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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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

list1 = list(str1)
ler1 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'

def subgroup(sample, list):
    x = []
    for i in range(sample, len(list)):
        x.append(list[i - sample:i])
    return x


soup = subgroup(9, list1)

for each in soup:
    if len(each) == 9:
        if each[0] not in ler1:
            if each[1] in ler1:
                if each[2] in ler1:
                    if each[3] in ler1:
                        if each[5] in ler1:
                            if each[6] in ler1:
                                if each[7] in ler1:
                                    if each[8] not in ler1:
                                        if each[4] not in ler1:
                                            if each[4] != '\n':
                                                print(each[4], end='')

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

def choice(list):
    x = []
    length = len(list1)
    for i in range(length):
        x.append(list[i:i+9])
    return x

soup = choice(list1)

跳转到最佳答案楼层

Newbie_XZW · 发表于 2020-11-28 17:30:57

帅哥美女们，快来帮忙看一下

suchocolate · 发表于 2020-11-28 18:18:18

题目是啥？

Newbie_XZW · 发表于 2020-11-28 20:50:24

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 · 发表于 2020-11-28 21:05:02

本帖最后由 suchocolate 于 2020-11-28 22:04 编辑

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

昨非 · 发表于 2020-11-29 00:44:43

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 · 发表于 2020-11-29 11:47:53

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 · 发表于 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

list1 = list(str1)
ler1 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'

def subgroup(sample, list):
    x = []
    for i in range(sample, len(list)):
        x.append(list[i - sample:i])
    return x


soup = subgroup(9, list1)

for each in soup:
    if len(each) == 9:
        if each[0] not in ler1:
            if each[1] in ler1:
                if each[2] in ler1:
                    if each[3] in ler1:
                        if each[5] in ler1:
                            if each[6] in ler1:
                                if each[7] in ler1:
                                    if each[8] not in ler1:
                                        if each[4] not in ler1:
                                            if each[4] != '\n':
                                                print(each[4], end='')

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

def choice(list):
    x = []
    length = len(list1)
    for i in range(length):
        x.append(list[i:i+9])
    return x

soup = choice(list1)

lijiachen · 发表于 2020-11-29 16:43:44

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 · 发表于 2020-11-29 22:11:57

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