小甲鱼20课课后作业问题

甜蜜难在

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qKWGc'''
length = len(a)
list = []
for i in range(length):
    if a[i] == '\n':
        continue
    if a[i].islower():
        if a[i-3:i].isupper() and a[i-4].islower():
            if a[i+1:i+4].isupper() and a[i+4].islower():
                list.append(a[i])
                i += 1
print(list)



根据20课课后第二题的要求，我写了如上代码，我是利用切片的方法进行判断。假如i是小写，那么判断它前面三个字符和后面三个字符范围内的字符是否是大写，由此来得出密码，但是当我执行了之后，结果显示如下：

['u', 'i', 'l', 'o', 'v', 'e', 'f', 'i', 's', 'd', 'h', 'c']

与小甲鱼老师的答案比较的话多了一个u和d字符。思考了很久都不知道是怎么回事，请问有大神能解答一下吗。？小白不胜感激。

AdiosSpike

我还没学到

傻眼貓咪

islower() 或 isupper() 最好一个一个比较，否则不准确，比如：

1. a = """
   
2. AASDFFF
   
3. HDTD
   
4. #$%^&**@@@
   
5. """
   
6. 
   
7. b = "ABC AABBSS JKHK"
   
8. 
   
9. print(a.isupper()) # True
   
10. print(b.isupper()) # True

复制代码

可以看出以上结果，是也 True 不是也 True 这不是我们想要的，我的代码为：

1. # 因为字符串太长，所以另外存放在 text.txt 中
   
2. 
   
3. def isValid(s: str):
   
4. 
   
5.     # 判断前第四个字符、后第四个字符、和中间字符是否都是小写
   
6.     # 如：x _ _ _ x _ _ _ x
   
7.     if not (s[0].islower() and s[-1].islower() and s[4].islower()): return False
   
8. 
   
9.     # 判断前三个字符串和后三个字符串是否是大写
   
10.     # 如：_ x x x _ x x x _
    
11.     for i in s[1:4]+s[5:8]:
    
12.         if not i.isupper():
    
13.             return False
    
14. 
    
15.     # 判断字符串是否有 9 个字符
    
16.     return len(s) == 9
    
17. 
    
18. f = open("text.txt", "r")
    
19. text = f.read()
    
20. 
    
21. length = len(text)
    
22. res = []
    
23. 
    
24. for i in range(length-4):
    
25.     if(isValid(text[i:i+9])):
    
26.         res.append(text[i+4])
    
27. 
    
28. print(''.join(res))

复制代码

甜蜜难在

傻眼貓咪 发表于 2022-1-7 20:50
islower() 或 isupper() 最好一个一个比较，否则不准确，比如：
可以看出以上结果，是也 True 不是也 True ...

但是我也是按照这个思路来判断的啊。先判断中间字母是否小写，然后再判断前面三位和后面三位是否为大写，再判断前第四位和后第四位是否为小写。而且在小甲鱼提供的字符串里面是没有字符的。所以还是不太明白为什么我执行出来会多两个字母。当然您的方法也是正确的。

傻眼貓咪

甜蜜难在 发表于 2022-1-8 16:26
但是我也是按照这个思路来判断的啊。先判断中间字母是否小写，然后再判断前面三位和后面三位是否为大写， ...

1. length = len(a)
   
2. list = []
   
3. for i in range(length):
   
4.     if a[i] == '\n':
   
5.         continue
   
6.     if a[i].islower():
   
7.         if a[i-3:i].isupper() and a[i-4].islower():
   
8.             if a[i+1:i+4].isupper() and a[i+4].islower():
   
9.                 list.append(a[i])
   
10.                 print(a[i-4:i+4]) # 试试加多这一行代码你就明白了
    
11.                 i += 1
    
12. print(list)

复制代码

甜蜜难在

傻眼貓咪 发表于 2022-1-8 16:52

我理解了，刚刚使用调试功能，找到了第一个u字母的索引位置，然后查看了前面和后面的字符，发现前面的三个字符中包含了一个\n字符，然后判断也是Ture，所以是这个原因导致了多了字母。万分感谢。！

甜蜜难在

我坚持用我的方法，然后经过我的倔强之后，通过方法先将字符串的'\n'全部删除掉，然后组成新字符串，然后再用我的思路来判断，现在结果终于和小甲鱼一样了。
a = []
with open(r"d:\Users\zhongsong\Desktop\string2.txt", "r") as f1:
    for each in f1.readlines():
        if each is not None:
            a.append(each.strip("\n"))
f1.close()
b = (''.join(a))
# print(b)
length = len(b)
list = []
for i in range(length):
    if b[i] == '\n':
        continue
    if b[i].islower():
        if b[i - 3:i].isupper() and b[i - 4].islower():
            if b[i + 1:i + 4].isupper() and b[i + 4].islower():
                list.append(b[i])
                i += 1
print(''.join(list))

总结下来，就是小甲鱼老师讲课的时候没有特别仔细，就像这种字符串里面有换行符的情况下也会判断为True，看来这个是个要点。需要大家规避一下。