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[已解决]小甲鱼20课课后作业问题

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发表于 2022-1-7 14:05:07 | 显示全部楼层 |阅读模式

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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。
最佳答案
2022-1-7 20:50:27
islower() 或 isupper() 最好一个一个比较,否则不准确,比如:
a = """
AASDFFF
HDTD
#$%^&**@@@
"""

b = "ABC AABBSS JKHK"

print(a.isupper()) # True
print(b.isupper()) # True
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:
# 因为字符串太长,所以另外存放在 text.txt 中

def isValid(s: str):

    # 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
    # 如:x _ _ _ x _ _ _ x
    if not (s[0].islower() and s[-1].islower() and s[4].islower()): return False

    # 判断前三个字符串和后三个字符串是否是大写
    # 如:_ x x x _ x x x _
    for i in s[1:4]+s[5:8]:
        if not i.isupper():
            return False

    # 判断字符串是否有 9 个字符
    return len(s) == 9

f = open("text.txt", "r")
text = f.read()

length = len(text)
res = []

for i in range(length-4):
    if(isValid(text[i:i+9])):
        res.append(text[i+4])

print(''.join(res))
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发表于 2022-1-7 18:10:15 | 显示全部楼层

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发表于 2022-1-7 20:50:27 | 显示全部楼层    本楼为最佳答案   
islower() 或 isupper() 最好一个一个比较,否则不准确,比如:
a = """
AASDFFF
HDTD
#$%^&**@@@
"""

b = "ABC AABBSS JKHK"

print(a.isupper()) # True
print(b.isupper()) # True
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:
# 因为字符串太长,所以另外存放在 text.txt 中

def isValid(s: str):

    # 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
    # 如:x _ _ _ x _ _ _ x
    if not (s[0].islower() and s[-1].islower() and s[4].islower()): return False

    # 判断前三个字符串和后三个字符串是否是大写
    # 如:_ x x x _ x x x _
    for i in s[1:4]+s[5:8]:
        if not i.isupper():
            return False

    # 判断字符串是否有 9 个字符
    return len(s) == 9

f = open("text.txt", "r")
text = f.read()

length = len(text)
res = []

for i in range(length-4):
    if(isValid(text[i:i+9])):
        res.append(text[i+4])

print(''.join(res))

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 楼主| 发表于 2022-1-8 16:26:50 | 显示全部楼层
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。
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发表于 2022-1-8 16:52:41 | 显示全部楼层
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, ...
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])
                print(a[i-4:i+4]) # 试试加多这一行代码你就明白了
                i += 1
print(list)
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 楼主| 发表于 2022-1-8 16:56:10 | 显示全部楼层

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。!
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 楼主| 发表于 2022-1-8 17:40:00 | 显示全部楼层
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。
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