[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

string = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
liststr= []
length = len(string)
for num0 in range(3,length-4):#从第三个开始索引
    numl = num0 - 3#密码左边三个数字
    numr = num0 + 4#密码右边三个数字
    if numr <= length -3:
        if (string[num0].islower()) and\
           (string[numl:num0].isupper()) and\
           (string[numl-1].islower()) and\
           (string[num0+1:numr].isupper()) and\
           (string[numr].islower()):
               #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
            liststr.append(string[num0])
print('密码是：',''.join(liststr))
            

最佳答案

月排行榜 / 总排行榜

yhhpf

2020-7-8 17:03:59

举个例子:

>>> s='''aaaa
bbbb
cccc'''
>>> s
'aaaa\nbbbb\ncccc'
>>> len(s)
14
>>> s[4]
'\n'
>>> s[4].isupper()
False
>>> s[4].islower()
False
>>> s[3:4].islower()
True

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：

for num0 in range(3, length - 4):  # 从第三个开始索引
    count += 1
    numl = num0 - 3  # 密码左边三个数字
    numr = num0 + 4  # 密码右边三个数字
    if numr <= length - 3:
        if (string[num0].islower()) and \
                (string[numl:num0].isupper()) and \
                (string[numl - 1].islower()) and \
                (string[num0 + 1:numr].isupper()) and \
                (string[numr].islower()) and \
                '\n' not in string[numl:numr]:     #加上这句判断是否存在\n换行符
            # 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
            liststr.append(string[num0])
print('密码是：', ''.join(liststr))

跳转到最佳答案楼层

yhhpf

举个例子:

>>> s='''aaaa
bbbb
cccc'''
>>> s
'aaaa\nbbbb\ncccc'
>>> len(s)
14
>>> s[4]
'\n'
>>> s[4].isupper()
False
>>> s[4].islower()
False
>>> s[3:4].islower()
True

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：

for num0 in range(3, length - 4):  # 从第三个开始索引
    count += 1
    numl = num0 - 3  # 密码左边三个数字
    numr = num0 + 4  # 密码右边三个数字
    if numr <= length - 3:
        if (string[num0].islower()) and \
                (string[numl:num0].isupper()) and \
                (string[numl - 1].islower()) and \
                (string[num0 + 1:numr].isupper()) and \
                (string[numr].islower()) and \
                '\n' not in string[numl:numr]:     #加上这句判断是否存在\n换行符
            # 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
            liststr.append(string[num0])
print('密码是：', ''.join(liststr))

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