[已解决]小甲魚課後練習題

TsaiYen

各位魚友悶~這題可以用切片解出來嗎? 我自己寫的code在最下面!!!!!!
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
str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
-------------------------------------------------------------------------------------------------------------------------
str2 = 'a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z'
str3 = 'A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z'
list1 = []
for each in str1:
    if each not in list1:
        if each in str2:
            idx = str1.index(each)
            a,b,c = str1[idx - 3 : idx]
            a, b, c in str3
            if each in str2:
                idx = str1.index(each)
                d,e,f = str1[idx + 1 : idx + 4]
                d, e, f in str3
                list1.append(each)
                print(each)

最佳答案

月排行榜 / 总排行榜

txxcat

2020-5-9 18:20:34

你更新的代码还是没有解决index只能得到第一个字母位置的问题，然后if each not in list1完全没必要，没有说密码不能有重复的字母，最后是 你判断前三后三是大写，却没有判断前三之前还后三之后那个是不是大写，这也会导致问题。根据你的思路改写了一下代码，可以跑出来，不过有一点不严谨的地方，看看你能找出来不。
还有个建议，判断大小写用isupper()和islow()更好。

1. list1 = []
   
2. lenth=len(str1)            #获取字符串长度
   
3. for idx in range(3,lenth-3):                #从第四个开始算到倒数第三个字符
   
4.     if str1[idx] in str2 and str1[idx-4] not in str3 and str1[idx+4] not in str3:     #除了判断是小写，加入判断前4后4的字符是否是大写的判断
   
5.         a,b,c = str1[idx - 3 : idx]
   
6.         d,e,f = str1[idx + 1 : idx + 4]
   
7.         if a in str3 and b in str3 and c in str3 and d in str3 and e in str3 and f in str3:
   
8.             list1.append(str1[idx])
   
9. print(''.join(list1))           #优化一下输出，其实不用列表，用字符相加更方便      

复制代码

跳转到最佳答案楼层

_2_

1. import re
   
2. str1="…" #这里自行补充
   
3. re.findall(r"(A-Z)\1\1(a-z)\1\1\1", str1)

复制代码

MIke_python小小

你這個繁體中文看的我很難受的

txxcat

说几个显性错误：
str1.index(each)获得索引值不靠谱，因为无论each出现多少次，只会返回each再str1中第一次出现的位置的值；
a, b, c in str3、a, b, c in str3，这两句没有任何意义，我猜你是想要：if a in str3 and b in str3 and c in str3:
str2、str3中的字母用逗号隔开没有必要，相反如果str1中有逗号反而容易误判。

TsaiYen

txxcat 发表于 2020-5-9 11:03
说几个显性错误：
str1.index(each)获得索引值不靠谱，因为无论each出现多少次，只会返回each再str1中第一 ...

這題用切片解的出來嗎

txxcat

TsaiYen 发表于 2020-5-9 12:01
這題用切片解的出來嗎

切片是解这题的必要操作，算法就各有不同了。

TsaiYen

txxcat 发表于 2020-5-9 12:06
切片是解这题的必要操作，算法就各有不同了。

只跑出 l  不知道哪裡出問題!!!!!

1. str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
   
2. lqNsYCyaQXBzrFUbkAUAWAKrDgDtAlGMBqWQhpEwquZqWZJpslUfMllCwWptqINjrOBTLuPzwvXNbLCx
   
3. oFRritKRpJgBOaGPZdkUzvYnvYmAlEsVmKRXqyQUOdCBqLYyboOYeAQNLnkuiDXCiNiksSSRpDMVQQgs
   
4. TmYThnppfKSmkpjjceGLaMOvYgsvNkGENKgGtUSzVPlLMeephDKrWGNpAxBqRiCnQIuKwDZurIRsznUp
   
5. xhstMWpHdZzqeEnttlAHiWbhbIJpwKBAGsFWthBiwBDKTFzIUamELZJFLbmqOmwBIYBJIofAUyxTLUEm
   
6. LcOXzjIHFcfUfEWccbWzhVuIZMdTVsOaNsLTpHjrvFaYTnJUrqaYnUmbobfOXXtkjKjFUEhVNlPWeaIY
   
7. uABNuEDKFWiUaqQiBHGsgEmDSKiJKLkkzFWlobXIyUlDzGcwDnEOwwWxgFpebsslmKItazIYctPROSJp
   
8. ImDHusQIkoFQSmuqwdcPNERvflfHtkGTrtbQXqZUsazoxmbVTPIuuxKvfALAaGnIuIZwczCcoIRMncBs
   
9. timFGHPJPaHMPTLjsBjIwbbzKyghaDTUStzJceaYAneywwpvdiUJifPHtXmAwZPChWsDmwonzeeQCEiB
   
10. xWsJLAtrdVEOBxEaYiSabHdUwlXlDOAKzZUwozJoBGvnFttDkXdYdjQSzzhTIPfYpWVFdiFIqSbyPZIi
    
11. VoXunBPbAJVMWluPcIfHmjPeHaTyClCvTOucEYEZXbVDiZMsfeuruDGMmQEsHutzvzVmWvCoPRguBiHh
    
12. taTeaguPmNuAxzYJLEcdoApfIFBaLUqettjTJNlHZYlJZRRoKXEvFnoUHdPjDuXuTAubGODkHAokpItw
    
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104. qKWGc'''
     
105. 
     
106. str2 = 'a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z'
     
107. str3 = 'A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z'
     
108. list1 = []           #創建空列表
     
109. for each in str1:    #把字串中的字符給each
     
110.     if each not in list1:          #判斷each是不是沒有在列表中(沒有的話)
     
111.         if each in str2:           #each 判斷是不是小寫字母
     
112.             idx = str1.index(each) #找出他在str1裡面的位移植丟給變數
     
113.             a,b,c = str1[idx - 3 : idx]
     
114.             d,e,f = str1[idx + 1 : idx + 4]
     
115.             if a in str3 and b in str3 and c in str3 and d in str3 and e in str3 and f in str3:
     
116.                 list1.append(each)
     
117. print(list1)

复制代码

TsaiYen

1. #修改過後只跑出一個 l
   
2. str1 = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
   
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105. qKWGc'''
     
106. 
     
107. str2 = 'abcdefghijklmnopqrstuvwxyz'
     
108. str3 = 'ABCDEFGHJIKLMNOPQRSTUVWXYZ'
     
109. list1 = []           #創建空列表
     
110. for each in str1:    #把字串中的字符給each
     
111.     if each not in list1:          #判斷each是不是沒有在列表中(沒有的話)
     
112.         if each in str2:           #each 判斷是不是小寫字母   
     
113.             idx = str1.index(each) #找出他在str1裡面的位移植丟給變數
     
114.             a,b,c = str1[idx - 3 : idx]
     
115.             d,e,f = str1[idx + 1 : idx + 4]
     
116.             if a in str3 and b in str3 and c in str3 and d in str3 and e in str3 and f in str3:
     
117.                 list1.append(each)
     
118. print(list1)
     
119.                         
     
120.             
     
121.         
     
122.         
     
123. 
     

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txxcat

你更新的代码还是没有解决index只能得到第一个字母位置的问题，然后if each not in list1完全没必要，没有说密码不能有重复的字母，最后是 你判断前三后三是大写，却没有判断前三之前还后三之后那个是不是大写，这也会导致问题。根据你的思路改写了一下代码，可以跑出来，不过有一点不严谨的地方，看看你能找出来不。
还有个建议，判断大小写用isupper()和islow()更好。

1. list1 = []
   
2. lenth=len(str1)            #获取字符串长度
   
3. for idx in range(3,lenth-3):                #从第四个开始算到倒数第三个字符
   
4.     if str1[idx] in str2 and str1[idx-4] not in str3 and str1[idx+4] not in str3:     #除了判断是小写，加入判断前4后4的字符是否是大写的判断
   
5.         a,b,c = str1[idx - 3 : idx]
   
6.         d,e,f = str1[idx + 1 : idx + 4]
   
7.         if a in str3 and b in str3 and c in str3 and d in str3 and e in str3 and f in str3:
   
8.             list1.append(str1[idx])
   
9. print(''.join(list1))           #优化一下输出，其实不用列表，用字符相加更方便      

复制代码

TsaiYen

txxcat 发表于 2020-5-9 18:20
你更新的代码还是没有解决index只能得到第一个字母位置的问题，然后if each not in list1完全没必要，没有 ...

想請問下不用這種思路還能用甚麼方法嗎?
for循環不是把str1一個個丟給each為甚麼還是只能得到第一個字母位置
萬分的感謝您!!!!!!!!!!!!!!!!!!!!!

txxcat

TsaiYen 发表于 2020-5-10 16:49
想請問下不用這種思路還能用甚麼方法嗎?
for循環不是把str1一個個丟給each為甚麼還是只能得到第一個字母 ...

遍历字符串可以获得每一个字符，但你要获得字符串的位置，使用了str1.index(each)，比如'a'这个字符，第一个a在第31个，获得的值是31，这是对的，但是后面的'a'呢，str1.index(each)还是会返回第一个'a'的位置，不会返回你希望的值。如果你希望遍历时获得索引值，还可以这么写：

1. idx=0
   
2. for each in str1:
   
3.     pass
   
4.     idx+=1

复制代码

这样可以获得正确的索引值，str1[idx]等于each，但是因为有索引值-3、+3、-4、+4的操作，要正确处理索引值溢出的问题，最后需要的代码会比我的方法要多不少。
这里有几种算法，其实大同小异：

1. lenth = len(str1)
   
2. for i in range(3,lenth-3):
   
3.     allupper=True                                 #定义一个标量标志全部是大写
   
4.     if str1[i].islower():
   
5.         for each in str1[i-3:i]+str1[i+1:i+4]:    #遍历前3个+后3个字符
   
6.             if not each.isupper():                #有1个不是大写就跳出判断，标志改为False
   
7.                 allupper=False
   
8.                 break
   
9.         if allupper and (str1[i-4].islower()) and (str1[i+4].islower()):
   
10.                 print(str1[i],end="")

复制代码

暴力简单去除\n法：

1. str1=str1.replace('\n','')
   
2. length = len(str1)
   
3. for i in range(3, length-3):
   
4.     if str1[i].islower() and  str1[i-3:i].isupper() and str1[i+1:i+4].isupper() and  not str1[i-4].isupper() and not str1[i+4].isupper():
   
5.         print(str1[i],end='')

复制代码

TsaiYen

txxcat 发表于 2020-5-10 20:51
遍历字符串可以获得每一个字符，但你要获得字符串的位置，使用了str1.index(each)，比如'a'这个字符，第 ...

OK感恩!!! 還在就學嗎?!好厲害

txxcat

TsaiYen 发表于 2020-5-11 02:49
OK感恩!!! 還在就學嗎?!好厲害

No，作为孩子她爹，对年轻人们充满了羡慕！

TsaiYen

是軟體工程師嗎哈哈！！

Pythonnewers

MIke_python小小 发表于 2020-5-9 10:26
你這個繁體中文看的我很難受的

哈哈,台灣人?

TsaiYen

對！