[已解决]代码不报错不给结果是什么情况？

影影影影影少丶

Python课后作业代码，写好后运行不给结果，也不报错。请问是什么情况？
运行结果如图，代码如下

def find1(str1):
    a = [] #建立空列表储存转化为大小写判断的flag
    b = '' #建议空字符串以便拼接密码
    for each in str1:
        if each.islower() == 1: #判断是否为小写
            flag = 1 #将小写标记为1
            a.append(flag) #将flag储存到列表a中
        else:
            flag = 0 #将大写标记为0
            a.append(flag) #将flag储存到列表a中
    i = 3 #按照题目要求，从第4个元素开始判断
    while i <= len(a)-3: #在倒数第4个元素处停止
        if i == 3 or i == (len(a)-3): #在第4个和倒数第4个处判断[0，0，0，1，0，0，0]的组合
            if a[i] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0:
                b = b + str1[i]
        else: #在其他位置上的元素判断[1，0，0，0，1，0，0，0，1]的组合
            if a[i] == 1 and a[i-4] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] == 1:
                b = b + str1[i]
    print('密码是' + c)
    
x = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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find1(x)

最佳答案

月排行榜 / 总排行榜

txxcat

2020-5-12 23:56:02

比较新奇的算法了，就是效率有点低，不过对于现在的电脑来说不算什么，先说两个显性错误：
1.while循环中i值没有变化，死循环了，加一条i+=1或者换成for循环；
2.变量c没有，是变量b。
然后是算法bug：
1.没有考虑非字母，这里主要是'\n'；
2.判断第四个字符和倒数第四个字符不是简单看前三后三，而是第四个是前三后四，倒数第四个是前四后三。
修正了代码，可以跑起来了，还有可以优化的地方再慢慢想吧：

def find1(str1):
    a = [] #建立空列表储存转化为大小写判断的flag
    b = '' #建议空字符串以便拼接密码
    for each in str1:
        if each.islower() == 1: #判断是否为小写
            flag = 1 #将小写标记为1
            a.append(flag) #将flag储存到列表a中
        elif each.isupper() == 1:
            flag = 0 #将大写标记为0
            a.append(flag) #将flag储存到列表a中
        else:
            flag = 2 #非字母标记为2
            a.append(flag) #将flag储存到列表a中
        
    i = 3 #按照题目要求，从第4个元素开始判断
    while i <= len(a)-3: #在倒数第4个元素处停止
        if i == 3 : #在第4个处判断[0，0，0，1，0，0，0，非0]的组合
            if a[i] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] != 0: 
                b = b + str1[i]
        if i == (len(a)-3): #在倒数第4个处判断[非0，0，0，0，1，0，0，0]的组合
            if a[i] == 1 and a[i-4] != 0 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0:
                b = b + str1[i]
        else: #在其他位置上的元素判断[非0，0，0，0，1，0，0，0，非0]的组合
            if a[i] == 1 and a[i-4] != 0 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] != 0:
                b = b + str1[i]
        i+=1
    print('密码是' + b)

跳转到最佳答案楼层

txxcat

比较新奇的算法了，就是效率有点低，不过对于现在的电脑来说不算什么，先说两个显性错误：
1.while循环中i值没有变化，死循环了，加一条i+=1或者换成for循环；
2.变量c没有，是变量b。
然后是算法bug：
1.没有考虑非字母，这里主要是'\n'；
2.判断第四个字符和倒数第四个字符不是简单看前三后三，而是第四个是前三后四，倒数第四个是前四后三。
修正了代码，可以跑起来了，还有可以优化的地方再慢慢想吧：

def find1(str1):
    a = [] #建立空列表储存转化为大小写判断的flag
    b = '' #建议空字符串以便拼接密码
    for each in str1:
        if each.islower() == 1: #判断是否为小写
            flag = 1 #将小写标记为1
            a.append(flag) #将flag储存到列表a中
        elif each.isupper() == 1:
            flag = 0 #将大写标记为0
            a.append(flag) #将flag储存到列表a中
        else:
            flag = 2 #非字母标记为2
            a.append(flag) #将flag储存到列表a中
        
    i = 3 #按照题目要求，从第4个元素开始判断
    while i <= len(a)-3: #在倒数第4个元素处停止
        if i == 3 : #在第4个处判断[0，0，0，1，0，0，0，非0]的组合
            if a[i] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] != 0: 
                b = b + str1[i]
        if i == (len(a)-3): #在倒数第4个处判断[非0，0，0，0，1，0，0，0]的组合
            if a[i] == 1 and a[i-4] != 0 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0:
                b = b + str1[i]
        else: #在其他位置上的元素判断[非0，0，0，0，1，0，0，0，非0]的组合
            if a[i] == 1 and a[i-4] != 0 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] != 0:
                b = b + str1[i]
        i+=1
    print('密码是' + b)

wuqramy

还在算吧

heidern0612

大致看了下，最后while的时候，i的变化呢？

i什么时候不满足while呢？

没退出的话，应该一直就循环，卡在循环里了吧？

你可以在if分支下面建个print，打印一句话。然后运行下，你会发现一直在打印这句话。

也就是说，你i没有小于你while条件的时候，自然就一直循环了。

你可以运行我这段代码试试，我就加了俩print,别的代码没动你的。

def find1(str1):
    a = [] #建立空列表储存转化为大小写判断的flag
    b = '' #建议空字符串以便拼接密码
    for each in str1:
        if each.islower() == 1: #判断是否为小写
            flag = 1 #将小写标记为1
            a.append(flag) #将flag储存到列表a中
        else:
            flag = 0 #将大写标记为0
            a.append(flag) #将flag储存到列表a中
    i = 3 #按照题目要求，从第4个元素开始判断
    while i <= len(a)-3: #在倒数第4个元素处停止
        if i == 3 or i == (len(a)-3): #在第4个和倒数第4个处判断[0，0，0，1，0，0，0]的组合
            if a[i] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0:
                b = b + str1[i]
                print("卡在if")
        else: #在其他位置上的元素判断[1，0，0，0，1，0，0，0，1]的组合
            if a[i] == 1 and a[i-4] == 1 and a[i-3] == 0 and a[i-2] == 0 and a[i-1] == 0 and a[i+1] == 0 and a[i+2] == 0 and a[i+3] == 0 and a[i+4] == 1:
                b = b + str1[i]
                print("卡在else")
    print('密码是' + c)
    
x = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
find1(x)

老八秘制

对的，没有出现下一个“>>>”应该是还在算

香蕉柠檬树

刷个存在感

影影影影影少丶

heidern0612 发表于 2020-5-13 07:37
大致看了下，最后while的时候，i的变化呢？

i什么时候不满足while呢？

感谢回复，确实卡在了if的位置。我都不知道啥时候不小心把i++给删掉了。我还一直没注意到。

影影影影影少丶

txxcat 发表于 2020-5-12 23:56
比较新奇的算法了，就是效率有点低，不过对于现在的电脑来说不算什么，先说两个显性错误：
1.while循环中i ...

感谢老哥，前面写了i += 1后面不小心删掉没注意到。最后的b和c也是后面整理代码出现的错误。算法bug也指出来了。太棒了！