Question

因此，我已经能够创建以下程序，该程序将两个字符串进行比较以查看它们是否是彼此的字谜。

def anagrams( string1, string2 ):
    if sorted(string1.lower()) == sorted(string2.lower()):
        return True
    else:
        return False

但是，我的问题是，如果两个输入字符串完全相同，我希望不返回True值。例如：

anagrams('silent','silent')

这将输出True，但我不希望它这样做，我应该进行哪些更改以实现此目的？

Answer 1

只需检查字符串是否不同：

@VaadinServletConfiguration

输出

foreach (Task t in tasks)
{
    Context  context = t.ConstructContext(condition);



    context.CommonProperty = "value";
    context.MethodToOverride();//Do different things based on the context type
}

public asbtract class Task
{
    //whatever might be in here
    public abstract Context ConstructContext();
}

public class TaskA : Task
{
    //NOTE: In my opinion condition should be internal to TaskA and no one else should know about and you should remove the parameter from the method
    // but I don't know where you get it from so I'm leaving it here
    public override Context ConstructContext(bool condition)
    {
        if (condition)
        {
            return new ContextA() { Task = this};
        }
        else
        {
            return new ContextA2() { Task = thid };
        }
    }
}

public class TaskB : Task
{
    public override Context ConstructContext(bool condition)
    {
        //ignore condition which implies it shouldn't really be passed

        return new ContextB() {Task = this};
    }
}

//etc...

您可以使用Counter而不是进行排序：

def anagrams(string1, string2):
    if sorted(string1.lower()) == sorted(string2.lower()) and string1.lower() != string2.lower():
        return True
    else:
        return False

result = anagrams('silent', 'silent')
print(result)

输出

False

更新

按照@RoadRunner的建议，您可以这样做：

from collections import Counter


def anagrams(string1, string2):
    if string1.lower() != string2.lower() and Counter(string1.lower()) == Counter(string2.lower()):
        return True
    else:
        return False

print(anagrams('silent', 'silent'))
print(anagrams('silent', 'sitlen'))

Answer 2

def anagrams( string1, string2 ):
  if sorted(string1.lower()) == sorted(string2.lower()) and string1.lower() != string2.lower():
    return True
  else:
    return False

print(anagrams('silent','silent'))

创建一个字谜检查器

2 个答案: