开发者

How to get the a string that is most repeated in a list

开发者 https://www.devze.com 2023-02-24 16:46 出处:网络
I have a lot of lists like the following: /html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[1]/div[2]/h4[1]

I have a lot of lists like the following:

/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[1]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[2]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[2]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[3]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[3]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[4]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[4]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[5]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[5]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[6]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[2]/div[1]/div[6]/div[1]/div[2]/ul[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[7]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[2]/div[1]/div[6]/div[1]/div[2]/ul[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[8]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[8]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[9]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[9]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[10]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[10]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[11]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[2]/div[1]/div[6]/div[1]/div[2]/ul[2]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[12]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[12]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[13]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[13]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[14]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[14]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[15]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[15]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[16]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[16]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[17]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[2]/div[1]/div[6]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[18]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[18]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[19]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[19]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[20]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[20]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[21]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[2]/div[1]/div[6]/div[1]/div[2]/ul[2]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[22]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[22]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[23]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[23]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[24]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[24]/div[2]/div[4]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[25]/div[2]/h4[1]
/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[25]/div[2]/div[4]

And I need to extract the portion that is most repeated in each line, which in this case is

/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2开发者_Python百科]/div[3]/ul[1]/li

What's the best way to do this?

I'm using C#/.net

thanks!


If I understand your question correctly, what you want is the longest common prefix of all lines. You could obtain it by doing something like that:

void Main()
{
    string path = @"D:\tmp\so5670107.txt";
    string[] lines = File.ReadAllLines(path);
    string prefix = LongestCommonPrefix(lines);
    Console.WriteLine(prefix);
}

static string LongestCommonPrefix(string a, string b)
{
    int length = 0;
    for (int i = 0; i < a.Length && i < b.Length; i++)
    {
        if (a[i] == b[i])
            length++;
        else
            break;
    }
    return a.Substring(0, length);
}

static string LongestCommonPrefix(IEnumerable<string> strings)
{
    return strings.Aggregate(LongestCommonPrefix);
}

The result is:

/html[1]/body[1]/div[5]/div[1]/div[2]/div[

(the expected result you give in the question seems incorrect, since there are lines that don't match it)

I chose a naive approach for the sake of simplicity, but of course there are more efficient ways of finding the longest common prefix between two strings (using a dichotomic search for instance)


You could do this with a loop. Assumption is that your list of strings is in a collection called paths:

var countByPath = new Dictionary<string, int>();
foreach (var path in paths)
{
    if (!countByPath.ContainsKey(path))
    {
        countByPath[path] = 1;
    }
    else
    {
        countByPath[path]++;
    }
}


The longest substring that is repeated in the list? Assumption is that your list of strings is in a collection called paths:

var currentChoice = "";
foreach (var path in paths)
{
    for (int i = path.Length; i > 0; i--)
    {
        var candidate = path.Substring(0, i);
        if (i > currentChoice.Length &&
            paths.Count(p => p.StartsWith(candidate)) > 1)
            currentChoice = candidate;
        else
            break;
    }
}
Console.WriteLine(currentChoice);

The result is then

/html[1]/body[1]/div[5]/div[1]/div[2]/div[3]/div[1]/div[3]/div[1]/div[2]/div[3]/ul[1]/li[10]

since it is repeated twice


There is already an algorithm for this. I can't remember what it's called, but if you are interested in language independent implementation. It works in the following way:

  1. Read first line
  2. Read second line. If second line is the same as first line, than increase counter by one, otherwise keep counter at zero.
  3. Carry on reading lines, if three lines are the same (i.e. repeat), than your counter will be 2. If next line is different to the previous three, than decrease counter by 1.

E.g.

String1 - Counter: 0 String1 - Counter: 1 (Store String1 in a variable) String1 - Counter: 2 (Store String1 in same variable) String2 - Counter: 1 (Still store String1 in variable)

I hope this makese sense. I did this at uni few years ago. Can't remember mathematician who came up with algorithm, but it's fairly old.

0

精彩评论

暂无评论...
验证码 换一张
取 消

关注公众号