如何从字符串内容中删除一些特殊单词？

Question

我有一些字符串包含表情符号图标的代码，例如:grinning:，:kissing_heart:或:bouquet:。我想处理它们以删除表情符号代码。

例如，给定：

  

你好：咧嘴笑：你好吗？：kissing_heart：你还好吗？：花束：

我想得到这个：

  你好，你好吗？你还好吗？

我知道我可以使用这段代码：

richTextBox2.Text = richTextBox1.Text.Replace(":kissing_heart:", "").Replace(":bouquet:", "").Replace(":grinning:", "").ToString();

但是，我必须删除856个不同的表情符号图标（使用此方法，将对Replace()进行856次调用）。有没有其他方法可以实现这个目标？

Answer 1

您可以使用正则表达式匹配:anything:之间的单词。使用Replace函数可以进行其他验证。

string pattern = @":(.*?):";
string input = "Hello:grinning: , how are you?:kissing_heart: Are you fine?:bouquet: Are you super fan, for example. :words not to replace:";
string output = Regex.Replace(input, pattern, (m) =>
{
    if (m.ToString().Split(' ').Count() > 1) // more than 1 word and other validations that will help preventing parsing the user text
    {
        return m.ToString();
    }
    return String.Empty;
}); // "Hello , how are you? Are you fine? Are you super fan, for example. :words not to replace:"

如果您不想使用使用lambda表达式的Replace，可以使用\w，如@ yorye-nathan所述，仅匹配单词。

string pattern = @":(\w*):";
string input = "Hello:grinning: , how are you?:kissing_heart: Are you fine?:bouquet: Are you super fan, for example. :words not to replace:";
string output = Regex.Replace(input, pattern, String.Empty); // "Hello , how are you? Are you fine? Are you super fan, for example. :words not to replace:"

Answer 2

string Text = "Hello:grinning: , how are you?:kissing_heart: Are you fine?:bouquet:";

我会这样解决

List<string> Emoj = new List<string>() { ":kissing_heart:", ":bouquet:", ":grinning:" };
Emoj.ForEach(x => Text = Text.Replace(x, string.Empty));

更新 - 参阅详细评论

另一种方法：只替换现有的Emojs

List<string> Emoj = new List<string>() { ":kissing_heart:", ":bouquet:", ":grinning:" };
var Matches = Regex.Matches(Text, @":(\w*):").Cast<Match>().Select(x => x.Value);
Emoj.Intersect(Matches).ToList().ForEach(x => Text = Text.Replace(x, string.Empty));

但我不确定这种短聊天串是否会有很大的不同，而且让代码易于阅读/维护更为重要。 OP的问题是关于减少冗余Text.Replace().Text.Replace()而不是最有效的解决方案。

Answer 3

我会结合使用已经建议的一些技术。首先，我将800多个表情符号字符串存储在数据库中，然后在运行时加载它们。使用HashSet将它们存储在内存中，这样我们就有了O（1）查找时间（非常快）。使用正则表达式从输入中提取所有可能的模式匹配，然后将每个模式匹配与我们的哈希表情符号进行比较，删除有效的表情符号并保留用户自己输入的任何非表情符号模式......

public class Program
{
    //hashset for in memory representation of emoji,
    //lookups are O(1), so very fast
    private HashSet<string> _emoji = null;

    public Program(IEnumerable<string> emojiFromDb)
    {
        //load emoji from datastore (db/file,etc)
        //into memory at startup
        _emoji = new HashSet<string>(emojiFromDb);
    }

    public string RemoveEmoji(string input)
    {
        //pattern to search for
        string pattern = @":(\w*):";
        string output = input;

        //use regex to find all potential patterns in the input
        MatchCollection matches = Regex.Matches(input, pattern);

        //only do this if we actually find the 
        //pattern in the input string...
        if (matches.Count > 0)
        {
            //refine this to a distinct list of unique patterns 
            IEnumerable<string> distinct = 
                matches.Cast<Match>().Select(m => m.Value).Distinct();

            //then check each one against the hashset, only removing
            //registered emoji. This allows non-emoji versions 
            //of the pattern to survive...
            foreach (string match in distinct)
                if (_emoji.Contains(match))
                    output = output.Replace(match, string.Empty);
        }

        return output;
    }
}

public class MainClass
{
    static void Main(string[] args)
    {
        var program = new Program(new string[] { ":grinning:", ":kissing_heart:", ":bouquet:" });
        string output = program.RemoveEmoji("Hello:grinning: :imadethis:, how are you?:kissing_heart: Are you fine?:bouquet: This is:a:strange:thing :to type:, but valid :nonetheless:");
        Console.WriteLine(output);
    }
}

结果是：

你好：imadethis：你好吗？你还好吗？这是：a：奇怪的：东西：输入：，  但有效：尽管如此：

Answer 4

您无需更换所有856表情符号。您只需要替换字符串中出现的那些。所以看看：

Finding a substring using C# with a twist

基本上你提取所有标记，即：和之间的字符串，然后用string.Empty（）替换它们

如果您担心搜索将返回非emojis的字符串，例如：其他一些文本：那么您可以进行哈希表查找以确保替换所述找到的令牌是合适的。

Answer 5

终于开始写点什么了。我结合了前面提到的几个想法，我们应该只在字符串上循环一次。根据这些要求，这听起来像是Linq的完美工作。

您应该缓存HashSet。除此之外，这具有O（n）性能，并且仅在列表上进行一次。对基准测试很有意思，但这很可能是最有效的解决方案。

这种方法非常简单。

• 首先在HashSet中加载所有Emoij，以便我们快速查找它们。
• 在input.Split(':')
处使用:拆分字符串
• 决定我们是否保留当前元素。
  • 如果最后一个元素匹配，请保留当前元素。
  • 如果最后一个元素不匹配，请检查当前元素是否匹配。
    • 如果是，请忽略它。 （这有效地从输出中删除了子串。）
    • 如果没有，请追加:并保留。
• 使用StringBuilder重新构建字符串。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
    static class Program
    {
        static void Main(string[] args)
        {
            ISet<string> emojiList = new HashSet<string>(new[] { "kissing_heart", "bouquet", "grinning" });

            Console.WriteLine("Hello:grinning: , ho:w: a::re you?:kissing_heart:kissing_heart: Are you fine?:bouquet:".RemoveEmoji(':', emojiList));
            Console.ReadLine();
        }

        public static string RemoveEmoji(this string input, char delimiter, ISet<string> emojiList)
        {
            StringBuilder sb = new StringBuilder();
            input.Split(delimiter).Aggregate(true, (prev, curr) =>
            {
                if (prev)
                {
                    sb.Append(curr);
                    return false;
                }
                if (emojiList.Contains(curr))
                {
                    return true;
                }
                sb.Append(delimiter);
                sb.Append(curr);
                return false;
            });
            return sb.ToString();
        }
    }
}

编辑：我使用Rx library做了一些很酷的事情，但后来意识到Aggregate是Rx中IEnumerable的{​​{1}}对应物，从而进一步简化了代码。< / p>

Answer 6

如果效率是一个问题，并且为了避免处理误报＆＃34;，请考虑使用StringBuilder重写字符串，同时跳过特殊的表情符号标记：

static HashSet<string> emojis = new HashSet<string>()
{
    "grinning",
    "kissing_heart",
    "bouquet"
};

static string RemoveEmojis(string input)
{
    StringBuilder sb = new StringBuilder();

    int length = input.Length;
    int startIndex = 0;
    int colonIndex = input.IndexOf(':');

    while (colonIndex >= 0 && startIndex < length)
    {
        //Keep normal text
        int substringLength = colonIndex - startIndex;
        if (substringLength > 0)
            sb.Append(input.Substring(startIndex, substringLength));

        //Advance the feed and get the next colon
        startIndex = colonIndex + 1;
        colonIndex = input.IndexOf(':', startIndex);

        if (colonIndex < 0) //No more colons, so no more emojis
        {
            //Don't forget that first colon we found
            sb.Append(':');
            //Add the rest of the text
            sb.Append(input.Substring(startIndex));
            break;
        }
        else //Possible emoji, let's check
        {
            string token = input.Substring(startIndex, colonIndex - startIndex);

            if (emojis.Contains(token)) //It's a match, so we skip this text
            {
                //Advance the feed
                startIndex = colonIndex + 1;
                colonIndex = input.IndexOf(':', startIndex);
            }
            else //No match, so we keep the normal text
            {
                //Don't forget the colon
                sb.Append(':');

                //Instead of doing another substring next loop, let's just use the one we already have
                sb.Append(token);
                startIndex = colonIndex;
            }
        }
    }

    return sb.ToString();
}

static void Main(string[] args)
{
    List<string> inputs = new List<string>()
    {
        "Hello:grinning: , how are you?:kissing_heart: Are you fine?:bouquet:",
        "Tricky test:123:grinning:",
        "Hello:grinning: :imadethis:, how are you?:kissing_heart: Are you fine?:bouquet: This is:a:strange:thing :to type:, but valid :nonetheless:"
    };

    foreach (string input in inputs)
    {
        Console.WriteLine("In  <- " + input);
        Console.WriteLine("Out -> " + RemoveEmojis(input));
        Console.WriteLine();
    }

    Console.WriteLine("\r\n\r\nPress enter to exit...");
    Console.ReadLine();
}

输出：

In  <- Hello:grinning: , how are you?:kissing_heart: Are you fine?:bouquet:
Out -> Hello , how are you? Are you fine?

In  <- Tricky test:123:grinning:
Out -> Tricky test:123

In  <- Hello:grinning: :imadethis:, how are you?:kissing_heart: Are you fine?:bouquet: This is:a:strange:thing :to type:, but valid :nonetheless:
Out -> Hello :imadethis:, how are you? Are you fine? This is:a:strange:thing :to type:, but valid :nonetheless:

Answer 7

使用我在下面提到的代码我认为使用此功能可以解决您的问题。

        string s = "Hello:grinning: , how are you?:kissing_heart: Are you fine?:bouquet:";

        string rmv = ""; string remove = "";
        int i = 0; int k = 0;
    A:
        rmv = "";
        for (i = k; i < s.Length; i++)
        {
            if (Convert.ToString(s[i]) == ":")
            {
                for (int j = i + 1; j < s.Length; j++)
                {
                    if (Convert.ToString(s[j]) != ":")
                    {
                        rmv += s[j];
                    }
                    else
                    {
                        remove += rmv + ",";
                        i = j;
                        k = j + 1;
                        goto A;
                    }
                }
            }
        }

        string[] str = remove.Split(',');
        for (int x = 0; x < str.Length-1; x++)
        {
            s = s.Replace(Convert.ToString(":" + str[x] + ":"), "");
        }
        Console.WriteLine(s);
        Console.ReadKey();

Answer 8

我会使用这样的扩展方法：

public static class Helper
{
   public static string MyReplace(this string dirty, char separator)
    {
        string newText = "";
        bool replace = false;

        for (int i = 0; i < dirty.Length; i++)
        {
            if(dirty[i] == separator) { replace = !replace ; continue;}
            if(replace ) continue;
            newText += dirty[i];
        }
        return newText;
    }

}

用法：

richTextBox2.Text = richTextBox2.Text.MyReplace（'：'）;

与Regex相比，此方法在性能方面表现更好

Answer 9

我会用'：'拆分文本，然后构建不包含找到的表情符号名称的字符串。

<html>
<body>
<input type="text" id="fname" onkeyup="myFunction()">
<input type="button" id="a" disabled = "true" value ="click me">

<script>
function myFunction() {

    var x = document.getElementById("fname");
    x.value = x.value.toUpperCase();
    if(x.value != "")
{
   document.getElementById('a').disabled=false;
}else {
document.getElementById('a').disabled=true;
}
}
</script>
</body>
</html>