计算没有标点符号的小写句子中短语的出现次数

Question

示例输入：

• 彼得警告彼得不要出海
• 阿尔弗雷德应该去海边
• 阿尔伯特警告皮特
• 我告诉过你不要去迪士尼乐园

示例输出：

• 3：去
• 2：出海
• 2：不去

可选：

• 不计算低于 2 的出现次数。

我的方法

public void zähleHäufigkeitWorte(string[] arr, int n)
{
    bool[] visited = new bool[n];

    // Traverse through array elements and 
    // count frequencies 
    for (int i = 0; i < n; i++)
    {

        // Skip this element if already processed 
        if (visited[i] == true)
            continue;

        // Count frequency 
        int count = 1;
        for (int j = i + 1; j < n; j++)
        {
            if (arr[i] == arr[j])
            {
                visited[j] = true;
                count++;
            }
        }
        wort.Add(arr[i],count);
    }
}

如何扩展以计算短语的出现次数？

Answer 1

这应该可以解决问题：

using System;
using System.Collections.Generic;

namespace SO68666232
{
    public class PhraseCounter
    {

        public static Dictionary<string, int> countWordAndPhraseFrequencies(string input)
        {
            var phraseDic = new Dictionary<string, int>();
            string[] tokens = input.Split(' ');
            int ntokens = tokens.Length;
            for (int start = 0; start < ntokens; start++)
            {
                for (int nwords = 1; nwords <= (ntokens- start) ; nwords++)
                {
                    ArraySegment<string> phraseTokens = new ArraySegment<string>(tokens, start, nwords);
                    string phrase = string.Join(" ", phraseTokens);
                    if (phraseDic.ContainsKey(phrase))
                    {
                        phraseDic[phrase]++;
                    }
                    else
                    {
                        phraseDic[phrase] = 1;
                    }
                }
            }
            // Remove elements with only 1 occurrence
            IEnumerable<string> keys = new List<string>(phraseDic.Keys);
            foreach (string key in keys)
            {
                if (phraseDic[key] <= 1)
                {
                    phraseDic.Remove(key);
                }
            }
            return phraseDic;
        }
    }
}

这可能不是最有效的方法，因为时间随着单词数的平方而增加，但它确实有效。

测试程序：

    class Program
    {
        const string sampleInput = "peter warns pete to not go to the sea alfred should go to the sea albert warns pete i told you to not go to disneyland";

        static void Main(string[] args)
        {
            var res = PhraseCounter.countWordAndPhraseFrequencies(sampleInput);
            foreach (string key in res.Keys)
            {
                Console.WriteLine("{0}: {1}", key, res[key]);
            }
            Console.WriteLine("Press any key to continue...");
            Console.ReadKey();
        }
    }

结果：

warns: 2
warns pete: 2
pete: 2
to: 5
to not: 2
to not go: 2
to not go to: 2
not: 2
not go: 2
not go to: 2
go: 3
go to: 3
go to the: 2
go to the sea: 2
to the: 2
to the sea: 2
the: 2
the sea: 2
sea: 2