计算数组中单词在句子中的次数

Question

我正在编写代码来计算数组中每个单词出现的次数以及出现的所有单词的总数。我设法创建了一个数组，允许用户添加他们希望检查的多个单词。但是，我正在努力寻找一种方法来单独计算每个单词在句子中出现的次数（我只能让它对数组的第一个元素起作用）。

我尝试了一个 for 循环，完成后它将移动到数组中的下一个元素，但它不会为下一个元素再次开始 for 循环，而是结束代码块。

int occurences = 0;

string[] words = new string[_wordCount];

for (int i = 0; i < words.Length; i++)
{
    Console.WriteLine("Type in the censored words you wish to be counted: ");
    words[i] = Console.ReadLine();

    if (_sentence.Contains(words[i]))
    {
        occurences++;
    }

    if (i > words.Length)
    {
        i++;
    }
}

Console.WriteLine("Number of censored word occurences: " + occurences);
return occurences;

Answer 1

您只需要像这样单独计算每个单词的出现次数：

int i, j, woccurence, occurences = 0;
string[] words, details = new string[_wordCount];
for (i = 0; i < words.Length; i++)
{
    Console.WriteLine("Type in the censored words you wish to be counted: ");
    words[i] = Console.ReadLine();
    woccurence = 0;
    details = _sentence.Split(' ');
    for (j = 0; j < details.Length; j++)
        if (details[j] == words[i])
            woccurences++;
    Console.WriteLine("Number of censored word occurences: " + woccurences);
    occurences += woccurences;
}
Console.WriteLine("Number of total censored words occurences: " + occurences);
return occurences;

详情

在循环的每个循环中，您使用woccurence 计算单词从头开始出现的次数并将其打印出来，然后将此值添加到总数中occurrence。

Answer 2

如果句子由空格分隔的单词组成，您可以使用 string.split 将其制成一个数组，然后您可以遍历该数组。

var sentenceArray = _sentence.Split(new Char [] {' '});

然后在你的主要单词循环中循环遍历句子数组。

Answer 3

我没有看到 _sentence 或 _wordCount 的设置位置，但请告诉我您是否想到了这一点。

警告：我假设用户会在提示时输入每个单词，单词之间有一个空格。您可能需要处理用户使用过多空格、逗号等的情况。

    static int GetWords()
    {
       int occurences = 0;
       int _wordCount = 0;

       string _sentence = "The quick brown fox jumped over the lazy dog.";
       string[] words = new string[_sentence.Length];

       Console.WriteLine("Type in the censored words you wish to be counted: ");
       string censoredWordString = Console.ReadLine();
       string[] censoredWords = censoredWordString.Split(' ');

       for (int i = 0; i < censoredWords.Length; i++)
       {                  
           if (_sentence.Contains(censoredWords[i]))
           {
              occurences++;
           }                       
       }

       Console.WriteLine("Number of censored word occurences: " + occurences);

       return occurences;
     }

Answer 4

你可以试试这个：

int occurences = 0;

string sentence = "This is a test sentence. This sentence is test. This sentence do nothing.";

var sentenceWords = new string(sentence.Where(c => !char.IsPunctuation(c)).ToArray()).Split(' ');

var wordsFound = new Dictionary<string, int>();

Console.WriteLine("Sentence = " + sentence);

while ( true)
{
  Console.WriteLine(Environment.NewLine);
  Console.WriteLine("Type in a censored word you wish to be counted (enter empty to end): ");
  string input = Console.ReadLine();
  if ( input == "" ) break;
  int count = sentenceWords.Count(word => word.ToLower() == input.ToLower());
  if ( count == 0 )
  {
    Console.WriteLine("Can't find \"" + input + "\".");
  }
  else
  {
    Console.WriteLine("Found " + count + " occurences of \"" + input + "\".");
    if ( !wordsFound.ContainsKey(input) )
      wordsFound.Add(input, count);
    occurences += count;
  }
}

Console.WriteLine(Environment.NewLine);
Console.WriteLine("Number of total censored words occurences: " + occurences);
foreach ( var item in wordsFound)
  Console.WriteLine("     " + item.Key + ": " + item.Value);

Console.ReadKey();

Answer 5

您只能使用 LINQ 实现这一目标

int totalWords = 0;
var sentence = "Don't cry because it's over, smile because it happened";

sentence.ToLower().Split(' ').GroupBy(x => x).ToList().ForEach(x=> {
    totalWords += x.Count();
    Console.WriteLine($"{x.Key}: {x.Count()}");
});

Console.WriteLine($"Total words: {totalWords}");

首先我们使用ToLower() se我们可以去掉lowers和uppers

然后我们用.Split(' ')空格分割

现在我们用GroupBy(x=>x)按每个单词分组

我们需要使用ToList() 来转换IGrouping<T>，这样我们就可以使用ForEach 来迭代结果

最后我们打印得到Key的结果，它引用了分组的​​对象，并使用Count()得到组包含的数量

Answer 6

就个人而言，我会使用字典。特别是使用string 和int 的<key, value>。

int occurances = 0;
string[] words = new string[_wordCount];
var results = new Dictionary<string, int>();
var splitSentence = _sentence.Split(' ').ToArray();


for(int i = 0; i < words.Length; i++)
{
    Console.WriteLine("Type in the censored words you wish to be counted: ");
    words[i] = Console.ReadLine();

    if(_sentence.Contains(words[i]))
    {
        if(!results.ContainsKey(words[i]))
        {
            results.Add(words[i], 0);
        }

        for(var j = 0; j < splitSentence.Length; j++)
        {
            if(splitSentence[j] == words[i])
            {
                 results[words[i]]++;
                 occurances++;
            }
        }
    }
}

对于字典，第一个“参数”必须是唯一的，因此任何时候再次出现一个单词，您只需首先检查该键是否存在，它只会添加到计数中（@987654326 中的 value @对）。

Answer 7

使用 LINQ 和正则表达式：

var sentence = "now is the time for all good men to come, to the aid of their country.";
var words = new[] { "time", "to" };

var wordsHS = words.ToHashSet();
var wordRE = new Regex(@"\w+", RegexOptions.Compiled);
var wordCounts = wordRE.Matches(sentence).Cast<Match>().Select(m => m.Value)
                       .Where(w => wordsHS.Contains(w))
                       .GroupBy(w => w)
                       .Select(wg => new { Word = wg.Key, Count = wg.Count() })
                       .ToList();
var total = wordCounts.Sum(wc => wc.Count);