如何通过删除字符串的非字母来计算单词频率？

Question

我有一个字符串：

var text = @"
I have a long string with a load of words,
and it includes new lines and non-letter characters.
I want to remove all of them and split this text to have one word per line, then I can count how many of each word exist."

go 关于删除所有非字母字符，然后将每个单词拆分到新行以便我可以存储和计算每个单词有多少的最佳方法是什么？

var words = text.Split(' ');

foreach(var word in words)
{
    word.Trim(',','.','-');
}

我尝试了各种方法，例如text.Replace(characters)用whitespace然后拆分。 我已经尝试过正则表达式（我不想使用它）。

我还尝试使用 StringBuilder class 从文本（字符串）中获取字符，并且仅在它是字母 az / AZ 时附加字符。

还尝试调用 sb.Replace 或 sb.Remove 我不想要的字符，然后再将它们存储在字典中。 但我似乎最终还是得到了我不想要的角色？

我尝试的一切，我似乎至少有一个我不想要的角色，并且无法完全弄清楚为什么它不起作用。

谢谢！

Answer 1

使用没有 RegEx 和 Linq 的扩展方法

static public class StringHelper
{
  static public Dictionary<string, int> CountDistinctWords(this string text)
  {
    string str = text.Replace(Environment.NewLine, " ");
    var words = new Dictionary<string, int>();
    var builder = new StringBuilder();
    char charCurrent;
    Action processBuilder = () =>
    {
      var word = builder.ToString();
      if ( !string.IsNullOrEmpty(word) )
        if ( !words.ContainsKey(word) )
          words.Add(word, 1);
        else
          words[word]++;
    };
    for ( int index = 0; index < str.Length; index++ )
    {
      charCurrent = str[index];
      if ( char.IsLetter(charCurrent) )
        builder.Append(charCurrent);
      else
      if ( !char.IsNumber(charCurrent) )
        charCurrent = ' ';
      if ( char.IsWhiteSpace(charCurrent) )
      {
        processBuilder();
        builder.Clear();
      }
    }
    processBuilder();
    return words;
  }
}

它解析所有字符，拒绝所有非字母，同时创建每个单词的字典，并计算出现次数。

测试

var result = text.CountDistinctWords();
Console.WriteLine($"Found {result.Count()} distinct words:");
Console.WriteLine();
foreach ( var item in result )
  Console.WriteLine($"{item.Key}: {item.Value}");

样品结果

Found 36 distinct words:

I: 3
have: 2
a: 2
long: 1
string: 1
with: 1
load: 1
of: 3
words: 1
and: 3
it: 1
includes: 1
new: 1
lines: 1
non: 1
letter: 1
characters: 1
want: 1
to: 2
remove: 1
all: 1
them: 1
split: 1
this: 1
text: 1
one: 1
word: 2
per: 1
line: 1
then: 1
can: 1
count: 1
how: 1
many: 1
each: 1
exist: 1

Answer 2

我确实相信使用字典计算频率的解决方案在性能和清晰度方面是最佳实践。 这是我的版本，与接受的答案略有不同（我使用String.Split()而不是遍历字符串的字符）：

var text = @"
    I have a long string with a load of words,
    and it includes new lines and non-letter characters.
    I want to remove all of them and split this text to have one word per line, then I       can count how many of each word exist.";

var words = text.Split(new [] {',', '.', '-', ' ', '\n', '\r' }, StringSplitOptions.RemoveEmptyEntries);

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

foreach (var word in words)
{
    if (freqByWord.ContainsKey(word))
    {
        freqByWord[word]++; // we found the same word
    }
    else
    {
        freqByWord.Add(word, 1); // we don't have this one yet
    }
}

foreach (var word in freqByWord.Keys)
{
    Console.WriteLine($"{word}: {freqByWord[word]}");
}

结果几乎相同：

I: 3
have: 2
a: 2
long: 1
string: 1
with: 1
load: 1
of: 3
words: 1
and: 3
it: 1
includes: 1
new: 1
lines: 1
non: 1
letter: 1
characters: 1
want: 1
to: 2
remove: 1
all: 1
them: 1
split: 1
this: 1
text: 1
one: 1
word: 2
per: 1
line: 1
then: 1
can: 1
count: 1
how: 1
many: 1
each: 1
exist: 1

Answer 3

使用不包括非字母字符的正则表达式。 这也将为您提供所有单词的集合。

var text = @"
I have a long string with a load of words,
and it includes new lines and non-letter characters.
I want to remove all of them and split this text to have one word per line, then I can count how many of each word exist.";

var words = Regex.Matches(text, @"[A-Za-z ]+").Cast<Match>().SelectMany(n => n.Value.Trim().Split(' '));
int wordCount = words.Count();