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."

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

var words = text.Split(' ');

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

我尝试了多种操作，例如将text.Replace(characters)与whitespace分开。我已经尝试过Regex（我不想使用）。

我还尝试过使用StringBuilder类从文本（字符串）中获取字符，并且仅在字母a-z / A-Z后面附加字符。

还尝试调用sb.Replace或sb.Remove将不需要的字符存储在字典中。但是我似乎仍然会遇到不需要的角色？

我尝试的所有内容，似乎都至少有一个我不想要的角色，并且无法弄清楚为什么它不起作用。

谢谢！

Answer 1

使用没有RegEx或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;
  }
}

它解析所有拒绝所有非字母的字符，同时为每个单词创建一个字典，统计出现次数。

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

我确实认为，就性能和清晰度而言，使用字典来计数频率的解决方案是最佳实践。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

使用正则表达式排除非字母字符。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();

如何通过删除字符串的非字母来计算单词的频率？

3 个答案: