3

有一些关于如何在 C# 中计算词频的好例子，但没有一个是全面的，我真的需要一个在 VB.NET 中。

我目前的方法仅限于每个频率计数一个单词。什么是最好的方法来改变这个，这样我就可以获得一个完全准确的词频列表？

wordFreq = New Hashtable()

Dim words As String() = Regex.Split(inputText, "(\W)")
    For i As Integer = 0 To words.Length - 1
        If words(i) <> "" Then
            Dim realWord As Boolean = True
            For j As Integer = 0 To words(i).Length - 1
                If Char.IsLetter(words(i).Chars(j)) = False Then
                    realWord = False
                End If
            Next j

            If realWord = True Then
                If wordFreq.Contains(words(i).ToLower()) Then
                    wordFreq(words(i).ToLower()) += 1
                Else
                    wordFreq.Add(words(i).ToLower, 1)
                End If
            End If
        End If
    Next

Me.wordCount = New SortedList

For Each de As DictionaryEntry In wordFreq
        If wordCount.ContainsKey(de.Value) = False Then
            wordCount.Add(de.Value, de.Key)
        End If
Next

我更喜欢实际的代码片段，但通用的“哦，是的......使用这个并运行那个”也可以。

4

4 回答 4

3

这可能是您正在寻找的：

    Dim Words = "Hello World ))))) This is a test Hello World"
    Dim CountTheWords = From str In Words.Split(" ") _
                        Where Char.IsLetter(str) _
                        Group By str Into Count()

我刚刚测试过它确实有效

编辑！我添加了代码以确保它只计算字母而不计算符号。

仅供参考：我发现一篇关于如何使用 LINQ 和目标 2.0 的文章，感觉有点脏，但它可能对某人有所帮助http://weblogs.asp.net/fmarguerie/archive/2007/09/05/linq-support-on -net-2-0.aspx

于 2008-10-22T04:48:49.953 回答
2 
Public Class CountWords

    Public Function WordCount(ByVal str As String) As Dictionary(Of String, Integer)
        Dim ret As Dictionary(Of String, Integer) = New Dictionary(Of String, Integer)

        Dim word As String = ""
        Dim add As Boolean = True
        Dim ch As Char

        str = str.ToLower
        For index As Integer = 1 To str.Length - 1 Step index + 1
            ch = str(index)
            If Char.IsLetter(ch) Then
                add = True
                word += ch
            ElseIf add And word.Length Then
                If Not ret.ContainsKey(word) Then
                    ret(word) = 1
                Else
                    ret(word) += 1
                End If
                word = ""
            End If
        Next

        Return ret
    End Function

End Class

然后，对于一个快速演示应用程序，创建一个带有一个名为 InputBox 的多行文本框、一个名为 OutputList 的列表视图和一个名为 CountBtn 的按钮的 winforms 应用程序。在列表视图中创建两列 - “Word”和“Freq”。选择“详细信息”列表类型。为 CountBtn 添加事件处理程序。然后使用此代码：

Imports System.Windows.Forms.ListViewItem

Public Class MainForm

    Private WordCounts As CountWords = New CountWords

    Private Sub CountBtn_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles CountBtn.Click
        OutputList.Items.Clear()
        Dim ret As Dictionary(Of String, Integer) = Me.WordCounts.WordCount(InputBox.Text)
        For Each item As String In ret.Keys
            Dim litem As ListViewItem = New ListViewItem
            litem.Text = item
            Dim csitem As ListViewSubItem = New ListViewSubItem(litem, ret.Item(item).ToString())

            litem.SubItems.Add(csitem)
            OutputList.Items.Add(litem)

            Word.Width = -1
            Freq.Width = -1
        Next
    End Sub
End Class

你做了一件可怕的事情让我用VB写这个，我永远不会原谅你。

:p

祝你好运！

编辑

修复了空白字符串错误和大小写错误

于 2008-10-22T05:16:59.727 回答
1

这可能会有所帮助：

自然语言处理的词频算法

于 2008-10-22T04:46:21.440 回答
1

非常接近，但 \w+ 是一个很好的匹配正则表达式（仅匹配单词字符）。

Public Function CountWords(ByVal inputText as String) As Dictionary(Of String, Integer)
    Dim frequency As New Dictionary(Of String, Integer)

    For Each wordMatch as Match in Regex.Match(inputText, "\w+")
        If frequency.ContainsKey(wordMatch.Value.ToLower()) Then
            frequency(wordMatch.Value.ToLower()) += 1
        Else
            frequency.Add(wordMatch.Value.ToLower(), 1)
        End If
    Next
    Return frequency
End Function
于 2008-10-22T05:23:33.287 回答