所以我正致力于文本挖掘项目,目前正在努力实现信息收益。我有一个数据,其中每行描述一个文档。所以新行字符会分割不同的文档。
我必须生成一个矩阵,其中列是所有文档中的所有不同单词,行是不同的文档。如果该文档中存在或不存在该单词,则该表中的每个单元格为1(真)或0(假)。 共有987个文件,总字数为22860,总字数为3680.所以3680个单词与22860个相比。这个运行缓慢但很好用。花费更多时间的循环是当我遍历单词列表的对象以生成矩阵时。见下文
注意:我已经删除了文档中所有重复的单词。
class word_list
{
public string word;
public List<bool> doc= new List<bool>();
};//class ends
private void button2_Click(object sender, EventArgs e)
{
//Convert the string into an array of words
string[] w1 = richTextBox1.Text.Trim().Split('\n',' ').Select(x => x.Trim().ToLower()).Distinct().ToArray(); //all distinct words
string[] rich_doc = richTextBox1.Text.Trim().Split('\n'); //all documents array
List<word_list> words = new List<word_list>();
richTextBox2.Text+=("no. of distict words: " + w1.Length + ", no. of docs " + rich_doc.Length);
for (int i = 0; i < w1.Length; i++)
{
word_list temp = new word_list();
temp.word = w1[i]; //temp has the current distict word as class object
for(int j=0;j<rich_doc.Length;j++)//traverse all doc array
{
temp.doc.Add(false);
List<string> doc_word = Regex.Split(rich_doc[j], @"\b").Distinct(StringComparer.CurrentCultureIgnoreCase).ToList();
//richTextBox2.Text += ("\n no. of words in this doc: " + doc_word.Count);
//richTextBox2.SelectionStart = richTextBox1.Text.Length;
//richTextBox2.Focus();
int doc_count = doc_word.Count; // number of docs
for (int k = 0; k < doc_count; k++)//All words in a doc are compared
{
if(doc_word[k].ToLower() == w1[i].ToLower())
{
temp.doc[temp.doc.Count-1]=true;
break;
}
}
}
if ((words.Count - 1)>=0)
richTextBox2.Text += ("\n word(" + words.Count + "/" + w1.Length + "): " + words[words.Count - 1].word);
richTextBox2.SelectionStart = richTextBox1.Text.Length;
richTextBox2.Focus();
words.Add(temp);
}
//generate matrix
int t = rich_doc.Length; //no. of docs
int word_count = words.Count;
richTextBox1.Text = "Doc";
foreach (word_list w in words)
{
richTextBox1.Text += "\t" + w.word;
}
richTextBox1.Text += "\n";
//This loop is slow
for (int i = 0; i < t; i++) //traverse through number of docs
{
richTextBox1.Text += i + 1;
for (int h = 0; h < word_count; h++)//traverse through each distinct word in the list
{
if (words[h].doc[i])
richTextBox1.Text += "\t1";
else
richTextBox1.Text += "\t0";
}
richTextBox1.Text += "\n";
}
}//end of button 2
答案 0 :(得分:2)
ta.speot.is是正确的。字符串应该使用StringBuilder
构建,例如使用Append
,并且只有在循环之后才将字符串分配给richTextBox1.Text
。代码如下所示:
//generate matrix
StringBuilder sb = new StringBuilder();
int t = rich_doc.Length; //no. of docs
int word_count = words.Count;
richTextBox1.Text = "Doc";
foreach (word_list w in words)
{
sb.Append("\t");
sb.Append(w.word);
}
sb.AppendLine();
//This loop is not slow anymore :)
for (int i = 0; i < t; i++) //traverse through number of docs
{
sb.Append(i + 1);
for (int h = 0; h < word_count; h++)//traverse through each distinct word in the list
{
if (words[h].doc[i])
sb.Append("\t1");
else
sb.Append("\t0");
}
sb.AppendLine();
}
richTextBox1.Text = sb.ToString();
编辑:以下是有价值的评论。更改RichEditBox.Text属性是这里最昂贵的操作。