c# 模拟线性回归的示例
作者:砖治不服 时间:2022-08-26 02:50:49
最近刚开始接触机器学习,在这里使用c#模拟一元线性回归,先上图看效果
因为源码中有一些控件是自己封装的,所以就不上传可运行的程序集了,贴出核心代码,以供参考,如有不对,请多多给予建议
private void ryButtonX1_Click(object sender, EventArgs e)
{
string[] xnum = richTextBox1.Text.Trim().Split(',');//x值
string[] ynum = richTextBox2.Text.Trim().Split(',');//y值
if (xnum.Length != ynum.Length)
{
MessageBox.Show("输入数据有误!");
return;
}
ryTextBoxX1.Text = xnum.Length+"";//个数
decimal xsum = 0;//x值求和
decimal ysum = 0;//y值求和
for(int i = 0; i < xnum.Length; i++)
{
xsum = xsum + ConvertExtend.ToDecimal(xnum[i],0);
ysum = ysum + ConvertExtend.ToDecimal(ynum[i], 0);
}
decimal xAve = ConvertExtend.ToDecimal(xsum / xnum.Length, 0);//x平均值
decimal yAve = ConvertExtend.ToDecimal(ysum / xnum.Length, 0);//y平均值
ryTextBoxX3.Text = string.Format("{0:N}", xAve);//保留两位小数
ryTextBoxX4.Text = string.Format("{0:N}", yAve);
decimal molecule = 0;//分子
decimal Denominator = 0;//分母
for (int i = 0; i < xnum.Length; i++)
{
molecule = molecule + (ConvertExtend.ToDecimal(xnum[i], 0) - xAve) * (ConvertExtend.ToDecimal(ynum[i], 0) - yAve);
Denominator = Denominator+(ConvertExtend.ToDecimal(xnum[i], 0) - xAve) * (ConvertExtend.ToDecimal(xnum[i], 0) - xAve);
}
ryTextBoxX2.Text = string.Format("{0:N}", molecule / Denominator);//斜率
ryTextBoxX5.Text = (yAve - (molecule / Denominator) * xAve)+"";//截距
if (ConvertExtend.ToDecimal(ryTextBoxX5.Text, 0) < 0)
{
ryTextBoxX6.Text = ryTextBoxX2.Text + "X" + ryTextBoxX5.Text;
}else
{
ryTextBoxX6.Text = ryTextBoxX2.Text + "X+" + ryTextBoxX5.Text;
}
#region 画点
chartLabTrend.ChartAreas[0].AxisX.CustomLabels.Clear();
List<decimal?> lx = new List<decimal?>();
List<decimal?> l1 = new List<decimal?>();
for (int i = 1; i <= xnum.Length; i++)
{
CustomLabel label1 = new CustomLabel();
if (xnum[i - 1] != "")
{
label1.Text = ConvertExtend.ToDecimal(xnum[i - 1],0).ToString();
label1.ToPosition = i * 2;
chartLabTrend.ChartAreas[0].AxisX.CustomLabels.Add(label1);
label1.GridTicks = GridTickTypes.Gridline;
lx.Add(i);
if (ynum[i - 1] == null)
{
l1.Add(null);
}
else
{
l1.Add(ConvertExtend.ToDecimal(ynum[i - 1],0));
}
}
}
chartLabTrend.Series[0].Points.DataBindXY(lx, l1);
#endregion
#region 画线
chartLabTrend.ChartAreas[0].AxisX.CustomLabels.Clear();
List<decimal?> lx1 = new List<decimal?>();
List<decimal?> l11 = new List<decimal?>();
for (int i = 1; i <= xnum.Length; i++)
{
CustomLabel label2 = new CustomLabel();
if (xnum[i - 1] != "")
{
label2.Text = ConvertExtend.ToDecimal(xnum[i - 1], 0).ToString();
label2.ToPosition = i * 2;
chartLabTrend.ChartAreas[0].AxisX.CustomLabels.Add(label2);
label2.GridTicks = GridTickTypes.Gridline;
lx1.Add(i);
if (ynum[i - 1] == null)
{
l11.Add(null);
}
else
{
l11.Add(ConvertExtend.ToDecimal(ConvertExtend.ToDecimal(xnum[i - 1],0)*molecule / Denominator
+ ConvertExtend.ToDecimal(ryTextBoxX5.Text,0), 0));
}
}
}
chartLabTrend.Series[1].Points.DataBindXY(lx1, l11);
#endregion
}
来源:https://www.cnblogs.com/cui0614/p/11293793.html
标签:c#,模拟,线性回归
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