Question

我已经使用Gradient下降算法堆积了一周。我使用Processing 2.2.1进行编码，我希望在收敛之前显示动画。我不擅长英语，所以很难描述这个问题。如果你在下面构建代码会很棒。

使用渐变下降与处理的线性回归

[步骤]

点击鼠标15次 - ＆gt; x，y鼠标坐标成为线性回归的输入数据
按键盘上的按钮
开始线性回归

[问题]

成本函数的价值......没有收敛，除此之外我认为价值是大的。参数theta0,1 ...需要花费很多时间来适应输入数据。

这是我的代码。我的算法在draw（）中有什么问题吗？

int num = 15;
double[] x = new double[num];
double[] y = new double[num];
float[] xx = new float[num];
float[] yy = new float[num];
double[] hy = new double[num];
float[] hy1 = new float[num];
float theta0=0;
float theta1, theta2=0;
float alpha=1.0E-4;
float pd0, pd1, pdd= 0;
int r=0;
int width=600, height=600;
int count, check= 0;
float temp0, temp1, temp2 =0;
boolean flag;
float xmin, hymin, xmax, hymax, ymax, ymin, hy1max, hy1min=0;
float xsum, ysum, xsd, ysd, xave, yave=0;

void setup() {
  frameRate(300);
  size(width, height);
  // scale(1, -1);
  noStroke();
  smooth();
  fill(255);
  theta0=0;
  theta1=0;
  //  theta2=0;
  pd0=0;
  pd1=0;
  pdd=0;
}


void draw() {

  background(0);
  coordinates();

  //Center
  translate(width/2, height/2);

  for (int ii = 0; ii<num; ii++) {
    if (x[ii]!=0&&y[ii]!=0) {
      fill(255);
      ellipse((float)x[ii], (float)-y[ii], 5, 5);
    }
  }

  //noLoop();
  if (flag==true) {

    for (int i =0; i<num; i++) {
      hy[i]=theta0+theta1*x[i];
    }

    for (int i =0; i<num; i++) {
      pd0+=(hy[i]-y[i]); 
      pd1+=(hy[i]-y[i])*x[i];
      pdd+=(hy[i]-y[i])*(hy[i]-y[i]);
    }

    pdd=pdd/(2.00*num);

    if ( !((theta0-temp0)<=1.0E-4 && (theta0-temp0)>=-1.0E-4)||((theta1-temp1)<=1.0E-4 && (theta1-temp1)>=-1.0E-4)) {
      if (check!=0)
        if (check<=500) 

      temp0=theta0-(alpha/int(num))*pd0;
      temp1=theta1-(alpha/int(num))*pd1;

      theta0=temp0;
      theta1=temp1;

      check++;
      println(pdd+","+check);

      for (int i =0; i<num; i++) {
        fill(255, 0, 0);
        println("x"+ x[i] +","+"y:"+y[i]+ "hy"+hy[i]+"cost F:"+pd0, pd1+"theta0,1:"+theta0+","+theta1+":alpha"+alpha+"check:"+check+"pdd:"+","+pdd);
        ellipse((int)x[i], (int)-hy[i], 3, 3);
      }
    }//if fin
    else {
      println("fin: pd is "+pd0, pd1+"Theta0"+theta0+", Theta1:"+theta1);
      for (int i =0; i<num; i++) {
        fill(255, 0, 0);       
        ellipse((int)x[i], (int)-hy[i], 5, 5);
      }

      noLoop();
    }
  }
}

void coordinates() {

  for (int i=-width; i<width; i+=20) {
    for (int j=-height; j<height; j+=20) {
      stroke(255, 2);
      strokeWeight(1/2);
      line(0, height/2+j, width, height/2+j);
      line(i, 0, i, height);
    }
  }
  noStroke();
  stroke(255, 0, 0);
  strokeWeight(2);
  line(0, height/2, width, height/2);
  stroke(0, 255, 0);
  line(width/2, 0, width/2, height);
  noStroke();
}

void mousePressed() {

  y[count] =   -mouseY+height/2;
  x[count] =   constrain(mouseX-width/2, -width/2, width/2);
  count++;
  if (count==num) {
    //flag=true;

    //normalization false
    for (int i =0; i<num; i++) {
      hy[i]=theta0+theta1*x[i];
      xsum+=x[i];
      ysum+=y[i];
    }

    for (int i =0; i<num; i++) {
      println(x[i]+","+y[i]);
    }

    println("COSTf"+pd0+","+pd1);
    println("COMPLETE");
  }
  if (count>num) {
    count = 0;
    // initsa();
    flag=false;
  }
}

void keyPressed() {
  if (key=='s') {
    background(0);
    for (int i=0; i<num; i++) {
      println(x[i]+","+y[i]);
    }
  }
  if (key=='q') {
    background(0);
    // initsa();
  }
  if (key=='w') {
    flag=true;
  }
  if (key=='d') {
    flag=false;
  }
  if (keyCode==UP) {
    alpha+=0.00001;
  }
  if (keyCode==DOWN) {
    alpha-=0.00001;
  }
}

Answer 1

关于使用梯度下降的多元线性回归的文章，http://melwin-jose.blogspot.com/2015/01/multivariate-linear-regression.html。来自HackerRank的sample code problem。我希望这会有所帮助。

（处理）梯度下降算法不起作用..（线性回归）

1 个答案: