如何计算多项式线性回归的误差？

Question

我正在计算我使用的训练数据的错误率。在

我想我计算错误了。公式如下：

y计算如下：

我在函数fitPoly(M)的第49行计算这个值。我相信我计算错了y(x(n))，但我不知道还能做什么。在

---

下面的例子是可验证的。在

import numpy as np
import matplotlib.pyplot as plt

dataTrain = [[2.362761180904257019e-01, -4.108125266714775847e+00],
[4.324296163702689988e-01,  -9.869308732049049127e+00],
[6.023323504115264404e-01,  -6.684279243433971729e+00],
[3.305079685397107614e-01,  -7.897042003779912278e+00],
[9.952423271981121200e-01,  3.710086310489402628e+00],
[8.308127402955634011e-02,  1.828266768673480147e+00],
[1.855495407116576345e-01,  1.039713135916495501e+00],
[7.088332047815845138e-01,  -9.783208407540947560e-01],
[9.475723071629885697e-01,  1.137746192425550085e+01],
[2.343475721257285427e-01,  3.098019704040922750e+00],
[9.338350584099475160e-02,  2.316408265530458976e+00],
[2.107903139601833287e-01,  -1.550451474833406396e+00],
[9.509966727520677843e-01,  9.295029459100994984e+00],
[7.164931165416982273e-01,  1.041025972594300075e+00],
[2.965557300301902011e-03,  -1.060607693351102121e+01]]

def strip(L, xt):
    ret = []
    for i in L:
        ret.append(i[xt])
    return ret

x1 = strip(dataTrain, 0)
y1 = strip(dataTrain, 1)

# HELP HERE

def getY(m, w, D):
    y = w[0]
    y += np.sum(w[1:] * D[:m])
    return y

# HELP ABOVE

def dataMatrix(X, M):
    Z = []
    for x in range(len(X)):
        row = []
        for m in range(M + 1):
            row.append(X[x][0] ** m)
        Z.append(row)
    return Z

def fitPoly(M):
    t = []
    for i in dataTrain:
        t.append(i[1])
    w, _, _, _ = np.linalg.lstsq(dataMatrix(dataTrain, M), t)
    w = w[::-1]
    errTrain = np.sum(np.subtract(t, getY(M, w, x1)) ** 2)/len(x1)
    print('errTrain: %s' % (errTrain))
    return([w, errTrain])

#fitPoly(8)

def plotPoly(w):
    plt.ylim(-15, 15)
    x, y = zip(*dataTrain)
    plt.plot(x, y, 'bo')
    xw = np.arange(0, 1, .001)
    yw = np.polyval(w, xw)
    plt.plot(xw, yw, 'r')

#plotPoly(fitPoly(3)[0])

def bestPoly():
    m = 0
    plt.figure(1)
    plt.xlim(0, 16)
    plt.ylim(0, 250)
    plt.xlabel('M')
    plt.ylabel('Error')
    plt.suptitle('Question 3: training and Test error')
    while m < 16:
        plt.figure(0)
        plt.subplot(4, 4, m + 1)
        plotPoly(fitPoly(m)[0])
        plt.figure(1)
        plt.plot(fitPoly(m)[1])
        #plt.plot(fitPoly(m)[2])
        m+= 1
    plt.figure(3)
    plt.xlabel('t')
    plt.ylabel('x')
    plt.suptitle('Question 3: best-fitting polynomial (degree = 8)')
    plotPoly(fitPoly(8)[0])
    print('Best M: %d\nBest w: %s\nTraining error: %s' % (8, fitPoly(8)[0], fitPoly(8)[1], ))

bestPoly()

Answer 1

更新：这个解决方案使用numpy的np.interp，它将把这些点作为一种“最佳匹配”连接起来。然后我们使用误差函数来找出插值线和每个多项式次数的预测y值之间的差异。在

import numpy as np
import matplotlib.pyplot as plt
import itertools

dataTrain = [
  [2.362761180904257019e-01, -4.108125266714775847e+00],
  [4.324296163702689988e-01,  -9.869308732049049127e+00],
  [6.023323504115264404e-01,  -6.684279243433971729e+00],
  [3.305079685397107614e-01,  -7.897042003779912278e+00],
  [9.952423271981121200e-01,  3.710086310489402628e+00],
  [8.308127402955634011e-02,  1.828266768673480147e+00],
  [1.855495407116576345e-01,  1.039713135916495501e+00],
  [7.088332047815845138e-01,  -9.783208407540947560e-01],
  [9.475723071629885697e-01,  1.137746192425550085e+01],
  [2.343475721257285427e-01,  3.098019704040922750e+00],
  [9.338350584099475160e-02,  2.316408265530458976e+00],
  [2.107903139601833287e-01,  -1.550451474833406396e+00],
  [9.509966727520677843e-01,  9.295029459100994984e+00],
  [7.164931165416982273e-01,  1.041025972594300075e+00],
  [2.965557300301902011e-03,  -1.060607693351102121e+01]
  ]

data = np.array(dataTrain)
data = data[data[:, 0].argsort()]

X,y = data[:, 0], data[:, 1]

fig,ax = plt.subplots(4, 4)
indices = list(itertools.product([0,1,2,3], repeat=2))
for i,loc in enumerate(indices, start=1):
  xx = np.linspace(X.min(), X.max(), 1000)
  yy = np.interp(xx, X, y)
  w = np.polyfit(X, y, i)
  y_pred = np.polyval(w, xx)
  ax[loc].scatter(X, y)
  ax[loc].plot(xx, y_pred)
  ax[loc].plot(xx, yy, 'r ')

  error = np.square(yy - y_pred).sum() / X.shape[0]
  print(error)

plt.show()

打印出来：

^{pr2}$

从视觉上看，它描绘出：

从这里开始，只需将这些错误保存到一个列表中并找到最小值。在

Answer 2

我可以贡献：

  def pol_y(x, w):
        y = 0; power = 0;
        for i in w:
            y += i*(x**power);
            power += 1;
        return y

M是隐式包含的，因为它是w的最后一个索引。所以如果w = [0, 0, 1]，那么{}与f（x）=x^2相同。在

如果要映射dataTrain的第一列：

^{pr2}$

误差可通过以下公式计算

vec_error = [(y1[i] - getY[i])**2 for i in range(0, len(y1)];
train_error = np.sum(vec_error)/len(y1);

希望这有帮助。在