我正在尝试编写两个演示局部加权线性回归的脚本。在第一个脚本中,我使用了Numpy来解决矩阵问题,代码如下:
trX = np.linspace(0, 1, 100)
trY= trX + np.random.normal(0,1,100)
xArr = []
yArr = []
for i in range(len(trX)):
xArr.append([1.0,float(trX[i])])
yArr.append(float(trY[i]))
xMat = mat(xArr);
yMat = mat(yArr).T
m = shape(xMat)[0]
weights = mat(eye((m)))
k = 0.01
yHat = zeros(m)
for i in range(m):
for j in range(m):
diffMat = xArr[i] - xMat[j,:]
weights[j,j] = exp(diffMat*diffMat.T/(-2.0*k**2))
xTx = xMat.T * (weights * xMat)
if linalg.det(xTx) == 0.0:
print("This matrix is singular, cannot do inverse")
ws = xTx.I * (xMat.T * (weights * yMat))
yHat[i] = xArr[i]*ws
plt.scatter(trX, trY)
plt.plot(trX, yHat, 'r')
plt.show()
如果运行上述脚本,结果如下:
![在此输入图片描述](https://istack.dev59.com/Gkpqh.webp)
trX = np.linspace(0, 1, 100)
trY= trX + np.random.normal(0,1,100)
sess = tf.Session()
xArr = []
yArr = []
for i in range(len(trX)):
xArr.append([1.0,float(trX[i])])
yArr.append(float(trY[i]))
xMat = mat(xArr);
yMat = mat(yArr).T
A_tensor = tf.constant(xMat)
b_tensor = tf.constant(yMat)
m = shape(xMat)[0]
weights = mat(eye((m)))
k = 0.01
yHat = zeros(m)
for i in range(m):
for j in range(m):
diffMat = xMat[i]- xMat[j,:]
weights[j,j] = exp(diffMat*diffMat.T/(-2.0*k**2))
weights_tensor = tf.constant(weights)
# Matrix inverse solution
wA = tf.matmul(weights_tensor, A_tensor)
tA_A = tf.matmul(tf.transpose(A_tensor), wA)
tA_A_inv = tf.matrix_inverse(tA_A)
wb = tf.matmul(weights_tensor, b_tensor)
tA_wb = tf.matmul(tf.transpose(A_tensor), wb)
solution = tf.matmul(tA_A_inv, tA_wb)
sol_val = sess.run(solution)
yHat[i] =sol_val[0][0]*xArr[i][1] + sol_val[1][0]
plt.scatter(trX, trY)
plt.plot(trX, yHat, 'r')
plt.show()
如果运行它:
两个结果之间的差异是由什么造成的?或者我的脚本有错误吗?请帮帮我。