Question

我已经调整了以下梯度下降算法，用于回归存储在数据[：，1]中的x变量的数据[：，4]中存储的y变量。然而，梯度下降似乎是分歧的。我很感激帮助我找出错误的地方。

#define the sum of squared residuals
ssquares <- function(x) 
  {
    t = 0
    for(i in 1:200)
      {
        t <- t + (data[i,4] - x[1] - x[2]*data[i,1])^2 
      }
    t/200
  }

# define the derivatives
derivative <- function(x) 
  {
    t1 = 0
    for(i in 1:200)
      {
        t1 <- t1 - 2*(data[i,4] - x[1] - x[2]*data[i,1]) 
      }
    t2 = 0
    for(i in 1:200)
      {
      t2 <- t2 - 2*data[i,1]*(data[i,4] - x[1] - x[2]*data[i,1]) 
      }
   c(t1/200,t2/200)
  }

# definition of the gradient descent method in 2D
gradient_descent <- function(func, derv, start, step=0.05, tol=1e-8) {
  pt1 <- start
  grdnt <- derv(pt1)
  pt2 <- c(pt1[1] - step*grdnt[1], pt1[2] - step*grdnt[2])
  while (abs(func(pt1)-func(pt2)) > tol) {
    pt1 <- pt2
    grdnt <- derv(pt1)
    pt2 <- c(pt1[1] - step*grdnt[1], pt1[2] - step*grdnt[2])
    print(func(pt2)) # print progress
  }
  pt2 # return the last point
}

# locate the minimum of the function using the Gradient Descent method
result <- gradient_descent(
  ssquares, # the function to optimize
  derivative, # the gradient of the function
  c(1,1), # start point of theplot_loss(simple_ex)  search 
  0.05, # step size (alpha)
  1e-8) # relative tolerance for one step

# display a summary of the results
print(result) # coordinate of fucntion minimum
print(ssquares(result)) # response of function minimum

Answer 1

您可以对目标/渐变函数进行矢量化以便更快地实现，因为您可以看到它实际上收敛于随机生成的数据，并且系数非常接近于在R中使用lm（）获得的系数：

ssquares <- function(x) {
  n <- nrow(data) # 200
  sum((data[,4] - cbind(1, data[,1]) %*% x)^2) / n
}

# define the derivatives
derivative <- function(x) {
  n <- nrow(data) # 200
  c(sum(-2*(data[,4] - cbind(1, data[,1]) %*% x)), sum(-2*(data[,1])*(data[,4] - cbind(1, data[,1]) %*% x))) / n
}

set.seed(1)
#data <- matrix(rnorm(800), nrow=200)

# locate the minimum of the function using the Gradient Descent method
result <- gradient_descent(
  ssquares, # the function to optimize
  derivative, # the gradient of the function
  c(1,1), # start point of theplot_loss(simple_ex)  search 
  0.05, # step size (alpha)
  1e-8) # relative tolerance for one step

# [1] 2.511904
# [1] 2.263448
# [1] 2.061456
# [1] 1.89721
# [1] 1.763634
# [1] 1.654984
# [1] 1.566592
# [1] 1.494668
# ...

# display a summary of the results
print(result) # coefficients obtained with gradient descent
#[1] -0.10248356  0.08068382

lm(data[,4]~data[,1])$coef # coefficients from R lm()
# (Intercept)   data[, 1] 
# -0.10252181  0.08045722 

# use new dataset, this time it takes quite sometime to converge, but the 
# values GD converges to are pretty accurate as you can see from below.
data <- read.csv('Advertising.csv') # with advertising data, removing the first rownames column

# locate the minimum of the function using the Gradient Descent method
result <- gradient_descent(
  ssquares, # the function to optimize
  derivative, # the gradient of the function
  c(1,1), # start point of theplot_loss(simple_ex)  search 
  0.00001, # step size (alpha), decreasing the learning rate
  1e-8) # relative tolerance for one step

# ...
# [1] 10.51364
# [1] 10.51364
# [1] 10.51364

print(result) # coordinate of fucntion minimum
[1] 6.97016852 0.04785365

lm(data[,4]~data[,1])$coef
(Intercept)   data[, 1] 
 7.03259355  0.04753664

为什么我的梯度下降在R中的回归失败？

1 个答案: