在这样的应用程序中,将要优化其值的参数(在您的情况下为cost、gamma 和epsilon)作为适应度函数的参数传递,然后运行模型拟合+评估函数并使用模型性能的度量作为适应度的度量。因此,目标函数的显式形式并不直接相关。
在下面的实现中,我使用 5 折交叉验证来估计给定参数集的 RMSE。特别是,由于包GA 最大化了适应度函数,我将给定参数值的适应度值写为减去交叉验证数据集的平均 rmse。因此,可以达到的最大适应度为零。
这里是:
library(e1071)
library(GA)
data(Ozone, package="mlbench")
Data <- na.omit(Ozone)
# Setup the data for cross-validation
K = 5 # 5-fold cross-validation
fold_inds <- sample(1:K, nrow(Data), replace = TRUE)
lst_CV_data <- lapply(1:K, function(i) list(
train_data = Data[fold_inds != i, , drop = FALSE],
test_data = Data[fold_inds == i, , drop = FALSE]))
# Given the values of parameters 'cost', 'gamma' and 'epsilon', return the rmse of the model over the test data
evalParams <- function(train_data, test_data, cost, gamma, epsilon) {
# Train
model <- svm(V4 ~ ., data = train_data, cost = cost, gamma = gamma, epsilon = epsilon, type = "eps-regression", kernel = "radial")
# Test
rmse <- mean((predict(model, newdata = test_data) - test_data$V4) ^ 2)
return (rmse)
}
# Fitness function (to be maximized)
# Parameter vector x is: (cost, gamma, epsilon)
fitnessFunc <- function(x, Lst_CV_Data) {
# Retrieve the SVM parameters
cost_val <- x[1]
gamma_val <- x[2]
epsilon_val <- x[3]
# Use cross-validation to estimate the RMSE for each split of the dataset
rmse_vals <- sapply(Lst_CV_Data, function(in_data) with(in_data,
evalParams(train_data, test_data, cost_val, gamma_val, epsilon_val)))
# As fitness measure, return minus the average rmse (over the cross-validation folds),
# so that by maximizing fitness we are minimizing the rmse
return (-mean(rmse_vals))
}
# Range of the parameter values to be tested
# Parameters are: (cost, gamma, epsilon)
theta_min <- c(cost = 1e-4, gamma = 1e-3, epsilon = 1e-2)
theta_max <- c(cost = 10, gamma = 2, epsilon = 2)
# Run the genetic algorithm
results <- ga(type = "real-valued", fitness = fitnessFunc, lst_CV_data,
names = names(theta_min),
min = theta_min, max = theta_max,
popSize = 50, maxiter = 10)
summary(results)
产生结果(对于我指定的参数值范围,可能需要根据数据进行微调):
GA results:
Iterations = 100
Fitness function value = -14.66315
Solution =
cost gamma epsilon
[1,] 2.643109 0.07910103 0.09864132