目标是使用grplasso 来挑选重要变数 就是使用group lasso 我的资料是response 是连续的 是房价 regressor 是 类别的 像是建物型态 ex 公寓大厦 高楼大厦 透天厝等等 我的资料类似长这样 因为有很多个变数 所以要使用group lasso 房价 建物型态 坪数 100 公寓大厦 20 但是我看不懂 那个 grplasso 的 index怎麽用 不知道可不可以请强者稍微示范一下 程式码该怎麽打 下面附上范例的code ## Use the Logistic Group Lasso on the splice data set data(splice) ## Define a list with the contrasts of the factors contr <- rep(list("contr.sum"), ncol(splice) - 1) names(contr) <- names(splice)[-1] ## Fit a logistic model fit.splice <- grplasso(y ~ ., data = splice, model = LogReg(), lambda = 20, contrasts = contr, standardize = TRUE) ## Perform the Logistic Group Lasso on a random dataset set.seed(79) n <- 50 ## observations p <- 4 ## variables ## First variable (intercept) not penalized, two groups having 2 degrees ## of freedom each index <- c(NA, 2, 2, 3, 3) 主要就是这行看不懂 看起来好像是 X1 X2 X3 X4 分别有2,2,3,3个 level 可是我去明明 x1 x2 x3 都是连续的阿 ## Create a random design matrix, including the intercept (first column) x <- cbind(1, matrix(rnorm(p * n), nrow = n)) colnames(x) <- c("Intercept", paste("X", 1:4, sep = ""))lambdamax 9 par <- c(0, 2.1, -1.8, 0, 0) prob <- 1 / (1 + exp(-x %*% par)) mean(pmin(prob, 1 - prob)) ## Bayes risk y <- rbinom(n, size = 1, prob = prob) ## binary response vector ## Use a multiplicative grid for the penalty parameter lambda, starting ## at the maximal lambda value lambda <- lambdamax(x, y = y, index = index, penscale = sqrt, model = LogReg()) * 0.9^(0:30) ## Fit the solution path on the lambda grid fit <- grplasso(x, y = y, index = index, lambda = lambda, model = LogReg(), penscale = sqrt, control = grpl.control(update.hess = "lambda", trace = 0)) ## Plot coefficient paths plot(fit) --

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