在运行以下R代码时:library(glmnet) library(ggplot2) # 生成5030的随机数据和30个变量 set.seed(1111) n <- 50 p <- 30 X <- matrix(runif(n * p), n, p) y <- rnorm(n) # 生成三组不同系数的线性模型 beta1 <- c(rep(1, 3), rep(0, p - 3)) beta2 <- c(rep(0, 10), rep(1, 3), rep(0, p - 13)) beta3 <- c(rep(0, 20), rep(1, 3), rep(0, p - 23)) y1 <- X %% beta1 + rnorm(n) y2 <- X %% beta2 + rnorm(n) y3 <- X %% beta3 + rnorm(n) # 设置交叉验证折数 k <- 10 # 设置不同的lambda值 lambda_seq <- 10^seq(10, -2, length.out = 100) # 执行交叉验证和岭回归,并记录CV error和Prediction error cv_error <- list() pred_error <- list() for (i in 1:3) { # 交叉验证 cvfit <- cv.glmnet(X, switch(i, y1, y2, y3), alpha = 0, lambda = lambda_seq, nfolds = k) cv_error[[i]] <- cvfit$cvm # 岭回归 fit <- glmnet(X, switch(i, y1, y2, y3), alpha = 0, lambda = lambda_seq) pred_error[[i]] <- apply(X, 2, function(x) { x_mat <- matrix(x, nrow = n, ncol = p, byrow = TRUE) pred <- predict(fit, newx = x_mat) pred <- t(pred) # 转置 mean((x_mat %% fit$beta - switch(i, y1, y2, y3))^2, na.rm = TRUE) # 修改此处 }) } # 绘制图形 par(mfrow = c(3, 2), mar = c(4, 4, 2, 1), oma = c(0, 0, 2, 0)) for (i in 1:3) { # CV error plot plot(log10(lambda_seq), cv_error[[i]], type = "l", xlab = expression(log10), ylab = "CV error", main = paste0("Model ", i)) abline(v = log10(cvfit$lambda.min), col = "red") # Prediction error plot plot(log10(lambda_seq), pred_error[[i]], type = "l", xlab = expression(log10), ylab = "Prediction error", main = paste0("Model ", i)) abline(v = log10(lambda_seq[which.min(pred_error[[i]])]), col = "red") }。发生了以下错误:Error in xy.coords(x, y, xlabel, ylabel, log) : 'x'和'y'的长度不一样。请对原代码进行修正
时间: 2023-09-22 17:05:44 浏览: 71
一个简单的R语言数据分析案例.pdf
这错误是因为在绘制图形时,x和y的长度不一致。在这段代码中,绘制函数的x和y参数都是cv_error[[i]],但是在第二个绘图中,应该将cv_error[[i]]替换为pred_error[[i]]。修改后的代码如下:
```
par(mfrow = c(3, 2), mar = c(4, 4, 2, 1), oma = c(0, 0, 2, 0))
for (i in 1:3) {
# CV error plot
plot(log10(lambda_seq), cv_error[[i]], type = "l", xlab = expression(log10), ylab = "CV error", main = paste0("Model ", i))
abline(v = log10(cvfit$lambda.min), col = "red")
# Prediction error plot
plot(log10(lambda_seq), pred_error[[i]], type = "l", xlab = expression(log10), ylab = "Prediction error", main = paste0("Model ", i))
abline(v = log10(lambda_seq[which.min(pred_error[[i]])]), col = "red")
}
```
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