线性回归中测试数据集的负 R 平方？

Question

我正在使用人工数据模拟线性回归，然后手动计算 RSE 和 R 平方。我对训练模型的样本数据集执行此操作，然后在样本外数据集上测试模型。样本外和样本内数据来自相同的正态分布，但种子不同。但是，当涉及到样本外数据集时，我的数字没有意义。你能帮我找到错误吗？

set.seed(1)
z1 <- rnorm(100)
z2 <- z1 ^ 2
error <- rnorm(100, sd = 0.25)
y1 <- 1 + 2 * z1 + error
data1 <- data.table(y1, z1, z2)
model_quad <- lm(y1 ~ z1 + z2, data1)
model_lin <- lm(y1 ~ z1, data1)

confint(model_lin)
confint(model_quad)

summary(model_lin)
summary(model_quad)

ggplot(data1) +
  geom_point(aes(x = z1, y = y1), color = "blue", size = 3) +
  geom_point(aes(x = z2, y = y1), color = "red", size = 3) +
  geom_line(stat = "smooth", method = lm, aes(x = z1, y = y1), color = "blue", size = 2, alpha = 0.5) +
  geom_line(stat = "smooth", method = lm, aes(x = z2, y = y1), color = "red", size = 2, alpha = 0.5) +
  geom_ribbon(stat = "smooth", method = lm, aes(x = z1, y = y1), fill = "blue", alpha = 0.1) +
  geom_ribbon(stat = "smooth", method = lm, aes(x = z2, y = y1), fill = "red", alpha = 0.1) 

set.seed(100)
z12 <- rnorm(100)
z22 <- z12 ^ 2
error2 <- rnorm(100, sd = 0.25)
y2 <- 1 + 2 * z12 + error2
data2 <- data.table(y2, z12, z22)

summary(model_lin)
summary(model_quad)

ggplot(data2) +
  geom_point(aes(x = z12, y = y2), color = "blue", size = 3) +
  geom_point(aes(x = z22, y = y2), color = "red", size = 3) +
  geom_line(stat = "smooth", method = lm, aes(x = z12, y = y2), color = "blue", size = 2, alpha = 0.5) +
  geom_line(stat = "smooth", method = lm, aes(x = z22, y = y2), color = "red", size = 2, alpha = 0.5) +
  geom_ribbon(stat = "smooth", method = lm, aes(x = z12, y = y2), fill = "blue", alpha = 0.1) +
  geom_ribbon(stat = "smooth", method = lm, aes(x = z22, y = y2), fill = "red", alpha = 0.1) +
  geom_abline(intercept = 0.99, slope = 1.999, size = 2, color = "yellow", alpha = 0.3)


predictions_in_sample_linear <-  predict(model_lin, data1)
predictions_in_sample_quadratic <- predict(model_quad, data1)
predictions_out_of_sample_linear <-  predict(model_lin, data2)
predictions_out_of_sample_quadratic <- predict(model_quad, data2)
TSE_in_sample <- (y1 - mean(y1)) %*% (y1 - mean(y1))
RSE_in_sample_linear <- (predictions_in_sample_linear - y1)  %*% (predictions_in_sample_linear - y1) 
RSE_in_sample_quadratic <- (predictions_in_sample_quadratic - y1)  %*% (predictions_in_sample_quadratic - y1) 
R_Square_in_sample_linear <- (TSE_in_sample - RSE_in_sample_linear) / TSE_in_sample
R_Square_in_sample_quadratic<- (TSE_in_sample - RSE_in_sample_quadratic) / TSE_in_sample
TSE_out_of_sample <- (y2 - mean(y2)) %*% (y2 - mean(y2))
RSE_out_of_sample_linear <- (predictions_out_of_sample_linear - y2)  %*% (predictions_out_of_sample_linear - y2) 
RSE_out_of_sample_quadratic <- (predictions_out_of_sample_quadratic - y2)  %*% (predictions_out_of_sample_quadratic - y2) 
R_Square_out_of_sample_linear <- (TSE_out_of_sample - RSE_out_of_sample_linear) / TSE_out_of_sample
R_Square_out_of_sample_quadratic<- (TSE_out_of_sample - RSE_out_of_sample_quadratic) / TSE_out_of_sample

predictions_in_sample_linear 
predictions_in_sample_quadratic 
predictions_out_of_sample_linear 
predictions_out_of_sample_quadratic 
TSE_in_sample 
RSE_in_sample_linear 
RSE_in_sample_quadratic 
R_Square_in_sample_linear 
R_Square_in_sample_quadratic
TSE_out_of_sample 
RSE_out_of_sample_linear 
RSE_out_of_sample_quadratic 
R_Square_out_of_sample_linear 
R_Square_out_of_sample_quadratic

这段代码returnsR_square在Out of Sample数据中是负的，这是荒谬的。

我们将不胜感激您的建议。

Answer 1

长问短答。你应该使用

data2 <- data.frame(y1 = y2, z1 = z12, z2 = z22)

这给出了

RSE_out_of_sample_linear
# 0.9902969

RSE_out_of_sample_quadratic
# 0.989241

线性回归中测试数据集的负 R 平方？

negative R-squared for a test dataset in linear regression?

r

model

regression

linear-regression

lm