计算模拟线性回归的 RMSE

Question

I am trying to calculate the RMSE for the simulated data. But the output gives NaN for the RMSE. Below is the code I am using.

library(caret)
RMSE <- function(x,y) sqrt(mean((x-y)^2))
sim.regression<-function(n.obs=200,coefficients=c(3,1.5,0,0,2,0,0,0),s.deviation=.1){
  
  n.var=length(coefficients)  
  M=matrix(0,ncol=n.var,nrow=n.obs)
  
  beta=as.matrix(coefficients)
  
  for (i in 1:n.var){
    M[,i]=rnorm(n.obs,0,1)
  }
  
  y=M %*% beta + rnorm(n.obs,0,s.deviation)
  
  train.data<-y[1:150]
  train.data<-data.frame(train.data)
  test.data<-y[151:200]
  test.data<-data.frame(test.data)
  prediction <- predict(lm(y~M),test.data)
  RMSE.data<-RMSE(prediction, test.data$y)
  
  return (list(x=M,y=y,coeff=coefficients, RMSE=RMSE.data))
  
}

set.seed(2000)
sim.regression(100)

Answer 1

Welcome to SO. There were few issues in the code:

• Assuming that you are trying to learn/predict 'y' based on 'M', you have to combine M and y and make a data frame.
• After that only, you should split first 150 for train and remaining for test.
• Then you train on train.data and predict on test.data
• Also, since you have hardcoded [1:150] and [150:200] for train-test split, you will have to pass 200 as in sim.regression(200).

Corrected code below:

library(caret)
RMSE <- function(x,y) sqrt(mean((x-y)^2))
sim.regression<-function(n.obs=200,coefficients=c(3,1.5,0,0,2,0,0,0),s.deviation=.1){
  
  n.var=length(coefficients)  
  M=matrix(0,ncol=n.var,nrow=n.obs)
  
  beta=as.matrix(coefficients)
  
  for (i in 1:n.var){
    M[,i]=rnorm(n.obs,0,1)
  }
  
  y=M %*% beta + rnorm(n.obs,0,s.deviation)
  data<-data.frame(M,y)
  train.data <- data[1:150,]
  test.data<-data[151:200,]
  prediction <- predict(lm(y~., data=train.data),test.data)
  RMSE.data<-RMSE(prediction, test.data$y)
  return (list(x=M,y=y,coeff=coefficients, RMSE=RMSE.data))
  
}

set.seed(2000)
sim.regression(200)

Prints:

$RMSE
0.0755869850491716