创建具有因子的矩阵lm模型

Question

I'd like to create 100 simulations and create a linear model for each model using a matrix. I used the following code, but got an error. What is the best way to accomplish this?

n=15;nsims=100
factor=matrix(as.factor(rep(1:5,3)),n, nsims)
sim=matrix(rnorm(nfactor*nsims,0,1),n,nsims)
> dim(factor)
[1]  15 100
> lm1=lm(sim~factor)
Error in `[[<-.data.frame`(`*tmp*`, i, value = c(1L, 2L, 3L, 4L, 5L, 1L,  : 
  replacement has 1500 rows, data has 15

Answer 1

I'm not clear what you want.

Create data:

n <- 15; nsims <- 100; nfactor <- 5
f <- matrix(as.factor(rep(1:nfactor,n/nfactor)), n, nsims)
set.seed(101)
sim <- matrix(rnorm(nfactor*nsims,0,1),n,nsims)

(In general it's bad practice to name variables with the names of existing functions ( factor ) ...)

res <- vector("list", nsims)
for (i in 1:nsims) {
   res[[i]] <- lm(y~x,
                  data=data.frame(y=sim[,i],x=f[,i]))
}

A for loop may seem clunky, but (as discussed in chapter 4 of the R Inferno ) it's often the clearest way to write your code ... If you really want to do it without a for loop, inu the general case where the columns of f might differ, you could do

res2 <- mapply(function(x,y) lm(y~x),
            split(f,col(f)),split(sim,col(sim)),SIMPLIFY=FALSE)

but I consider that less clear than the for loop ...

update : I don't know whether it was on purpose or not, but every column of your factor matrix is identical. Thus you can put the single input variable on the right-hand side of the formula and the matrix response on the left-hand side:

all(apply(f[,-1],2,identical,f[,1]))  ## all TRUE
res2 <- lm(sim~f[,1])