I have an example dataset below.
train<-data.frame(x1 = c(4,5,6,4,3,5), x2 = c(4,2,4,0,5,4), x3 = c(1,1,1,0,0,1),
x4 = c(1,0,1,1,0,0), x5 = c(0,0,0,1,1,1))
Suppose I want to create separate models for column x3, x4, x5 based on column x1 and x2. For example
lm1 <- lm(x3 ~ x1 + x2)
lm2 <- lm(x4 ~ x1 + x2)
lm3 <- lm(x5 ~ x1 + x2)
I want to then take these models and apply them to a testing set using predict, and then create a matrix that has each model outcome as a column.
test <- data.frame(x1 = c(4,3,2,1,5,6), x2 = c(4,2,1,6,8,5))
p1 <- predict(lm1, newdata = test)
p2 <- predict(lm2, newdata = test)
p3 <- predict(lm3, newdata = test)
final <- cbind(p1, p2, p3)
This is a simplified version where you can do it step by step, the actual data is far too large. Is there a way to create a function or use a for statement to combine this into one or two steps?
I had an inclination to close your question as a duplicate to Fitting a linear model with multiple LHS, but sadly the prediction issue is not addressed over there. On the other hand, Prediction of 'mlm' linear model object from lm() talks about prediction, but is a little bit far off your situation, as you work with formula interface instead of matrix interface.
I did not manage to locate a perfect duplicate target in "mlm" tag. So I think it a good idea to contribute another answer for this tag. As I said in linked questions, predict.mlm does not support se.fit, and at the moment, this is also a missing issue in "mlm" tag. So I would take this chance to fill such gap.
Here is a function to get standard error of prediction:
f <- function (mlmObject, newdata) {
## model formula
form <- formula(mlmObject)
## drop response (LHS)
form[[2]] <- NULL
## prediction matrix
X <- model.matrix(form, newdata)
Q <- forwardsolve(t(qr.R(mlmObject$qr)), t(X))
## unscaled prediction standard error
unscaled.se <- sqrt(colSums(Q ^ 2))
## residual standard error
sigma <- sqrt(colSums(residuals(mlmObject) ^ 2) / mlmObject$df.residual)
## scaled prediction standard error
tcrossprod(unscaled.se, sigma)
}
For your given example, you can do
## fit an `mlm`
fit <- lm(cbind(x3, x4, x5) ~ x1 + x2, data = train)
## prediction (mean only)
pred <- predict(fit, newdata = test)
# x3 x4 x5
#1 0.555956679 0.38628159 0.60649819
#2 0.003610108 0.47653430 0.95848375
#3 -0.458483755 0.48014440 1.27256318
#4 -0.379061372 -0.03610108 1.35920578
#5 1.288808664 0.12274368 0.17870036
#6 1.389891697 0.46570397 0.01624549
## prediction error
pred.se <- f(fit, newdata = test)
# [,1] [,2] [,3]
#[1,] 0.1974039 0.3321300 0.2976205
#[2,] 0.3254108 0.5475000 0.4906129
#[3,] 0.5071956 0.8533510 0.7646849
#[4,] 0.6583707 1.1077014 0.9926075
#[5,] 0.5049637 0.8495959 0.7613200
#[6,] 0.3552794 0.5977537 0.5356451
We can verify that f is correct:
## `lm1`, `lm2` and `lm3` are defined in your question
predict(lm1, test, se.fit = TRUE)$se.fit
# 1 2 3 4 5 6
#0.1974039 0.3254108 0.5071956 0.6583707 0.5049637 0.3552794
predict(lm2, test, se.fit = TRUE)$se.fit
# 1 2 3 4 5 6
#0.3321300 0.5475000 0.8533510 1.1077014 0.8495959 0.5977537
predict(lm3, test, se.fit = TRUE)$se.fit
# 1 2 3 4 5 6
#0.2976205 0.4906129 0.7646849 0.9926075 0.7613200 0.5356451