Rolling regression and prediction with lm() and pr

Question

I need to apply lm() to an enlarging subset of my dataframe dat, while making prediction for the next observation. For example, I am doing:

fit model      predict
----------     -------
dat[1:3, ]     dat[4, ]
dat[1:4, ]     dat[5, ]
    .             .
    .             .
dat[-1, ]      dat[nrow(dat), ]

I know what I should do for a particular subset (related to this question: predict() and newdata - How does this work?). For example to predict the last row, I do

dat1 = dat[1:(nrow(dat)-1), ]
dat2 = dat[nrow(dat), ]

fit = lm(log(clicks) ~ log(v1) + log(v12), data=dat1)
predict.fit = predict(fit, newdata=dat2, se.fit=TRUE)

How can I do this automatically for all subsets, and potentially extract what I want into a table?

• From fit, I'd need the summary(fit)$adj.r.squared;
• From predict.fit I'd need predict.fit$fit value.

Thanks.

Answer 1

I just made up some random data to use for this example. I'm calling the object data because that was what it was called in the question at the time that I wrote this solution (call it anything you like).

(Efficient) Solution

data <- data.frame(v1=rnorm(100),v2=rnorm(100),clicks=rnorm(100))

data1 = data[1:(nrow(data)-1), ]
data2 = data[nrow(data), ]

for(i in 3:nrow(data)){
  nam  <- paste("predict", i, sep = "")
  nam1 <- paste("fit", i, sep = "")
  nam2 <- paste("summary_fit", i, sep = "")

  fit = lm(clicks ~ v1 + v2, data=data[1:i,])
  tmp  <- predict(fit, newdata=data2, se.fit=TRUE)
  tmp1 <- fit
  tmp2 <- summary(fit)
  assign(nam, tmp)
  assign(nam1, tmp1)
  assign(nam2, tmp2)
}

All of the results you want will be stored in the data objects this creates.

For example:

> summary_fit10$r.squared
[1] 0.3087432

You mentioned in the comments that you'd like a table of results. You can programmatically create tables of results from the 3 types of output files like this:

rm(data,data1,data2,i,nam,nam1,nam2,fit,tmp,tmp1,tmp2)
frames <- ls()

frames.fit     <- frames[1:98] #change index or use pattern matching as needed
frames.predict <- frames[99:196]
frames.sum     <- frames[197:294]

fit.table <- data.frame(intercept=NA,v1=NA,v2=NA,sourcedf=NA)
for(i in 1:length(frames.fit)){
  tmp <- get(frames.fit[i])
  fit.table              <- rbind(fit.table,c(tmp$coefficients[[1]],tmp$coefficients[[2]],tmp$coefficients[[3]],frames.fit[i]))
}

fit.table

> fit.table
             intercept                   v1                   v2 sourcedf
2  -0.0647017971121678     1.34929652763687   -0.300502017324518    fit10
3  -0.0401617893034109   -0.034750571912636  -0.0843076273486442   fit100
4   0.0132968863522573     1.31283604433593   -0.388846211083564    fit11
5   0.0315113918953643     1.31099122173898   -0.371130010135382    fit12
6    0.149582794027583    0.958692838785998   -0.299479715938493    fit13
7  0.00759688947362175    0.703525856001948   -0.297223988673322    fit14
8    0.219756240025917    0.631961979610744   -0.347851129205841    fit15
9     0.13389223748979    0.560583832333355   -0.276076134872669    fit16
10   0.147258022154645    0.581865844000838   -0.278212722024832    fit17
11  0.0592160359650468    0.469842498721747   -0.163187274356457    fit18
12   0.120640756525163    0.430051839741539   -0.201725012088506    fit19
13   0.101443924785995     0.34966728554219   -0.231560038360121    fit20
14  0.0416637001406594    0.472156988919337   -0.247684504074867    fit21
15 -0.0158319749710781    0.451944113682333   -0.171367482879835    fit22
16 -0.0337969739950376    0.423851304105399   -0.157905431162024    fit23
17  -0.109460218252207     0.32206642419212   -0.055331391802687    fit24
18  -0.100560410735971    0.335862465403716  -0.0609509815266072    fit25
19  -0.138175283219818    0.390418411384468  -0.0873106257144312    fit26
20  -0.106984355317733    0.391270279253722  -0.0560299858019556    fit27
21 -0.0740684978271464    0.385267011513678  -0.0548056844433894    fit28

Answer 2

(Efficient) solution

This is what you can do:

p <- 3  ## number of parameters in lm()
n <- nrow(dat) - 1

## a function to return what you desire for subset dat[1:x, ]
bundle <- function(x) {
  fit <- lm(log(clicks) ~ log(v1) + log(v12), data = dat, subset = 1:x, model = FALSE)
  pred <- predict(fit, newdata = dat[x+1, ], se.fit = TRUE)
  c(summary(fit)$adj.r.squared, pred$fit, pred$se.fit)
  }

## rolling regression / prediction
result <- t(sapply(p:n, bundle))
colnames(result) <- c("adj.r2", "prediction", "se")

Note I have done several things inside the bundle function:

• I have used subset argument for selecting a subset to fit
• I have used model = FALSE to not save model frame hence we save workspace

Overall, there is no obvious loop, but sapply is used.

• Fitting starts from p, the minimum number of data required to fit a model with p coefficients;
• Fitting terminates at nrow(dat) - 1, as we at least need the final column for prediction.

---

Test

Example data (with 30 "observations")

dat <- data.frame(clicks = runif(30, 1, 100), v1 = runif(30, 1, 100),
                  v12 = runif(30, 1, 100))

Applying code above gives results (27 rows in total, truncated output for 5 rows)

            adj.r2 prediction        se
 [1,]          NaN   3.881068       NaN
 [2,]  0.106592619   3.676821 0.7517040
 [3,]  0.545993989   3.892931 0.2758347
 [4,]  0.622612495   3.766101 0.1508270
 [5,]  0.180462206   3.996344 0.2059014

The first column is the adjusted-R.squared value for fitted model, while the second column is the prediction. The first value for adj.r2 is NaN, because the first model we fit has 3 coefficients for 3 data points, hence no sensible statistics is available. The same happens to se as well, as the fitted line has no 0 residuals, so prediction is done without uncertainty.