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Block bootstrap from subject list

I'm trying to efficiently implement a block bootstrap technique to get the distribution of regression coefficients. The main outline is as follows.

I have a panel data set, and say firm and year are the indices. For each iteration of the bootstrap, I wish to sample n subjects with replacement. From this sample, I need to construct a new data frame that is an rbind() stack of all the observations for each sampled subject, run the regression, and pull out the coefficients. Repeat for a bunch of iterations, say 100.

  • Each firm can potentially be selected multiple times, so I need to include it data multiple times in each iteration's data set.
  • Using a loop and subset approach, like below, seems computationally burdensome.
  • Note that for my real data frame, n, and the number iterations is much larger than the example below.

My thoughts initially are to break the existing data frame into a list by subject using the split() command. From there, use

sample(unique(df1$subject),n,replace=TRUE)

to get the new list, then perhaps implement quickdf from the plyr package to construct a new data frame.

Example slow code:

require(plm)
data("Grunfeld", package="plm")

firms = unique(Grunfeld$firm)
n = 10
iterations = 100
mybootresults=list()

for(j in 1:iterations){

  v = sample(length(firms),n,replace=TRUE)
  newdata = NULL

  for(i in 1:n){
    newdata = rbind(newdata,subset(Grunfeld, firm == v[i]))
  }

  reg1 = lm(value ~ inv + capital, data = newdata)
  mybootresults[[j]] = coefficients(reg1)

}

mybootresults = as.data.frame(t(matrix(unlist(mybootresults),ncol=iterations)))
names(mybootresults) = names(reg1$coefficients)
mybootresults

  (Intercept)      inv    capital
1    373.8591 6.981309 -0.9801547
2    370.6743 6.633642 -1.4526338
3    528.8436 6.960226 -1.1597901
4    331.6979 6.239426 -1.0349230
5    507.7339 8.924227 -2.8661479
...
...
like image 625
baha-kev Avatar asked Aug 12 '12 04:08

baha-kev


1 Answers

How about something like this:

myfit <- function(x, i) {
   mydata <- do.call("rbind", lapply(i, function(n) subset(Grunfeld, firm==x[n])))
   coefficients(lm(value ~ inv + capital, data = mydata))
}

firms <- unique(Grunfeld$firm)

b0 <- boot(firms, myfit, 999)
like image 142
seancarmody Avatar answered Oct 20 '22 12:10

seancarmody