More efficient option to append()?

length(w)

count(w)

This takes a few seconds to do 1 million lines, and remains explicit/for loop form

numberofSalaryBands = 1000000 # 2000000
x ? ? ? ?= sample(1:15,numberofSalaryBands, replace=T)
y ? ? ? ?= sample((1:10)*1000, numberofSalaryBands, replace=T)
df ? ? ? = data.frame(x,y)
finalN ? = sum(df$x)
myVar ? ?= rep(NA, finalN)
outIndex = 1
i ? ? ? ?= 1
for (i in 1:numberofSalaryBands) {
? ? ? ?kount = df$x[i]
? ? ? ?myVar[outIndex:(outIndex+kount-1)] = rep(df$y[i], kount) # Make x[i] copies of value y[i]
? ? ? ?outIndex = outIndex+kount
}
head(myVar)
plyr::count(myVar)


On Aug 18, 2011, at 12:17 AM, Alex Ruiz Euler wrote:

Dear R community,

I have a 2 million by 2 matrix that looks like this:

x<-sample(1:15,2000000, replace=T)
y<-sample(1:10*1000, 2000000, replace=T)
? ? ?x ? ? y
[1,] 10 ?4000
[2,] ?3 ?1000
[3,] ?3 ?4000
[4,] ?8 ?6000
[5,] ?2 ?9000
[6,] ?3 ?8000
[7,] ?2 10000
(...)


The first column is a population expansion factor for the number in the
second column (household income). I want to expand the second column
with the first so that I end up with a vector beginning with 10
observations of 4000, then 3 observations of 1000 and so on. In my mind
the natural approach would be to create a NULL vector and append the
expansions:

myvar<-NULL
myvar<-append(myvar, replicate(x[1],y[1]), 1)

for (i in 2:length(x)) {
myvar<-append(myvar,replicate(x[i],y[i]),sum(x[1:i])+1)
}

to end with a vector of sum(x), which in my real database corresponds
to 22 million observations.

This works fine --if I only run it for the first, say, 1000
observations. If I try to perform this on all 2 million observations
it takes long, way too long for this to be useful (I left it running
11 hours yesterday to no avail).


I know R performs well with operations on relatively large vectors. Why
is this so inefficient? And what would be the smart way to do this?

Thanks in advance.
Alex