How to rewrite this Stata code in R? - r

Question

One of the things Stata does well is the way it constructs new variables (see example below). How to do this in R?
foreach i in A B C D {
forval n=1990/2000 {
local m = 'n'-1
# create new columns from existing ones on-the-fly
generate pop'i''n' = pop'i''m' * (1 + trend'n')
}
}

Answer 1

DONT do it in R. The reason its messy is because its UGLY code. Constructing lots of variables with programmatic names is a BAD THING. Names are names. They have no structure, so do not try to impose one on them. Decent programming languages have structures for this - rubbishy programming languages have tacked-on 'Macro' features and end up with this awful pattern of constructing variable names by pasting strings together. This is a practice from the 1970s that should have died out by now. Don't be a programming dinosaur.
For example, how do you know how many popXXXX variables you have? How do you know if you have a complete sequence of pop1990 to pop2000? What if you want to save the variables to a file to give to someone. Yuck, yuck yuck.
Use a data structure that the language gives you. In this case probably a list.

Answer 2

Both Spacedman and Joshua have very valid points. As Stata has only one dataset in memory at any given time, I'd suggest to add the variables to a dataframe (which is also a kind of list) instead of to the global environment (see below).
But honestly, the more R-ish way to do so, is to keep your factors factors instead of variable names.
I make some data as I believe it is in your R version now (at least, I hope so...)
Data <- data.frame(
popA1989 = 1:10,
popB1989 = 10:1,
popC1989 = 11:20,
popD1989 = 20:11
)
Trend <- replicate(11,runif(10,-0.1,0.1))
You can then use the stack() function to obtain a dataframe where you have a factor pop and a numeric variable year
newData <- stack(Data)
newData$pop <- substr(newData$ind,4,4)
newData$year <- as.numeric(substr(newData$ind,5,8))
newData$ind <- NULL
Filling up the dataframe is then quite easy :
for(i in 1:11){
tmp <- newData[newData$year==(1988+i),]
newData <- rbind(newData,
data.frame( values = tmp$values*Trend[,i],
pop = tmp$pop,
year = tmp$year+1
)
)
}
In this format, you'll find most R commands (selections of some years, of a single population, modelling effects of either or both, ...) a whole lot easier to perform later on.
And if you insist, you can still create a wide format with unstack()
unstack(newData,values~paste("pop",pop,year,sep=""))
Adaptation of Joshua's answer to add the columns to the dataframe :
for(L in LETTERS[1:4]) {
for(i in 1990:2000) {
new <- paste("pop",L,i,sep="") # create name for new variable
old <- get(paste("pop",L,i-1,sep=""),Data) # get old variable
trend <- Trend[,i-1989] # get trend variable
Data <- within(Data,assign(new, old*(1+trend)))
}
}

Answer 3

Assuming popA1989, popB1989, popC1989, popD1989 already exist in your global environment, the code below should work. There are certainly more "R-like" ways to do this, but I wanted to give you something similar to your Stata code.
for(L in LETTERS[1:4]) {
for(i in 1990:2000) {
new <- paste("pop",L,i,sep="") # create name for new variable
old <- get(paste("pop",L,i-1,sep="")) # get old variable
trend <- get(paste("trend",i,sep="")) # get trend variable
assign(new, old*(1+trend))
}
}

Answer 4

Assuming you have population data in vector pop1989
and data for trend in trend.
require(stringr)# because str_c has better default for sep parameter
dta <- kronecker(pop1989,cumprod(1+trend))
names(dta) <- kronecker(str_c("pop",LETTERS[1:4]),1990:2000,str_c)