I'm a Stata user that's transitioning to R and there's one Stata crutch that I find hard to give up. This is because I don't know how to do the equivalent with R's "apply" functions.
In Stata, I often generate a local macro list of stubnames and then loop over that list, calling on variables whose names are built off of those stubnames.
For a simple example, imagine that I have the following dataset:
study_id year varX06 varX07 varX08 varY06 varY07 varY08
1 6 50 40 30 20.5 19.8 17.4
1 7 50 40 30 20.5 19.8 17.4
1 8 50 40 30 20.5 19.8 17.4
2 6 60 55 44 25.1 25.2 25.3
2 7 60 55 44 25.1 25.2 25.3
2 8 60 55 44 25.1 25.2 25.3
and so on...
I want to generate two new variables, varX and varY that take on the values of varX06 and varY06 respectively when year is 6, varX07 and varY07 respectively when year is 7, and varX08 and varY08 respectively when year is 8.
The final dataset should look like this:
study_id year varX06 varX07 varX08 varY06 varY07 varY08 varX varY
1 6 50 40 30 20.5 19.8 17.4 50 20.5
1 7 50 40 30 20.5 19.8 17.4 40 19.8
1 8 50 40 30 20.5 19.8 17.4 30 17.4
2 6 60 55 44 25.1 25.2 25.3 60 25.1
2 7 60 55 44 25.1 25.2 25.3 55 25.2
2 8 60 55 44 25.1 25.2 25.3 44 25.3
and so on...
To clarify, I know that I can do this with melt and reshape commands - essentially converting this data from wide to long format, but I don't want to resort to that. That's not the intent of my question.
My question is about how to loop over a local macro list of stubnames in R and I'm just using this simple example to illustrate a more generic dilemma.
In Stata, I could generate a local macro list of stubnames:
local stub varX varY
And then loop over the macro list. I can generate a new variable varX or varY and replace the new variable value with the value of varX06 or varY06 (respectively) if year is 6 and so on.
foreach i of local stub {
display "`i'"
gen `i'=.
replace `i'=`i'06 if year==6
replace `i'=`i'07 if year==7
replace `i'=`i'08 if year==8
}
The last section is the section that I find hardest to replicate in R. When I write 'x'06, Stata takes the string "varX", concatenates it with the string "06" and then returns the value of the variable varX06. Additionally, when I write 'i', Stata returns the string "varX" and not the string "'i'".
How do I do these things with R?
I've searched through Muenchen's "R for Stata Users", googled the web, and searched through previous posts here at StackOverflow but haven't been able to find an R solution.
I apologize if this question is elementary. If it's been answered before, please direct me to the response.
Thanks in advance,
Tara
Well, here's one way. Columns in R data frames can be accessed using their character names, so this will work:
# create sample dataset
set.seed(1) # for reproducible example
df <- data.frame(year=as.factor(rep(6:8,each=100)), #categorical variable
varX06 = rnorm(300), varX07=rnorm(300), varX08=rnorm(100),
varY06 = rnorm(300), varY07=rnorm(300), varY08=rnorm(100))
# you start here...
years <- unique(df$year)
df$varX <- unlist(lapply(years,function(yr)df[df$year==yr,paste0("varX0",yr)]))
df$varY <- unlist(lapply(years,function(yr)df[df$year==yr,paste0("varY0",yr)]))
print(head(df),digits=4)
# year varX06 varX07 varX08 varY06 varY07 varY08 varX varY
# 1 6 -0.6265 0.8937 -0.3411 -0.70757 1.1350 0.3412 -0.6265 -0.70757
# 2 6 0.1836 -1.0473 1.5024 1.97157 1.1119 1.3162 0.1836 1.97157
# 3 6 -0.8356 1.9713 0.5283 -0.09000 -0.8708 -0.9598 -0.8356 -0.09000
# 4 6 1.5953 -0.3836 0.5422 -0.01402 0.2107 -1.2056 1.5953 -0.01402
# 5 6 0.3295 1.6541 -0.1367 -1.12346 0.0694 1.5676 0.3295 -1.12346
# 6 6 -0.8205 1.5122 -1.1367 -1.34413 -1.6626 0.2253 -0.8205 -1.34413
For a given yr, the anonymous function extracts the rows with that yr and column named "varX0" + yr (the result of paste0(...). Then lapply(...) "applies" this function for each year, and unlist(...) converts the returned list into a vector.
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