Passing variable names to function in R

Question

I have a subset function that takes in an object of a user defined class, a condition passed to the function, and adds that condition as an attribute of the object.

subset.survey.data.frame <- function(x, condition, drop=FALSE, inside=FALSE) {
  if(inside) {
    condition_call <- deparse(substitute(condition, env=parent.frame(n=1)))
  } 
  else {
    condition_call <- substitute(condition)
  }

  x[["user_conditions"]] <- unique(c(x[["user_conditions"]],list(condition_call)))
  cat("Subset Conditions have been added to SDF")
  x
}

I can call this function as:

sdf <- subset.survey.data.frame(sdf,dsex =="Male")

This adds dsex == "Male" in user_conditions attribute.

However, if I want to call it from within another function and a loop, it passed v1 and v2, instead of the actual variable names.

 for(i in 1:length(lvls)) {
    v1 <- rhs_vars[1]
    v2 <- lvls[i]
    print(v1) #"dsex"
    print(v2) #"Male"
    dsdf <- subset.survey.data.frame(sdf,  v1 == v2, inside=T)

How can I modify the subset function so that I can get the names of v1 and v2 and then add the condition to the object?

Here is what SDF, lvls, and rhs_vars looks like

sdf <- list(user_conditions = list(),default_conditions = list(default_conditions) ,data =    data_Laf, weights=weights, pvvars=pvs, fileDescription = f)

Here, data_Laf is an LaF object (http://cran.r-project.org/web/packages/LaF/index.html), weights, pvs, and f are all lists.

rhs_vars <- rhs.vars(y ~ dsex + b017451) # from formula.tools package
> rhs_vars
[1] "dsex"    "b017451"

lvls is the levels of a column in a dataframe

lvls <- levels(data[,rhs_vars[1]])
"Male" "Female"

Here is a working example:

 default_conditions= quote(rptsamp=="Reporting sample")
 sdf <- list(user_conditions = list(),default_conditions = list(default_conditions))
 class(sdf) <- "Userdefined"

subset.survey.data.frame <- function(x, condition, drop=FALSE, inside=FALSE) {
  if(inside) {
    condition_call <- deparse(substitute(condition, env=parent.frame(n=1)))
  } 
  else {
    condition_call <- substitute(condition)
  }

  x[["user_conditions"]] <- unique(c(x[["user_conditions"]],list(condition_call)))
  cat("Subset Conditions have been added to X")
  x
}
sdf <- subset.survey.data.frame(sdf,dsex =="Male")
print(sdf)
#This gives the correct answer and adds dsex == "Male" to user conditions


#Creating some sample data
dsex =c('1','2','1','1','2','1','1','2','1')
b017451 <- sample(c(1:100), 9)
y <- rep(10, 9)
data <- data.frame(dsex, y, b017451)
data[,'dsex'] <- factor(data[,'dsex'], levels=c("1", "2"), labels=c('Male','Female'))
require(formula.tools)
rhs_vars <- rhs.vars(y ~ dsex + b017451)
lvls <- levels(data[,rhs_vars[1]])

for(i in 1:length(lvls)) {
    v1 <- rhs_vars[1]
    v2 <- lvls[i]
    print(v1) #"dsex"
    print(v2) #"Male"
    dsdf <- subset.survey.data.frame(sdf,  v1 == v2, inside=F)
    print(dsdf)
#this doesnt give the correct answer and adds v1 == v2 to user conditions

    break
}

If I understand your question correctly, substitute should get you what you are looking for. — nrussell
– nrussell, Commented Jan 13, 2015 at 19:24
Would you mind adding your other objects (sdf, lvls, rhs_vars)? — nrussell
– nrussell, Commented Jan 13, 2015 at 20:11
How about default_conditions --> Error: object 'default_conditions' not found ? Try starting a clean session and run only the code included in your question to make sure that you add all of the necessary objects to make this problem reproducible. — nrussell
– nrussell, Commented Jan 13, 2015 at 20:39

MrFlick · Accepted Answer · 2015-01-13 20:59:39Z

1

As @nrussell was alluding to, substitute should help you build your expressions. Then you just need to evalute them. Here's a simple example

v1 <- quote(cyl)
v2 <- 6
eval(substitute(subset(mtcars, v1==v2), list(v1=v1, v2=v2)))

If your v1 is a character class, you can convert it to a symbol vi as.name() because you need a symbol and not a character for the expression to work.

v1 <- "cyl"
v2 <- 6
eval(substitute(subset(mtcars, v1==v2), list(v1=as.name(v1), v2=v2)))

answered Jan 13, 2015 at 20:59

MrFlick

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BrodieG · Accepted Answer · 2015-01-13 21:02:32Z

1

If you're controlling the "inside" parameter, then isn't it as simple as:

if(inside) condition_call = call(substitute(condition[[1]]), as.name(condition[[2]]), condition[[3]])

This of course assumes people are only using binary conditions, but you can extend the above logic.

answered Jan 13, 2015 at 21:02

BrodieG

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