R passing data.table parameters through function calls

Question

so if I have a data.table defined as:

> dt <- data.table (x=c(1,2,3,4), y=c("y","n","y","m"), z=c("pickle",3,8,"egg"))

    > dt
        x   y        z 
    1:  1   y   pickle
    2:  2   n        3
    3:  3   y        8
    4:  4   m      egg

And a variable

    fn <- "z"

I get that I can pull a column from the data.table by the following:

    > dt[,fn, with=FALSE]

What I don't know how to do is the data.table equivalent of the following:

    > factorFunction <- function(df, fn) {
      df[,fn] <- as.factor(df[,fn])
      return(df)
     }

If I set fn="x" and call factorFunction(data.frame(dt),fn) it works just fine.

So I try it with a data.table, but this doesn't work

    > factorFunction <- function(dt, fn) {
      dt[,fn, with=FALSE] <- as.factor(dt[,fn, with=FALSE])
      return(dt)
     }

Error in sort.list(y) : 'x' must be atomic for 'sort.list' Have you called 'sort' on a list?

By the way, here's one (very unidiomatic) way: dt[,fn] <- as.factor(dt[,fn, with=FALSE][[1]]) It's very close to what you've written, I think. — Frank
– Frank, Commented Jun 18, 2015 at 19:52

akrun · Accepted Answer · 2015-06-18 17:41:02Z

4

You can try

 dt[,(fn):= factor(.SD[[1L]]),.SDcols=fn]

If there are multiple columns, use lapply(.SD, factor)

Wrapping it in a function

factorFunction <- function(df, fn) {
 df[, (fn):= factor(.SD[[1L]]), .SDcols=fn]
 }

 str(factorFunction(dt, fn))
 #Classes ‘data.table’ and 'data.frame':    4 obs. of  3 variables:
 #$ x: num  1 2 3 4
 #$ y: chr  "y" "n" "y" "m"
 #$ z: Factor w/ 4 levels "3","8","egg",..: 4 1 2 3

edited Jun 18, 2015 at 17:41

answered Jun 18, 2015 at 17:32

akrun

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1 Comment

akrun Over a year ago

@DavidWagle Glad to know that it works. We are specifying the columns to be used as subset in .SDcols and do the operation in .SD[[1L]]. Here, I used 1L to convert the list to vector. More generally, it would be lapply(.SD, yourfuncton)

Roland · Accepted Answer · 2015-06-18 17:46:11Z

3

Similar to @akrun's answer:

class(dt[[fn]])
#[1] "character"

setFactor <- function(DT, col) {
  #change the column type by reference
  DT[, c(col) := factor(DT[[col]])]
  invisible(NULL)
  }

setFactor(dt, fn)
class(dt[[fn]])
#[1] "factor"

answered Jun 18, 2015 at 17:46

Roland

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1 Comment

Frank Over a year ago

Or just use set: setFactor <- function(DT, col) set(DT, j = col, value = factor(DT[[col]]) )

Frank · Accepted Answer · 2015-06-18 20:03:28Z

2

I don't recommend this, since it's very unidiomatic:

factorFunction <- function(df,col){
  df[,col] <- factor(df[[col]])
  df
}

The upside is that it works in both base R and data.table:

df <- setDF(copy(dt))

class(df[[fn]]) # character
df <- factorFunction(df,fn)
class(df[[fn]]) # factor

class(dt[[fn]]) # character
dt <- factorFunction(dt,fn)
class(dt[[fn]]) # factor

answered Jun 18, 2015 at 20:03

Frank

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