“If You Were an R Function, What Function Would You Be?”

February 26, 2019

(This article was first published on R – Win-Vector Blog, and kindly contributed to R-bloggers)

We’ve been getting some good uptake on our piping in R article announcement.

The article is necessarily a bit technical. But one of its key points comes from the observation that piping into names is a special opportunity to give general objects the following personality quiz: “If you were an R function, what function would you be?”

So our question is: can we add a meaningful association between the two deepest concepts in R objects (or references to them) and functions?

We think the answer is a resounding “yes!”

The following example (adapted from the paper) should help illustrate the idea.

Suppose we had simple linear model.

data_use <- base::sample(c("train", "test"), 
   nrow(mtcars), replace = TRUE)
mtcars_train <- mtcars[data_use == "train", , drop = FALSE]
mtcars_test <- mtcars[data_use == "test", , drop = FALSE]
model <- lm(mpg ~ disp  + wt, data = mtcars_train)

Now if “model” were an R function, what function would it be? One possible answer is: it would be predict.lm(). It would be nice if “model(mtcars_test)” meant “predict(model, data = mtcars_test)“. Or, if we accept the pipe notation “mtcars_test %.>% model” as an approximate substitute for (note: not an equivalent of) “model(mtcars_test)” we can make that happen.

The “%.>%” is the wrapr dot arrow pipe. It can be made to ask the question “If you were an R function, what function would you be?” as follows.

First a bit of preparation, we tell R‘s S3 class system how to answer the question.

apply_right.lm <-
           right_arg_name) {
            newdata = pipe_left_arg)

And now we can treat any reference to an object of class “lm” as a pipe destination or function.

mtcars_test %.>% model

And we see our results.

#          Mazda RX4       Mazda RX4 Wag      Hornet 4 Drive          Duster 360            Merc 280 
#          23.606199           22.518582           20.477232           18.347774           20.062914 
#          Merc 280C          Merc 450SE  Cadillac Fleetwood Lincoln Continental            Fiat 128 
#          20.062914           16.723133           10.506642            9.836894           25.888019 
#   Dodge Challenger         AMC Javelin       Porsche 914-2        Lotus Europa      Ford Pantera L 
#          18.814401           19.261396           25.892974           28.719255           20.108134 
#      Maserati Bora 
#          18.703696 

Notice we didn’t have to alter model or wrap it in a function. This solution can be used again and again in many different circumstances.

To leave a comment for the author, please follow the link and comment on their blog: R – Win-Vector Blog.

R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: Data science, Big Data, R jobs, visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more...

If you got this far, why not subscribe for updates from the site? Choose your flavor: e-mail, twitter, RSS, or facebook...

Comments are closed.

Search R-bloggers


Never miss an update!
Subscribe to R-bloggers to receive
e-mails with the latest R posts.
(You will not see this message again.)

Click here to close (This popup will not appear again)