Should I use %$% instead of %>%?

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Recently I have found the %$% pipe operator, but I am missing the point regarding its difference with %>% and if it could completely replace it.

Motivation to use %$%

  • The operator %$% could replace %>% in many cases:
mtcars %>% summary()
mtcars %$% summary(.)
mtcars %>% head(10)
mtcars %$% head(.,10)
  • Apparently, %$% is more usable than %>%:
mtcars %>% plot(.$hp, .$mpg) # Does not work
mtcars %$% plot(hp, mpg)     # Works
  • Implicitly fills the built-in data argument:
mtcars %>% lm(mpg ~ hp, data = .)
mtcars %$% lm(mpg ~ hp)
  • Since % and $ are next to each other in the keyboard, inserting %$% is more convenient than inserting %>%.

Documentation

We find the following information in their respective help pages.

(?magrittr::`%>%`):

Description:

     Pipe an object forward into a function or call expression.

Usage:

     lhs %>% rhs

(?magrittr::`%$%`):

Description:

     Expose the names in ‘lhs’ to the ‘rhs’ expression. This is useful
     when functions do not have a built-in data argument.

Usage:

     lhs %$% rhs

I was not able to understand the difference between the two pipe operators. Which is the difference between piping an object and exposing a name? But, in the rhs of %$%, we are able to get the piped object with the ., right?

Should I start using %$% instead of %>%? Which problems could I face doing so?

magrittr r
2 Answers

In addition to the provided comments:

%$% also called the Exposition pipe vs. %>%:

This is a short summary of this article https://towardsdatascience.com/3-lesser-known-pipe-operators-in-tidyverse-111d3411803a

"The key difference in using %$% or %>% lies in the type of arguments of used functions."

One advantage, and as far as I can understand it, for me the only one to use %$% over %>% is the fact that we can avoid repetitive input of the dataframe name in functions that have no data as an argument.

For example the lm() has a data argument. In this case we can use both %>% and %$% interchangeable.

But in functions like the cor() which has no data argument:

mtcars %>% cor(disp, mpg) # Will give an Error

cor(mtcars$disp, mtcars$mpg)

is equivalent to

mtcars %$% cor(disp, mpg)

And note to use %$% pipe operator you have to load library(magrittr)

Update: on OPs comment: The pipe independent which one allows us to transform machine or computer language to a more readable human language.

ggplot2 is special. ggplot2 is not internally consistent. ggplot1 had a tidier API then ggplot2

Pipes would work with ggplot1: library(ggplot1) mtcars %>% ggplot(list( x= mpg, y = wt)) %>% ggpoint() %>% ggsave("mtcars.pdf", width= 8 height = 6)

In 2016 Wick Hadley said: "ggplot2 newver would have existed if I'd discovered the pipe 10 years earlier!" https://www.youtube.com/watch?v=K-ss_ag2k9E&list=LL&index=9

user_133216470

No, you shouldn't use %$% routinely. It is like using the with() function, i.e. it exposes the component parts of the LHS when evaluating the RHS. But it only works when the value on the left has names like a list or dataframe, so you can't always use it. For example,

library(magrittr)
x <- 1:10
x %>% mean()
#> [1] 5.5
x %$% mean()
#> Error in eval(substitute(expr), data, enclos = parent.frame()): numeric 'envir' arg not of length one

Created on 2022-02-06 by the reprex package (v2.0.1.9000)

You'd get a similar error with x %$% mean(.).

Even when the LHS has names, it doesn't automatically put the . argument in the first position. For example,

mtcars %>% nrow()
#> [1] 32
mtcars %$% nrow()
#> Error in nrow(): argument "x" is missing, with no default

Created on 2022-02-06 by the reprex package (v2.0.1.9000)

In this case mtcars %$% nrow(.) would work, because mtcars has names.

Your example involving .$hp and .$mpg is illustrating one of the oddities of magrittr pipes. Because the . is only used in expressions, not alone as an argument, it is passed as the first argument as well as being passed in those expressions. You can avoid this using braces, e.g.

mtcars %>% {plot(.$hp, .$mpg)}
user_25543300
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