Understand how does new pipe work
Include libraries:
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library(magrittr)
library(dplyr)
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##
## Attaching package: 'dplyr'
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## The following objects are masked from 'package:stats':
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## filter, lag
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## The following objects are masked from 'package:base':
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## intersect, setdiff, setequal, union
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New pipe |> forces to call function directly while magittr pipe %>% does not need ()
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abc <-3
abc |> sqrt()
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One more example.
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mtcars %>%
select(where(is.numeric)) %>%
na.omit %>%
sapply(mean)
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## mpg cyl disp hp drat wt qsec
## 20.090625 6.187500 230.721875 146.687500 3.596563 3.217250 17.848750
## vs am gear carb
## 0.437500 0.406250 3.687500 2.812500
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mtcars |>
select(where(is.numeric)) |>
na.omit () |>
sapply(mean)
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## mpg cyl disp hp drat wt qsec
## 20.090625 6.187500 230.721875 146.687500 3.596563 3.217250 17.848750
## vs am gear carb
## 0.437500 0.406250 3.687500 2.812500
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But for some cases |> will not work:
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# works
1:10 %>% (call("sum"))
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# won't work
# 1:10 |> (call("sum"))
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For lazy typers both of them support lambda syntax. Here is minor differences:
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a <- 10
# regular lambda function
a %>% (function(x) x^2)
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a |> (function(x) x^2)()
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# short one
a %>% (\(.) .^2)
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# strange one
a %>% {a^2}
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# so, it won't work...
# a |> {a^2}
# a |> {a^2}()
# ... but if you wish so much
.d <- `{`
a |> .d(a^2)
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Magrittr pipe allows us to use . operator to send the data
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sample(1:10) %>% paste0(LETTERS[.])
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## [1] "8H" "1A" "3C" "2B" "9I" "7G" "4D" "5E" "10J" "6F"
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# This can be avoided:
rnorm(100) %>% {c(min(.), mean(.), max(.))} %>% floor
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# won't work either
# rnorm(100) |> .d(c(min(.), mean(.), max(.)))
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Lambda expressions
One example
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iris %>%
{
size <- sample(1:10, size = 1)
rbind(head(., size), tail(., size))
}
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## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1 5.1 3.5 1.4 0.2 setosa
## 2 4.9 3.0 1.4 0.2 setosa
## 3 4.7 3.2 1.3 0.2 setosa
## 4 4.6 3.1 1.5 0.2 setosa
## 5 5.0 3.6 1.4 0.2 setosa
## 6 5.4 3.9 1.7 0.4 setosa
## 145 6.7 3.3 5.7 2.5 virginica
## 146 6.7 3.0 5.2 2.3 virginica
## 147 6.3 2.5 5.0 1.9 virginica
## 148 6.5 3.0 5.2 2.0 virginica
## 149 6.2 3.4 5.4 2.3 virginica
## 150 5.9 3.0 5.1 1.8 virginica
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Using . operator re assign data to variable and perform calculations
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iris %>%
{
my_data <- .
size <- sample(1:10, size = 1)
rbind(head(my_data, size), tail(my_data, size))
}
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## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 1 5.1 3.5 1.4 0.2 setosa
## 2 4.9 3.0 1.4 0.2 setosa
## 3 4.7 3.2 1.3 0.2 setosa
## 4 4.6 3.1 1.5 0.2 setosa
## 5 5.0 3.6 1.4 0.2 setosa
## 6 5.4 3.9 1.7 0.4 setosa
## 7 4.6 3.4 1.4 0.3 setosa
## 144 6.8 3.2 5.9 2.3 virginica
## 145 6.7 3.3 5.7 2.5 virginica
## 146 6.7 3.0 5.2 2.3 virginica
## 147 6.3 2.5 5.0 1.9 virginica
## 148 6.5 3.0 5.2 2.0 virginica
## 149 6.2 3.4 5.4 2.3 virginica
## 150 5.9 3.0 5.1 1.8 virginica
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Building unary functions with %>%
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trig_fest <- . %>% tan %>% cos %>% sin
one2ten <- 1:10
one2ten %>% trig_fest
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## [1] 0.0133878 -0.5449592 0.8359477 0.3906486 -0.8257855 0.8180174
## [7] 0.6001744 0.7640323 0.7829771 0.7153150
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# same as:
sin(cos(tan(1:10)))
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## [1] 0.0133878 -0.5449592 0.8359477 0.3906486 -0.8257855 0.8180174
## [7] 0.6001744 0.7640323 0.7829771 0.7153150
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# or:
one2ten |> trig_fest()
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## [1] 0.0133878 -0.5449592 0.8359477 0.3906486 -0.8257855 0.8180174
## [7] 0.6001744 0.7640323 0.7829771 0.7153150
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How to send more than one argument into function with pipe?
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# normal way
plot(mtcars$hp, mtcars$mpg)
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lm(mtcars$mpg ~ mtcars$hp)
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##
## Call:
## lm(formula = mtcars$mpg ~ mtcars$hp)
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## Coefficients:
## (Intercept) mtcars$hp
## 30.09886 -0.06823
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# or
lm(mpg ~ hp, data = mtcars)
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##
## Call:
## lm(formula = mpg ~ hp, data = mtcars)
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## Coefficients:
## (Intercept) hp
## 30.09886 -0.06823
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# magrittr pipe way if function has data argument
mtcars %>%
lm(mpg ~ hp, data = .)
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##
## Call:
## lm(formula = mpg ~ hp, data = .)
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## Coefficients:
## (Intercept) hp
## 30.09886 -0.06823
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# but this won't work
# mtcars %>% plot(.$mpg, mtcars$hp)
# but we can override the defaulter behavior using `{` function
mtcars %>% {plot(.$mpg, .$hp)}
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# or use with function
with(mtcars, plot(hp, mpg))
# another solution:
mtcars |> with(plot(hp, mpg))
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# true hacker way
mtcars %$% lm(mpg ~ hp + disp)
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##
## Call:
## lm(formula = mpg ~ hp + disp)
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## Coefficients:
## (Intercept) hp disp
## 30.73590 -0.02484 -0.03035
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A couple more lambda examples from new 4.1 R update.
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# the shortcut for anonymous functions \(x) {} is the same as function(x) {}:
mtcars |> (\(x) {x[which.max(x$mpg), ]})()
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## mpg cyl disp hp drat wt qsec vs am gear carb
## Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.9 1 1 4 1
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mtcars |> (\(.) {.[which.max(.$mpg), ]})()
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## mpg cyl disp hp drat wt qsec vs am gear carb
## Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.9 1 1 4 1
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# Curly bracets for the first expression, {}(), work too:
mtcars |> {\(x) {x[which.max(x$mpg), ]}}()
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## mpg cyl disp hp drat wt qsec vs am gear carb
## Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.9 1 1 4 1
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# same as
mtcars %>% (\(.) {.[which.max(.$mpg), ]})
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## mpg cyl disp hp drat wt qsec vs am gear carb
## Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.9 1 1 4 1
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# lets plot
mtcars %>% plot(.$hp)
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# won't work
# mtcars |> plot(.$hp) # error
# Getting to the solution
# verbosely
mtcars |> (function(.) plot(.$hp, .$mpg))()
# using the anonymous function shortcut for emulation the dot syntax
mtcars |> (\(.) plot(.$hp, .$mpg))()
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# or more readable
mtcars |> (\(data) plot(data$hp, data$mpg))()
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Alex Romanko
included in R