Is there a way to write pipelined functions in R where the result of one function passes immediately into the next? I'm coming from F# and really appreciated this ability but have not found how to do it in R. It should be simple but I can't find how. In F# it would look something like this:
let complexFunction x =
x |> square
|> add 5
|> toString
In this case the input would be squared, then have 5 added to it and then converted to a string. I'm wanting to be able to do something similar in R but don't know how. I've searched for how to do something like this but have not come across anything. I'm wanting this for importing data because I typically have to import it and then filter. Right now I do this in multiple steps and would really like to be able to do something the way you would in F# with pipelines.
Here is a functional programming approach using Reduce. It is in fact an example from ?Reduce
square <- function(x) x^2
add_5 <- function(x) x+5
x <- 1:5
## Iterative function application:
Funcall <- function(f, ...) f(...)
Reduce(Funcall, list(as.character, add_5, square,x), right = TRUE)
## [1] "6" "9" "14" "21" "30"
Or even more simply using the functional package and Compose
This is nice as it will create the function for you
library(functional)
do_stuff <- Compose(square,add_5,as.character )
do_stuff(1:5)
## [1] "6" "9" "14" "21" "30"
I note that I would not consider either of these approaches idiomatically R ish (if that is even a phrase)
We can use Compose from the functional package to create our own binary operator that does something similar to what you want
# Define our helper functions
square <- function(x){x^2}
add5 <- function(x){x + 5}
# functional contains Compose
library(functional)
# Define our binary operator
"%|>%" <- Compose
# Create our complexFunction by 'piping' our functions
complexFunction <- square %|>% add5 %|>% as.character
complexFunction(1:5)
#[1] "6" "9" "14" "21" "30"
# previously had this until flodel pointed out
# that the above was sufficient
#"%|>%" <- function(fun1, fun2){ Compose(fun1, fun2) }
I guess we could technically do this without requiring the functional package - but it feels so right using Compose for this task.
"%|>%" <- function(fun1, fun2){
function(x){fun2(fun1(x))}
}
complexFunction <- square %|>% add5 %|>% as.character
complexFunction(1:5)
#[1] "6" "9" "14" "21" "30"
I think that you might just want to write a function to do the steps you desire.
complexFunction <- function(x) {
as.character(x^2 + 5)
}
Then just call complexFunction(x).
Edit to show what R is doing internally (#mnel) -- The way R parses the and evaluates as.character(x^2 + 5) does what you want
You can use codetools to investigate what R to see how the values are being passed to eachother
flattenAssignment(quote(as.character(x^2+5)))
[[1]]
[[1]][[1]]
x
[[1]][[2]]
`*tmp*`^2
[[1]][[3]]
`*tmp*` + 5
[[2]]
[[2]][[1]]
`as.character<-`(`*tmp*`, value = `*tmpv*`)
[[2]][[2]]
`+<-`(`*tmp*`, 5, value = `*tmpv*`)
[[2]][[3]]
`^<-`(x, 2, value = `*tmpv*`)
Or you can get the Lisp style representation to see how it is parsed (and the results passed)
showTree(quote(as.character(x^2+5)))
(as.character (+ (^ x 2) 5))
Since this question was asked, the magrittr pipe has become enormously popular in R. So your example would be:
library (magrittr)
fx <- function (x) {
x %>%
`^` (2) %>%
`+` (5) %>%
as.character ()
}
Note that the backquote notation is because I'm literally using R's built-in functions and I need to specially quote them to use them in this manner. More normally-named functions (like exp or if I'd created a helper function add) wouldn't need backquotes and would appear more like your example.
Note also that %>% passes the incoming value as the first argument to the next function automatically, though you can change this. Note also that an R function returns the last value calculated so I don't need to return or assign the calculation in order to return it.
This is a lot like the nice special operators defined by other answers, but it uses a particular function that's widely used in R now.
This works for R pretty similar in F#:
"%|>%" <- function(x, fun){
if(is.function(x)) {
function(...) fun(x(...))
} else {
fun(x)
}
}
Related
I am looking for the reverse of get().
Given an object name, I wish to have the character string representing that object extracted directly from the object.
Trivial example with foo being the placeholder for the function I am looking for.
z <- data.frame(x=1:10, y=1:10)
test <- function(a){
mean.x <- mean(a$x)
print(foo(a))
return(mean.x)}
test(z)
Would print:
"z"
My work around, which is harder to implement in my current problem is:
test <- function(a="z"){
mean.x <- mean(get(a)$x)
print(a)
return(mean.x)}
test("z")
The old deparse-substitute trick:
a<-data.frame(x=1:10,y=1:10)
test<-function(z){
mean.x<-mean(z$x)
nm <-deparse(substitute(z))
print(nm)
return(mean.x)}
test(a)
#[1] "a" ... this is the side-effect of the print() call
# ... you could have done something useful with that character value
#[1] 5.5 ... this is the result of the function call
Edit: Ran it with the new test-object
Note: this will not succeed inside a local function when a set of list items are passed from the first argument to lapply (and it also fails when an object is passed from a list given to a for-loop.) You would be able to extract the ".Names"-attribute and the order of processing from the structure result, if it were a named vector that were being processed.
> lapply( list(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a # This "a" and the next one in the print output are put in after processing
$a[[1]]
[1] "X" "" "1L]]" # Notice that there was no "a"
$b
$b[[1]]
[1] "X" "" "2L]]"
> lapply( c(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a
$a[[1]] # but it's theoretically possible to extract when its an atomic vector
[1] "structure(c(4, 5), .Names = c(\"a\", \"b\"))" ""
[3] "1L]]"
$b
$b[[1]]
[1] "structure(c(4, 5), .Names = c(\"a\", \"b\"))" ""
[3] "2L]]"
deparse(quote(var))
My intuitive understanding
In which the quote freeze the var or expression from evaluation
and the deparse function which is the inverse of parse function makes that freezed symbol back to String
Note that for print methods the behavior can be different.
print.foo=function(x){ print(deparse(substitute(x))) }
test = list(a=1, b=2)
class(test)="foo"
#this shows "test" as expected
print(test)
#this (just typing 'test' on the R command line)
test
#shows
#"structure(list(a = 1, b = 2), .Names = c(\"a\", \"b\"), class = \"foo\")"
Other comments I've seen on forums suggests that the last behavior is unavoidable. This is unfortunate if you are writing print methods for packages.
To elaborate on Eli Holmes' answer:
myfunc works beautifully
I was tempted to call it within another function (as discussed in his Aug 15, '20 comment)
Fail
Within a function, coded directly (rather than called from an external function), the deparse(substitute() trick works well.
This is all implicit in his answer, but for the benefit of peeps with my degree of obliviousness, I wanted to spell it out.
an_object <- mtcars
myfunc <- function(x) deparse(substitute(x))
myfunc(an_object)
#> [1] "an_object"
# called within another function
wrapper <- function(x){
myfunc(x)
}
wrapper(an_object)
#> [1] "x"
I came across this example from an R tutorial recently and I found this syntax really odd because it hints towards normal order reduction where arguments are wrapped / delayed. In applicative order reduction something like this should result in all the strings printing.
switch(grade,
"A" = print("Great"),
"B" = print("Good"),
"C" = print("Ok"),
"D" = print("Bad"),
"F" = print("Terrible"),
print("No Such Grade"))
Was wondering if anyone is privy and familiar with how R implements this?
Arguments to functions, including switch, are passed as promises which are forced, i.e. evaluated, only when actually used. See https://cran.r-project.org/doc/manuals/R-ints.html#Argument-evaluation
A promise has several parts. Its value slot is filled in the first time it is forced (i.e. accessed). Until then it just exists as unevaluated code and its environment as well as components which indicate that it has not been evaluated.
f does not force x:
library(pryr)
f <- function(x) promise_info(x)
f(3+pi)
giving:
$code
3 + pi
$env
<environment: R_GlobalEnv>
$evaled
[1] FALSE
$value
NULL
g forces x:
g <- function(x) { force(x); promise_info(x) }
g(3 + pi)
giving:
$code
3 + pi
$env
NULL
$evaled
[1] TRUE
$value
[1] 6.141593
I am looking for the reverse of get().
Given an object name, I wish to have the character string representing that object extracted directly from the object.
Trivial example with foo being the placeholder for the function I am looking for.
z <- data.frame(x=1:10, y=1:10)
test <- function(a){
mean.x <- mean(a$x)
print(foo(a))
return(mean.x)}
test(z)
Would print:
"z"
My work around, which is harder to implement in my current problem is:
test <- function(a="z"){
mean.x <- mean(get(a)$x)
print(a)
return(mean.x)}
test("z")
The old deparse-substitute trick:
a<-data.frame(x=1:10,y=1:10)
test<-function(z){
mean.x<-mean(z$x)
nm <-deparse(substitute(z))
print(nm)
return(mean.x)}
test(a)
#[1] "a" ... this is the side-effect of the print() call
# ... you could have done something useful with that character value
#[1] 5.5 ... this is the result of the function call
Edit: Ran it with the new test-object
Note: this will not succeed inside a local function when a set of list items are passed from the first argument to lapply (and it also fails when an object is passed from a list given to a for-loop.) You would be able to extract the ".Names"-attribute and the order of processing from the structure result, if it were a named vector that were being processed.
> lapply( list(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a # This "a" and the next one in the print output are put in after processing
$a[[1]]
[1] "X" "" "1L]]" # Notice that there was no "a"
$b
$b[[1]]
[1] "X" "" "2L]]"
> lapply( c(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a
$a[[1]] # but it's theoretically possible to extract when its an atomic vector
[1] "structure(c(4, 5), .Names = c(\"a\", \"b\"))" ""
[3] "1L]]"
$b
$b[[1]]
[1] "structure(c(4, 5), .Names = c(\"a\", \"b\"))" ""
[3] "2L]]"
deparse(quote(var))
My intuitive understanding
In which the quote freeze the var or expression from evaluation
and the deparse function which is the inverse of parse function makes that freezed symbol back to String
Note that for print methods the behavior can be different.
print.foo=function(x){ print(deparse(substitute(x))) }
test = list(a=1, b=2)
class(test)="foo"
#this shows "test" as expected
print(test)
#this (just typing 'test' on the R command line)
test
#shows
#"structure(list(a = 1, b = 2), .Names = c(\"a\", \"b\"), class = \"foo\")"
Other comments I've seen on forums suggests that the last behavior is unavoidable. This is unfortunate if you are writing print methods for packages.
To elaborate on Eli Holmes' answer:
myfunc works beautifully
I was tempted to call it within another function (as discussed in his Aug 15, '20 comment)
Fail
Within a function, coded directly (rather than called from an external function), the deparse(substitute() trick works well.
This is all implicit in his answer, but for the benefit of peeps with my degree of obliviousness, I wanted to spell it out.
an_object <- mtcars
myfunc <- function(x) deparse(substitute(x))
myfunc(an_object)
#> [1] "an_object"
# called within another function
wrapper <- function(x){
myfunc(x)
}
wrapper(an_object)
#> [1] "x"
I am looking for the reverse of get().
Given an object name, I wish to have the character string representing that object extracted directly from the object.
Trivial example with foo being the placeholder for the function I am looking for.
z <- data.frame(x=1:10, y=1:10)
test <- function(a){
mean.x <- mean(a$x)
print(foo(a))
return(mean.x)}
test(z)
Would print:
"z"
My work around, which is harder to implement in my current problem is:
test <- function(a="z"){
mean.x <- mean(get(a)$x)
print(a)
return(mean.x)}
test("z")
The old deparse-substitute trick:
a<-data.frame(x=1:10,y=1:10)
test<-function(z){
mean.x<-mean(z$x)
nm <-deparse(substitute(z))
print(nm)
return(mean.x)}
test(a)
#[1] "a" ... this is the side-effect of the print() call
# ... you could have done something useful with that character value
#[1] 5.5 ... this is the result of the function call
Edit: Ran it with the new test-object
Note: this will not succeed inside a local function when a set of list items are passed from the first argument to lapply (and it also fails when an object is passed from a list given to a for-loop.) You would be able to extract the ".Names"-attribute and the order of processing from the structure result, if it were a named vector that were being processed.
> lapply( list(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a # This "a" and the next one in the print output are put in after processing
$a[[1]]
[1] "X" "" "1L]]" # Notice that there was no "a"
$b
$b[[1]]
[1] "X" "" "2L]]"
> lapply( c(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a
$a[[1]] # but it's theoretically possible to extract when its an atomic vector
[1] "structure(c(4, 5), .Names = c(\"a\", \"b\"))" ""
[3] "1L]]"
$b
$b[[1]]
[1] "structure(c(4, 5), .Names = c(\"a\", \"b\"))" ""
[3] "2L]]"
deparse(quote(var))
My intuitive understanding
In which the quote freeze the var or expression from evaluation
and the deparse function which is the inverse of parse function makes that freezed symbol back to String
Note that for print methods the behavior can be different.
print.foo=function(x){ print(deparse(substitute(x))) }
test = list(a=1, b=2)
class(test)="foo"
#this shows "test" as expected
print(test)
#this (just typing 'test' on the R command line)
test
#shows
#"structure(list(a = 1, b = 2), .Names = c(\"a\", \"b\"), class = \"foo\")"
Other comments I've seen on forums suggests that the last behavior is unavoidable. This is unfortunate if you are writing print methods for packages.
To elaborate on Eli Holmes' answer:
myfunc works beautifully
I was tempted to call it within another function (as discussed in his Aug 15, '20 comment)
Fail
Within a function, coded directly (rather than called from an external function), the deparse(substitute() trick works well.
This is all implicit in his answer, but for the benefit of peeps with my degree of obliviousness, I wanted to spell it out.
an_object <- mtcars
myfunc <- function(x) deparse(substitute(x))
myfunc(an_object)
#> [1] "an_object"
# called within another function
wrapper <- function(x){
myfunc(x)
}
wrapper(an_object)
#> [1] "x"
I am looking for the reverse of get().
Given an object name, I wish to have the character string representing that object extracted directly from the object.
Trivial example with foo being the placeholder for the function I am looking for.
z <- data.frame(x=1:10, y=1:10)
test <- function(a){
mean.x <- mean(a$x)
print(foo(a))
return(mean.x)}
test(z)
Would print:
"z"
My work around, which is harder to implement in my current problem is:
test <- function(a="z"){
mean.x <- mean(get(a)$x)
print(a)
return(mean.x)}
test("z")
The old deparse-substitute trick:
a<-data.frame(x=1:10,y=1:10)
test<-function(z){
mean.x<-mean(z$x)
nm <-deparse(substitute(z))
print(nm)
return(mean.x)}
test(a)
#[1] "a" ... this is the side-effect of the print() call
# ... you could have done something useful with that character value
#[1] 5.5 ... this is the result of the function call
Edit: Ran it with the new test-object
Note: this will not succeed inside a local function when a set of list items are passed from the first argument to lapply (and it also fails when an object is passed from a list given to a for-loop.) You would be able to extract the ".Names"-attribute and the order of processing from the structure result, if it were a named vector that were being processed.
> lapply( list(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a # This "a" and the next one in the print output are put in after processing
$a[[1]]
[1] "X" "" "1L]]" # Notice that there was no "a"
$b
$b[[1]]
[1] "X" "" "2L]]"
> lapply( c(a=4,b=5), function(x) {nm <- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a
$a[[1]] # but it's theoretically possible to extract when its an atomic vector
[1] "structure(c(4, 5), .Names = c(\"a\", \"b\"))" ""
[3] "1L]]"
$b
$b[[1]]
[1] "structure(c(4, 5), .Names = c(\"a\", \"b\"))" ""
[3] "2L]]"
deparse(quote(var))
My intuitive understanding
In which the quote freeze the var or expression from evaluation
and the deparse function which is the inverse of parse function makes that freezed symbol back to String
Note that for print methods the behavior can be different.
print.foo=function(x){ print(deparse(substitute(x))) }
test = list(a=1, b=2)
class(test)="foo"
#this shows "test" as expected
print(test)
#this (just typing 'test' on the R command line)
test
#shows
#"structure(list(a = 1, b = 2), .Names = c(\"a\", \"b\"), class = \"foo\")"
Other comments I've seen on forums suggests that the last behavior is unavoidable. This is unfortunate if you are writing print methods for packages.
To elaborate on Eli Holmes' answer:
myfunc works beautifully
I was tempted to call it within another function (as discussed in his Aug 15, '20 comment)
Fail
Within a function, coded directly (rather than called from an external function), the deparse(substitute() trick works well.
This is all implicit in his answer, but for the benefit of peeps with my degree of obliviousness, I wanted to spell it out.
an_object <- mtcars
myfunc <- function(x) deparse(substitute(x))
myfunc(an_object)
#> [1] "an_object"
# called within another function
wrapper <- function(x){
myfunc(x)
}
wrapper(an_object)
#> [1] "x"