R: Testing function in interactive session - r

I am not sure if it is possible, but I would like to be able to grab the default argument values of a function and test them and the code within my functions without having to remove the commas (this is especially useful in the case when there are many arguments).
In effect, I want to be able to have commas when sending arguments into the function but not have those commas if I copy and paste the arguments and run them by themselves.
For example:
foo=function(
x=1,
y=2,
z=3
) {
bar(x,y,z)
}
Now to test pieces of the function outside of the code block, copy and paste
x=1,
y=2,
z=3
bar(x,y,z)
But this gives an error because there is a comma after x=1
Perhaps I am not asking the right question. If this is strange, what is the preferred method for debugging functions?
Please note, just posted nearly identical question in Julia.

If you want to programmatically get at the arguments of a function and their default values, you can use formals:
fxn = function(a, b, d = 2) {}
formals(fxn)
# $a
#
#
# $b
#
#
# $d
# [1] 2
I suppose if you wanted to store the default value of every argument to your function into a variable of that name (which is what you're asking about in the OP), then you could do this with a for loop:
info <- formals(fxn)
for (varname in names(info)) {
assign(varname, info[[varname]])
}
a
# Error: argument "a" is missing, with no default
b
# Error: argument "b" is missing, with no default
d
# [1] 2
So for arguments without default values, you'll need to provide a value after running this code (as you would expect). For arguments with defaults, they'll now be set.

Related

Directly calling an anonymous function

I tried to create a function and then immediately call it.
function(x){x+1}(3)
This produces some strange result. Fortunately, I already know where I went wrong. I should have let the function statement be evaluated first before attempting to call it.
(function(x){x+1})(3)
# 4
However, I am confused as to what the first line of code actually evaluates to. Is someone able to explain what is going on in the R code below?
a <- function(x){x+1}(3)
a
# function(x){x+1}(3)
class(a)
# [1] "function"
a(3)
# Error in a(3) : attempt to apply non-function
a()
# Error in a() : argument "x" is missing, with no default
(a)
# function(x){x+1}(3)
# <bytecode: 0x128b52c50>
# everything in brackets on the right don't seem to be evaluated
function(x){x+1}(1)(2)(a,b,c)[1:4,d:5,,,][seq_along(letters)]
# function(x){x+1}(1)(2)(a,b,c)[1:4,d:5,,,][seq_along(letters)]
(function(x){x+1}(1)(2)(a,b,c)[1:4,d:5,,,][seq_along(letters)])
# function(x){x+1}(1)(2)(a,b,c)[1:4,d:5,,,][seq_along(letters)]
((function(x){x+1}(1)(2)(a,b,c)[1:4,d:5,,,][seq_along(letters)]))
# function(x){x+1}(1)(2)(a,b,c)[1:4,d:5,,,][seq_along(letters)]
Actually the curly braces are a call, like a function.
The first line is equivalent to
function(x)`{`(x+1)()
As the function is not evaluated, it is not known wether the bracket "call" return a function, thus is valid code.
For example:
{mean}(1:2)
# 1.5
`{`(mean)(1:2)
# 1.5
When you run a function definition but do not evaluate it, you actually see the function definition (formals and body) as output.
See ?"{" for more detail.

What's the difference between substitute and quote in R

In the official docs, it says:
substitute returns the parse tree for the (unevaluated) expression
expr, substituting any variables bound in env.
quote simply returns its argument. The argument is not evaluated and
can be any R expression.
But when I try:
> x <- 1
> substitute(x)
x
> quote(x)
x
It looks like both quote and substitute returns the expression that's passed as argument to them.
So my question is, what's the difference between substitute and quote, and what does it mean to "substituting any variables bound in env"?
Here's an example that may help you to easily see the difference between quote() and substitute(), in one of the settings (processing function arguments) where substitute() is most commonly used:
f <- function(argX) {
list(quote(argX),
substitute(argX),
argX)
}
suppliedArgX <- 100
f(argX = suppliedArgX)
# [[1]]
# argX
#
# [[2]]
# suppliedArgX
#
# [[3]]
# [1] 100
R has lazy evaluation, so the identity of a variable name token is a little less clear than in other languages. This is used in libraries like dplyr where you can write, for instance:
summarise(mtcars, total_cyl = sum(cyl))
We can ask what each of these tokens means: summarise and sum are defined functions, mtcars is a defined data frame, total_cyl is a keyword argument for the function summarise. But what is cyl?
> cyl
Error: object 'cyl' not found
It isn't anything! Well, not yet. R doesn't evaluate it right away, but treats it as an expression to be parsed later with some parse tree that is different than the global environment your command line is working in, specifically one where the columns of mtcars are defined. Somewhere in the guts of dplyr, something like this is happening:
> substitute(cyl, mtcars)
[1] 6 6 4 6 8 ...
Suddenly cyl means something. That's what substitute is for.
So what is quote for? Well sometimes you want your lazily-evaluated expression to be represented somewhere else before it's evaluated, i.e. you want to display the actual code you're writing without any (or only some) values substituted. The docs you quoted explain this is common for "informative labels for data sets and plots".
So, for example, you could create a quoted expression, and then both print the unevaluated expression in your chart to show how you calculated and actually calculate with the expression.
expr <- quote(x + y)
print(expr) # x + y
eval(expr, list(x = 1, y = 2)) # 3
Note that substitute can do this expression trick also while giving you the option to parse only part of it. So its features are a superset of quote.
expr <- substitute(x + y, list(x = 1))
print(expr) # 1 + y
eval(expr, list(y = 2)) # 3
Maybe this section of the documentation will help somewhat:
Substitution takes place by examining each component of the parse tree
as follows: If it is not a bound symbol in env, it is unchanged. If it
is a promise object, i.e., a formal argument to a function or
explicitly created using delayedAssign(), the expression slot of the
promise replaces the symbol. If it is an ordinary variable, its value
is substituted, unless env is .GlobalEnv in which case the symbol is
left unchanged.
Note the final bit, and consider this example:
e <- new.env()
assign(x = "a",value = 1,envir = e)
> substitute(a,env = e)
[1] 1
Compare that with:
> quote(a)
a
So there are two basic situations when the substitution will occur: when we're using it on an argument of a function, and when env is some environment other than .GlobalEnv. So that's why you particular example was confusing.
For another comparison with quote, consider modifying the myplot function in the examples section to be:
myplot <- function(x, y)
plot(x, y, xlab = deparse(quote(x)),
ylab = deparse(quote(y)))
and you'll see that quote really doesn't do any substitution.
Regarding your question why GlobalEnv is treated as an exception for substitute, it is just a heritage of S. From The R language definition (https://cran.r-project.org/doc/manuals/r-release/R-lang.html#Substitutions):
The special exception for substituting at the top level is admittedly peculiar. It has been inherited from S and the rationale is most likely that there is no control over which variables might be bound at that level so that it would be better to just make substitute act as quote.

'=' vs. '<-' as a function argument in R

I am a beginner so I'd appreciate any thoughts, and I understand that this question might be too basic for some of you.
Also, this question is not about the difference between <- and =, but about the way they get evaluated when they are part of the function argument. I read this thread, Assignment operators in R: '=' and '<-' and several others, but I couldn't understand the difference.
Here's the first line of code:
My objective is to get rid of variables in the environment. From reading the above thread, I would believe that <- would exist in the user workspace, so there shouldn't be any issue with deleting all variables.
Here is my code and two questions:
Question 1
First off, this code doesn't work.
rm(ls()) #throws an error
I believe this happens because ls() returns a character vector, and rm() expects an object name. Am I correct? If so, I would appreciate if someone could guide me how to get object names from character array.
Question 2
I googled this topic and found that this code below deletes all variables.
rm(list = ls())
While this does help me, I am unsure why = is used instead of <-. If I run the following code, I get an error Error in rm(list <- ls()) : ... must contain names or character strings
rm(list <- ls())
Why is this? Can someone please guide me? I'd appreciate any help/guidance.
I read this thread, Assignment operators in R: '=' and '<-' and several others, but I couldn't understand the difference.
No wonder, since the answers there are actually quite confusing, and some are outright wrong. Since that’s the case, let’s first establish the difference between them before diving into your actual question (which, it turns out, is mostly unrelated):
<- is an assignment operator
In R, <- is an operator that performs assignment from right to left, in the current scope. That’s it.
= is either an assignment operator or a distinct syntactic token
=, by contrast, has several meanings: its semantics change depending on the syntactic context it is used in:
If = is used inside a parameter list, immediately to the right of a parameter name, then its meaning is: “associate the value on the right with the parameter name on the left”.
Otherwise (i.e. in all other situations), = is also an operator, and by default has the same meaning as <-: i.e. it performs assignment in the current scope.
As a consequence of this, the operators <- and = can be used interchangeably1. However, = has an additional syntactic role in an argument list of a function definition or a function call. In this context it’s not an operator and cannot be replaced by <-.
So all these statements are equivalent:
x <- 1
x = 1
x[5] <- 1
x[5] = 1
(x <- 1)
(x = 1)
f((x <- 5))
f((x = 5))
Note the extra parentheses in the last example: if we omitted these, then f(x = 5) would be interpreted as a parameter association rather than an assignment.
With that out of the way, let’s turn to your first question:
When calling rm(ls()), you are passing ls() to rm as the ... parameter. Ronak’s answer explains this in more detail.
Your second question should be answered by my explanation above: <- and = behave differently in this context because the syntactic usage dictates that rm(list = ls()) associates ls() with the named parameter list, whereas <- is (as always) an assignment operator. The result of that assignment is then once again passed as the ... parameter.
1 Unless somebody changed their meaning: operators, like all other functions in R, can be overwritten with new definitions.
To expand on my comment slightly, consider this example:
> foo <- function(a,b) b+1
> foo(1,b <- 2) # Works
[1] 3
> ls()
[1] "b" "foo"
> foo(b <- 3) # Doesn't work
Error in foo(b <- 3) : argument "b" is missing, with no default
The ... argument has some special stuff going on that restricts things a little further in the OP's case, but this illustrates the issue with how R is parsing the function arguments.
Specifically, when R looks for named arguments, it looks specifically for arg = val, with an equals sign. Otherwise, it is parsing the arguments positionally. So when you omit the first argument, a, and just do b <- 1, it thinks the expression b <- 1 is what you are passing for the argument a.
If you check ?rm
rm(..., list = character(),pos = -1,envir = as.environment(pos), inherits = FALSE)
where ,
... - the objects to be removed, as names (unquoted) or character strings (quoted).
and
list - a character vector naming objects to be removed.
So, if you do
a <- 5
and then
rm(a)
it will remove the a from the global environment.
Further , if there are multiple objects you want to remove,
a <- 5
b <- 10
rm(a, b)
This can also be written as
rm(... = a, b)
where we are specifying that the ... part in syntax takes the arguments a and b
Similarly, when we want to specify the list part of the syntax, it has to be given by
rm(list = ls())
doing list <- ls() will store all the variables from ls() in the variable named list
list <- ls()
list
#[1] "a" "b" "list"
I hope this is helpful.

Passing character as variable name in R

I am trying to solve an implicit equation in R using multiroot function from package rootSolve.
I am reading the implicit equation from a text file using parse. Also, the variable to be solved for is read from a text file as a character.
For using multiroot,
multiroot(function, initial_guess, ....))
we have to generate a function from the read equation. I did this by
fun <- function(op) {fun <- eval(expr.im)}
op = as.name(opim.names)
where expr.im is the read implicit equation as an expression from the text file, and opim.names is the variable to be solved for, as character.
But the problem arises when I pass the variable op to be solved as a symbol to the function. It gives an error saying that the object
"variable to be solved for" not found.
I think that the variable symbol is not being passed correctly in the function.
Please tell me how to do it correctly.
Since a lot of stuff is going on in my code, I cannot post the whole thing here.
Let me just state a small example for it.
var.name = "x1" # This is what I read from the text file #
var.sym = as.name(var.name)
func <- function(var.sym){
func = x1^2 # the expression x1^2 is also read from a text file #
} # I am trying to solve the implicit equation x1^2 = 0 #
initial_guess = 1
root = multiroot(f=func, start = initial_guess)
As requested by nicola here's what I want -
I have a text file giving me the name of the variable and its initial guess.
I read the variable name (say "x") and the initial guess value (say 1) into variables var (character) and guess(numeric).
I also have another text file containing the following equation -
x^3-1
I read this as an expression in the variable expr.
I want to find the solution to the implicit equation expr.
(The text files can have different names of variables and correspondingly an implicit expression in another file)
As you know, for using the multiroot function, we need to have a function.
The problem is I am not able to pass the variable name stored in var to the function.
Any further clarification will be given if asked.
You can build your function in the following way.
#input: function expression, variable names and initial guess
expr<-"x^3-1"
var.name<-"x"
initial.guess<-2
#we build an "empty" function
func<-function() {}
#now we set the formal arguments and the body of the function
formals(func)<-eval(parse(text=paste0("alist(",paste(var.name,collapse="=,"),"=)")))
body(func)<-parse(text=expr)
#we can see that func is correctly defined
func
#function (x)
#x^3 - 1
#now we can call multiroot
multiroot(func,initial.guess)
#$root
#[1] 1
#$f.root
#[1] 3.733019e-08
#$iter
#[1] 6
#$estim.precis
#[1] 3.733019e-08
You need a little more care if you are dealing with function of more than one variable. See ?multiroot to check how to pass the arguments. Basically you have to pass just one argument which is a vector of all the function arguments. It shouldn't be difficult if you take some time to see how I managed to build func. If you are exclusively dealing with one variable function, you should use the base uniroot function.
Not able to understand the description fully. But to answer the heading, you can try this procedure-
a = "random_string"
b = "a"
eval(parse(text = b))
[1] "random_string"

How to retrieve formals of a primitive function?

For the moment, at least, this is an exercise in learning for me, so the actual functions or their complexity is not the issue. Suppose I write a function whose argument list includes some input variables and a function name, passed as a string. This function then calculates some variables internally and "decides" how to feed them to the function name I've passed in.
For nonprimitive functions, I can do (for this example, assume non of my funcname functions have any arguments other than at most (x,y,z). If they did, I'd have to write some code to search for matching names(formals(get(funcname))) so as not to delete the other arguments):
foo <- function (a,b,funcname) {
x <- 2*a
y <- a+3*b
z <- -b
formals(get(funcname)) <- list(x=x, y=y, z=z)
bar <- get(funcname)()
return(bar)
}
And the nice thing is, even if the function funcname will execute without error even if it doesn't use x, y or z (so long as there are no other args that don't have defaults) .
The problem with "primitive" functions is I don't know any way to find or modify their formals. Other than writing a wrapper, e.g. foosin <-function(x) sin(x), is there a way to set up my foo function to work with both primitive and nonprimitive function names as input arguments?
formals(args(FUN)) can be used to get the formals of a primitive function.
You could add an if statement to your existing function.
> formals(sum)
# NULL
> foo2 <- function(x) {
if(is.primitive(x)) formals(args(x)) else formals(x)
## formals(if(is.primitive(x)) args(x) else x) is another option
}
> foo2(sum)
# $...
#
#
# $na.rm
# [1] FALSE
#
> foo2(with)
# $data
#
#
# $expr
#
#
# $...
Building on Richard S' response, I ended up doing the following. Posted just in case anyone else ever tries do things as weird as I do.
EDIT: I think more type-checking needs to be done. It's possible that coleqn could be
the name of an object, in which case get(coleqn) will return some data. Probably I need
to add a if(is.function(rab)) right after the if(!is.null(rab)). (Of course, given that I wrote the function for my own needs, if I was stupid enough to pass an object, I deserve what I get :-) ).
# "coleqn" is the input argument, which is a string that could be either a function
# name or an expression.
rab<-tryCatch(get(coleqn),error=function(x) {} )
#oops, rab can easily be neither NULL nor a closure. Damn.
if(!is.null(rab)) {
# I believe this means it must be a function
# thanks to Richard Scriven of SO for this fix to handle primitives
# we are not allowed to redefine primitive's formals.
qq <- list(x=x,y=y,z=z)
# matchup the actual formals names
# by building a list of valid arguments to pass to do.call
argk<-NULL
argnames<-names(formals(args(coleqn)))
for(j in 1:length(argnames) )
argk[j]<-which(names(qq)==argnames[1] )
arglist<-list()
for(j in 1:length(qq) )
if(!is.na(argk[j])) arglist[[names(qq)[j]]]<-qq[[j]]
colvar<- do.call(coleqn,arglist)
} else {
# the input is just an expression (string), not a function
colvar <- eval(parse(text=coleqn))
}
The result is an object generated either by the expression or the function just created, using variables internal to the main function (which is not shown in this snippet)

Resources