After really annoying and long debugging, I found that apply() does not pass arguments to functions via ...! Here is a simplified example (this is a simplified example to demonstrate the error):
A <- matrix(0, nrow = 2, ncol = 2)
apply(A, 1, function (x, ...) { cat("argument arg1: "); print(arg1); }, arg1 = 10)
# Argument arg1: Error in print(arg1) (from #1) : object 'arg1' not found
Do you have any idea why or what to do with this? Workaround is obvious, to list all arguments instead of using ..., which is anoying since I use this as a wrapper for other more complex functions. Any thoughts?
The problem isn't that that argument is not being passed to the function (it is), the problem is you are not "catching" it via a named parameter. This works for example
apply(A, 1, function (x, ..., arg1) { cat("argument arg1: "); print(arg1); }, arg1 = 10)
And we can use that variable as arg1 in the function because we caught it. Otherwise it's left inside the ... so you can pass it along to another function. For example we can just pass everything to list like this...
apply(A, 1, function (x, ...) { print(list(...)) }, arg1 = 10)
So since your function uses ... those values that aren't named stay "in" the dots. In order to get them out you need to capture them as arguments.
When you want to pass more than one argument, you may use mappy() instead.
mapply(your_function, arg1, arg2, argn
Related
How can a function inspect and modify the arguments of a call that it received as argument?
Application: A user feeds a call to function a as an argument to function b, but they forget to specify one of the required arguments of a. How can function b detect the problem and fix it?
In this minimal example, function a requires two arguments:
a <- function(arg1, arg2) {
return(arg1 + arg2)
}
Function b accepts a call and an argument. The commented lines indicate what I need to do:
b <- function(CALL, arg3) {
# 1. check if `arg2` is missing from CALL
# 2. if `arg2` is missing, plug `arg3` in its place
# 3. return evaluated call
CALL
}
Expected behavior:
b(CALL = a(arg1 = 1, arg2 = 2), arg3 = 3)
> 3
b(CALL = a(arg1 = 1), arg3 = 3)
> 4
The second call currently fails because the user forgot to specify the required arg2 argument. How can function b fix this mistake automatically?
Can I exploit lazy evaluation to modify the call to a before it is evaluated? I looked into rlang::modify_call but couldn't figure it out.
Here's a method that would work
b <- function(CALL, arg3) {
scall <- substitute(CALL)
stopifnot(is.call(scall)) #check that it's a call
lcall <- as.list(scall)
if (!"arg2" %in% names(lcall)) {
lcall <- c(lcall, list(arg2 = arg3))
}
eval.parent(as.call(lcall))
}
We use substitute() to grab the unevaluated version the CALL parameter. We convert it to a list so we can modify it. Then we append to the list another list with the parameter name/value we want. Finally, we turn the list back into a call and then evaluate that call in the environment of the caller rather than in the function body itself.
If you wanted to use rlang::modify_call and other rlang functions you could use
b <- function(CALL, arg3) {
scall <- rlang::enquo(CALL)
stopifnot(rlang::quo_is_call(scall))
if (!"arg2" %in% names(rlang::quo_get_expr(scall))) {
scall <- rlang::call_modify(scall, arg2=arg3)
}
rlang::eval_tidy(scall, env = rlang::caller_env())
}
I don't see why fancy language manipulation is needed. The problem is what to do when a, which requires 2 arguments, is supplied only 1. Wrapping it with b, which has a default value for the 2nd argument, solves this.
b <- function(arg1, arg2=42)
{
a(arg1, arg2)
}
b(1)
# [1] 43
b(1, 2)
# [1] 3
Let for example function g be defined by g(x):=x+1.
I want to programm a function f which can take a arbitrary function h(a_1,...,a_n) (a_1,...,a_n being the arguments) and returns the function g(h). So that
f(h)(a_1=1,...,a_n=n) works and returns the same as g(h(a_1=1,...,a_n=n)).
So we need something like
f <- (h){
- get the arguments of h and put them in a list/vector arg(I found functions that do that)
- return a function ´f(h)´ that has the elements of arg as arguments. (I am not sure how to do that)
}
I'm not sure I understand your question since what you wrote seems ok but is that what you are looking for?
somelistorvector <- list(a = 1, b = 2)
fct <- function(arg){
arg[[1]] + arg[[2]] # arg[["a"]] + arg[["b"]] could also work
}
fct(somelistorvector)
[1] 3
Also are the arguments always going to be a and b or element 1 and 2 ?
How can I use function call in a for loop where one variable is the current iteration of the loop? My current code (very simplified) looks like this:
funCall <- call('FUN', arg1 = 10, arg2 = get('i'))
for(i in 1:x){
ans[i] <- eval(funCall)
}
Given a FUN like arg1 * arg2 the evaluated function should return 10 for i = 1, 20 for i = 2 etc.
Using get('i') I get the error that object 'i' is not found and I don't understand
1) whyget is evaluated right away inside call
2) how I can implement this callto get the correct i
edit
The solution to the problem is to use another call() inside the call like
call('FUN', arg1 = 10, arg2 = call('get','i')) so the get()function is only evaluated inside the loop.
In the function shown below, there is no return. However, after executing it, I can confirm that the value entered d normally.
There is no return. Any suggestions in this regard will be appreciated.
Code
#installed plotly, dplyr
accumulate_by <- function(dat, var) {
var <- lazyeval::f_eval(var, dat)
lvls <- plotly:::getLevels(var)
dats <- lapply(seq_along(lvls), function(x) {
cbind(dat[var %in% lvls[seq(1, x)], ], frame = lvls[[x]])
})
dplyr::bind_rows(dats)
}
d <- txhousing %>%
filter(year > 2005, city %in% c("Abilene", "Bay Area")) %>%
accumulate_by(~date)
In the function, the last assignment is creating 'dats' which is returned with bind_rows(dats) We don't need an explicit return statement. Suppose, if there are two objects to be returned, we can place it in a list
In some languages like python, for memory efficiency, generators are used which will yield instead of creating the whole output in memory i.e. Consider two functions in python
def get_square(n):
result = []
for x in range(n):
result.append(x**2)
return result
When we run it
get_square(4)
#[0, 1, 4, 9]
The same function can be written as a generator. Instead of returning anything,
def get_square(n):
for x in range(n):
yield(x**2)
Running the function
get_square(4)
#<generator object get_square at 0x0000015240C2F9E8>
By casting with list, we get the same output
list(get_square(4))
#[0, 1, 4, 9]
There is always a return :) You just don't have to be explicit about it.
All R expressions return something. Including control structures and user-defined functions. (Control-structures are just functions, by the way, so you can just remember that everything is a value or a function call, and everything evaluates to a value).
For functions, the return value is the last expression evaluated in the execution of the function. So, for
f <- function(x) 2 + x
when you call f(3) you will invoke the function + with two parameters, 2 and x. These evaluate to 2 and 3, respectively, so `+`(2, 3) evaluates to 5, and that is the result of f(3).
When you call the return function -- and remember, this is a function -- you just leave the control-flow of a function early. So,
f <- function(x) {
if (x < 0) return(0)
x + 2
}
works as follows: When you call f, it will call the if function to figure out what to do in the first statement. The if function will evaluate x < 0 (which means calling the function < with parameters x and 0). If x < 0 is true, if will evaluate return(0). If it is false, it will evaluate its else part (which, because if has a special syntax when it comes to functions, isn't shown, but is NULL). If x < 0 is not true, f will evaluate x + 2 and return that. If x < 0 is true, however, the if function will evaluate return(0). This is a call to the function return, with parameter 0, and that call will terminate the execution of f and make the result 0.
Be careful with return. It is a function so
f <- function(x) {
if (x < 0) return;
x + 2
}
is perfectly valid R code, but it will not return when x < 0. The if call will just evaluate to the function return but not call it.
The return function is also a little special in that it can return from the parent call of control structures. Strictly speaking, return isn't evaluated in the frame of f in the examples above, but from inside the if calls. It just handles this special so it can return from f.
With non-standard evaluation this isn't always the case.
With this function
f <- function(df) {
with(df, if (any(x < 0)) return("foo") else return("bar"))
"baz"
}
you might think that
f(data.frame(x = rnorm(10)))
should return either "foo" or "bar". After all, we return in either case in the if statement. However, the if statement is evaluated inside with and it doesn't work that way. The function will return baz.
For non-local returns like that, you need to use callCC, and then it gets more technical (as if this wasn't technical enough).
If you can, try to avoid return completely and rely on functions returning the last expression they evaluate.
Update
Just to follow up on the comment below about loops. When you call a loop, you will most likely call one of the built-in primitive functions. And, yes, they return NULL. But you can write your own, and they will follow the rule that they return the last expression they evaluate. You can, for example, implement for in terms of while like this:
`for` <- function(itr_var, seq, body) {
itr_var <- as.character(substitute(itr_var))
body <- substitute(body)
e <- parent.frame()
j <- 1
while (j < length(seq)) {
assign(x = itr_var, value = seq[[j]], envir = e)
eval(body, envir = e)
j <- j + 1
}
"foo"
}
This function, will definitely return "foo", so this
for(i in 1:5) { print(i) }
evalutes to "foo". If you want it to return NULL, you have to be explicit about it (or just let the return value be the result of the while loop -- if that is the primitive while it returns NULL).
The point I want to make is that functions return the last expression they evaluate has to do with how the functions are defined, not how you call them. The loops use non-standard evaluation, so the last expression in the loop body you provide them might be the last value they evaluate and might not. For the primitive loops, it is not.
Except for their special syntax, there is nothing magical about loops. They follow the rules all functions follow. With non-standard evaluation it can get a bit tricky to work out from a function call what the last expression they will evaluate might be, because the function body looks like it is what the function evaluates. It is, to a degree, if the function is sensible, but the loop body is not the function body. It is a parameter. If it wasn't for the special syntax, and you had to provide loop bodies as normal parameters, there might be less confusion.
I'm trying to convert the following to pointfree style: a function that partially applies a value to the transformer function add before passing in the collection to be iterated over. (Using Ramda.js)
R.compose(
R.map,
R.add
)(1, [1,2,3])
The problem is that R.add is arity 2, as is R.map. I want the application order to be as follows:
add(1)
map(add(1))
map(add(1), [1,2,3])
[add(1,1), add(1,2), add(1,3)]
But what happens instead is this:
add(1, [1,2,3])
map(add(1, [1,2,3]))
<partially applied map, waiting for collection>
Anyone know of a way to specify this behavior?
A plain compose or pipe won't do this because either will absorb all the arguments supplied into the first function. Ramda includes two additional functions that help with this, converge and useWith. In this case useWith is the one that will help:
useWith(map, [add, identity])(1, [1, 2, 3]); //=> [2, 3, 4]
While identity is not absolutely required here, it gives the generated function the correct arity.
Figured it out. If anyone's curious, here's the gist. (You can try it in the console on RamdaJS.com.)
0) For a baseline, here's the pointed version.
func0 = x => R.map(R.add(x))
addOne = func0(1)
addOne([1,2,3]) // [2,3,4]
1) Here's the pointfree core, but it has the ordering problem from the question above.
func1 = R.compose(R.map, R.add)
addOne = func1(1)
addOne([1,2,3]) // [2,3,4])
func1(1, [1,2,3]) // function
2) If the composition is unary (arity 1) 2 invocations are needed apply all params.
func2 = R.unary(R.compose(R.map, R.add))
addOne = func2(1)
addOne([1,2,3]) // [2,3,4])
3) We want one invocation to apply both params, so we uncurry 2.
func3 = R.uncurryN(2, func2)
func3(1, [1,2,3]) // [2,3,4])
4) To prove func2 is composable, let's double the results.
func4 = R.compose(
R.map(R.multiply(2)),
R.uncurryN(2, func2)
)
func4(1, [1,2,3]) // [4,6,8])
5) Substitution gives us a completely pointfree function.
func5 = R.compose(
R.map(R.multiply(2)),
R.uncurryN(2, R.unary(R.compose(
R.map,
R.add
)))
)
func5(1, [1,2,3]) // [4,6,8])