Goal
I am trying to create a function in R to replicate the functionality of a homonymous MATLAB function which returns the number of arguments that were passed to a function.
Example
Consider the function below:
addme <- function(a, b) {
if (nargin() == 2) {
c <- a + b
} else if (nargin() == 1) {
c <- a + a
} else {
c <- 0
}
return(c)
}
Once the user runs addme(), I want nargin() to basically look at how many parameters were passed―2 (a and b), only 1 (a) or none―and calculate c accordingly.
What I have tried
After spending a lot of time messing around with environments, this is the closest I ever got to a working solution:
nargin <- function() {
length(as.list(match.call(envir = parent.env(environment()))))
}
The problem with this function is that it always returns 0, and the reason why is that I think it's looking at its own environment instead of its parent's (in spite of my attempt of throwing in a parent.env there).
I know I can use missing() and args() inside addme() to achieve the same functionality, but I'll be needing this quite a few other times throughout my project, so wrapping it in a function is definitely something I should try to do.
Question
How can I get nargin() to return the number of arguments that were passed to its parent function?
You could use
nargin <- function() {
if(sys.nframe()<2) stop("must be called from inside a function")
length(as.list(sys.call(-1)))-1
}
Basically you just use sys.call(-1) to go up the call stack to the calling function and get it's call and then count the number of elements and subtract one for the function name itself.
Good day,
I am a beginner and trying to understand why I am getting the error below.
I am trying to create a function that would return 0 or 1 based on column values in data set.
LT = function(Lost.time) {
For (i in 1:dim(df)) {
if (df$Lost.time > 0) {
x = 1
}
else {
x = 0
}
return(x)
}
}
Error: no function to return from, jumping to top level In addition: Warning
message: In if (df$Lost.time > 0) { : the condition has length > 1 and only
the first element will be used> } Error: unexpected '}' in "}"
There are a couple of mistakes in the code:
R is case sensitive. Use for instead of For.
If you are looping over the entries in df$Lost.time, the individual elements should be addressed within the loop using df$Lost.time[i]. However, a loop is not necessary for this task.
An else statement should not begin on a new line of the code. The parser cannot know that the if statement is not finished after the first block. If the else statement is enclosed in curly braces like in } else { there will be no problem in this sense.
The parameter passed to the function is not suitable. Maybe you could pass df, instead of Lost.time, but it may be necessary to rewrite parts of the function.
The use of 1:dim(df) in the for loop should work, but it will trigger a warning message. It is better to use 1:nrow(df).
Those are syntax problems. However, the main issue is probably what has been addressed in the answer by #TimBiegeleisen: In the loop you are checking for each of the ̀nrow(df) elements of df$Lost.time whether a specific condition is fulfilled. It therefore does not seem to make sense to have a single binary result as return value. The purpose of the function should be clarified before it is implemented.
An alternative to this function could be constructed in a one-liner with ifelse.
It is not clear what you actually want to return in your function. return can only be called once, after which it will return a single value and the function will terminate.
If you want to get a vector which will contain 1 or 0 depending on whether a given row in your data frame has Lost.time > 0, then the following one liner should do the trick:
x <- as.numeric(df$Lost.time > 0)
If loops are used for writing a function indices should be used for each element.
Create a variable(x) in your dataframe, if the statements goes true it prints 1 else 0
LT = function(Lost.time) {
for (i in 1:dim(df)) {
if (as.numeric(df$Lost.time[i]) > 0) {
df$x[i] <- 1
}else{
df$x[i] <- 0
}
}
}
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 10 years ago.
I have a function of the following form:
a<-function(a,b,c){
if (a<0){ s<<-1;d<<-0; c
}
else { nest<-function(b,c){
if(b=0){ s<<-2; c}
else { 0
}
}
Is it possible to do it this way?
It's not particularly clear what you're trying to do. If a<0 then the two global assignments will take place and the function will return the value of c. Otherwise (once you change b=0 to b==0 and add enough closing brackets) the function will invisibly return the function nest you defined. It won't set s globally, because the function nest just gets defined, not called.
a<-function(a,b,c) {
if (a<0){
s<<-1;d<<-0; c
} else {
nest<-function(b,c){
if(b==0){ s<<-2; c} else { 0 }
}
}
}
First case:
a(-1,1,1)
## [1] 1
s
## [1] 1
d
## [1] 0
Second case:
x <- a(2,1,1)
x
## function(b,c){
## if(b==0){ s<<-2; c} else { 0 }
## }
## <environment: 0x998e634>
It seems more likely that you meant:
a <- function(a,b,c) {
if (a<0){
s<<-1;d<<-0; c
} else {
if (b==0) { s<<-2; c} else { 0 }
}
}
Which
if a<0 will set s and d globally and return c
if a>0 and b==0 will set s globally and return c
if a>0 and b!=0 will just return 0
By the way, it's a little confusing (although legal) to call your function a and to give it a parameter called a.
So the narrow answer to "can I do this?" is "yes, if you fix your syntax". The broad answer is "I'm not sure what you're trying to do, what you're doing doesn't seem very sensible. If you can express the logic of what you're trying to do more clearly (i.e. what are the desired outcomes of different inputs?) then we can help you come up with code that embodies that logic."
Not sure what you try to do, but I assume that the function you would like to have does certain things depending on the values of the variables a, b, and c. You can do that with a function as the one below:
a <- function(a,b,c){
IF(a<0){
# do some stuff
}else{
# do some other stuff
} #close if statement
} #close function statement
In other words, there is no problem in using a if statement within a function.
It is ok to have nested if statments,
e.g.
myfun <- function(x){
if(x < 0){
print("less") }else{
if(x>7) {
print("more") }else{
print("middle")
}
}
}
myfun(-1)
[1] "less"
> myfun(8)
[1] "more"
> myfun(6)
[1] "middle"
But in your second part you are saying if condition then name a function, not call the function. Obviously with choice I wouldn't use a function like the one above though, we can do better than that. Remember as well a useful tool to cut down on some of this is ifelse
A while back I got rebuked by Simon Urbanek from the R core team (I believe) for recommending a user to explicitly calling return at the end of a function (his comment was deleted though):
foo = function() {
return(value)
}
instead he recommended:
foo = function() {
value
}
Probably in a situation like this it is required:
foo = function() {
if(a) {
return(a)
} else {
return(b)
}
}
His comment shed some light on why not calling return unless strictly needed is a good thing, but this was deleted.
My question is: Why is not calling return faster or better, and thus preferable?
Question was: Why is not (explicitly) calling return faster or better, and thus preferable?
There is no statement in R documentation making such an assumption.
The main page ?'function' says:
function( arglist ) expr
return(value)
Is it faster without calling return?
Both function() and return() are primitive functions and the function() itself returns last evaluated value even without including return() function.
Calling return() as .Primitive('return') with that last value as an argument will do the same job but needs one call more. So that this (often) unnecessary .Primitive('return') call can draw additional resources.
Simple measurement however shows that the resulting difference is very small and thus can not be the reason for not using explicit return. The following plot is created from data selected this way:
bench_nor2 <- function(x,repeats) { system.time(rep(
# without explicit return
(function(x) vector(length=x,mode="numeric"))(x)
,repeats)) }
bench_ret2 <- function(x,repeats) { system.time(rep(
# with explicit return
(function(x) return(vector(length=x,mode="numeric")))(x)
,repeats)) }
maxlen <- 1000
reps <- 10000
along <- seq(from=1,to=maxlen,by=5)
ret <- sapply(along,FUN=bench_ret2,repeats=reps)
nor <- sapply(along,FUN=bench_nor2,repeats=reps)
res <- data.frame(N=along,ELAPSED_RET=ret["elapsed",],ELAPSED_NOR=nor["elapsed",])
# res object is then visualized
# R version 2.15
The picture above may slightly difffer on your platform.
Based on measured data, the size of returned object is not causing any difference, the number of repeats (even if scaled up) makes just a very small difference, which in real word with real data and real algorithm could not be counted or make your script run faster.
Is it better without calling return?
Return is good tool for clearly designing "leaves" of code where the routine should end, jump out of the function and return value.
# here without calling .Primitive('return')
> (function() {10;20;30;40})()
[1] 40
# here with .Primitive('return')
> (function() {10;20;30;40;return(40)})()
[1] 40
# here return terminates flow
> (function() {10;20;return();30;40})()
NULL
> (function() {10;20;return(25);30;40})()
[1] 25
>
It depends on strategy and programming style of the programmer what style he use, he can use no return() as it is not required.
R core programmers uses both approaches ie. with and without explicit return() as it is possible to find in sources of 'base' functions.
Many times only return() is used (no argument) returning NULL in cases to conditially stop the function.
It is not clear if it is better or not as standard user or analyst using R can not see the real difference.
My opinion is that the question should be: Is there any danger in using explicit return coming from R implementation?
Or, maybe better, user writing function code should always ask: What is the effect in not using explicit return (or placing object to be returned as last leaf of code branch) in the function code?
If everyone agrees that
return is not necessary at the end of a function's body
not using return is marginally faster (according to #Alan's test, 4.3 microseconds versus 5.1)
should we all stop using return at the end of a function? I certainly won't, and I'd like to explain why. I hope to hear if other people share my opinion. And I apologize if it is not a straight answer to the OP, but more like a long subjective comment.
My main problem with not using return is that, as Paul pointed out, there are other places in a function's body where you may need it. And if you are forced to use return somewhere in the middle of your function, why not make all return statements explicit? I hate being inconsistent. Also I think the code reads better; one can scan the function and easily see all exit points and values.
Paul used this example:
foo = function() {
if(a) {
return(a)
} else {
return(b)
}
}
Unfortunately, one could point out that it can easily be rewritten as:
foo = function() {
if(a) {
output <- a
} else {
output <- b
}
output
}
The latter version even conforms with some programming coding standards that advocate one return statement per function. I think a better example could have been:
bar <- function() {
while (a) {
do_stuff
for (b) {
do_stuff
if (c) return(1)
for (d) {
do_stuff
if (e) return(2)
}
}
}
return(3)
}
This would be much harder to rewrite using a single return statement: it would need multiple breaks and an intricate system of boolean variables for propagating them. All this to say that the single return rule does not play well with R. So if you are going to need to use return in some places of your function's body, why not be consistent and use it everywhere?
I don't think the speed argument is a valid one. A 0.8 microsecond difference is nothing when you start looking at functions that actually do something. The last thing I can see is that it is less typing but hey, I'm not lazy.
This is an interesting discussion. I think that #flodel's example is excellent. However, I think it illustrates my point (and #koshke mentions this in a comment) that return makes sense when you use an imperative instead of a functional coding style.
Not to belabour the point, but I would have rewritten foo like this:
foo = function() ifelse(a,a,b)
A functional style avoids state changes, like storing the value of output. In this style, return is out of place; foo looks more like a mathematical function.
I agree with #flodel: using an intricate system of boolean variables in bar would be less clear, and pointless when you have return. What makes bar so amenable to return statements is that it is written in an imperative style. Indeed, the boolean variables represent the "state" changes avoided in a functional style.
It is really difficult to rewrite bar in functional style, because it is just pseudocode, but the idea is something like this:
e_func <- function() do_stuff
d_func <- function() ifelse(any(sapply(seq(d),e_func)),2,3)
b_func <- function() {
do_stuff
ifelse(c,1,sapply(seq(b),d_func))
}
bar <- function () {
do_stuff
sapply(seq(a),b_func) # Not exactly correct, but illustrates the idea.
}
The while loop would be the most difficult to rewrite, because it is controlled by state changes to a.
The speed loss caused by a call to return is negligible, but the efficiency gained by avoiding return and rewriting in a functional style is often enormous. Telling new users to stop using return probably won't help, but guiding them to a functional style will payoff.
#Paul return is necessary in imperative style because you often want to exit the function at different points in a loop. A functional style doesn't use loops, and therefore doesn't need return. In a purely functional style, the final call is almost always the desired return value.
In Python, functions require a return statement. However, if you programmed your function in a functional style, you will likely have only one return statement: at the end of your function.
Using an example from another StackOverflow post, let us say we wanted a function that returned TRUE if all the values in a given x had an odd length. We could use two styles:
# Procedural / Imperative
allOdd = function(x) {
for (i in x) if (length(i) %% 2 == 0) return (FALSE)
return (TRUE)
}
# Functional
allOdd = function(x)
all(length(x) %% 2 == 1)
In a functional style, the value to be returned naturally falls at the ends of the function. Again, it looks more like a mathematical function.
#GSee The warnings outlined in ?ifelse are definitely interesting, but I don't think they are trying to dissuade use of the function. In fact, ifelse has the advantage of automatically vectorizing functions. For example, consider a slightly modified version of foo:
foo = function(a) { # Note that it now has an argument
if(a) {
return(a)
} else {
return(b)
}
}
This function works fine when length(a) is 1. But if you rewrote foo with an ifelse
foo = function (a) ifelse(a,a,b)
Now foo works on any length of a. In fact, it would even work when a is a matrix. Returning a value the same shape as test is a feature that helps with vectorization, not a problem.
It seems that without return() it's faster...
library(rbenchmark)
x <- 1
foo <- function(value) {
return(value)
}
fuu <- function(value) {
value
}
benchmark(foo(x),fuu(x),replications=1e7)
test replications elapsed relative user.self sys.self user.child sys.child
1 foo(x) 10000000 51.36 1.185322 51.11 0.11 0 0
2 fuu(x) 10000000 43.33 1.000000 42.97 0.05 0 0
____EDIT __________________
I proceed to others benchmark (benchmark(fuu(x),foo(x),replications=1e7)) and the result is reversed... I'll try on a server.
My question is: Why is not calling return faster
It’s faster because return is a (primitive) function in R, which means that using it in code incurs the cost of a function call. Compare this to most other programming languages, where return is a keyword, but not a function call: it doesn’t translate to any runtime code execution.
That said, calling a primitive function in this way is pretty fast in R, and calling return incurs a minuscule overhead. This isn’t the argument for omitting return.
or better, and thus preferable?
Because there’s no reason to use it.
Because it’s redundant, and it doesn’t add useful redundancy.
To be clear: redundancy can sometimes be useful. But most redundancy isn’t of this kind. Instead, it’s of the kind that adds visual clutter without adding information: it’s the programming equivalent of a filler word or chartjunk).
Consider the following example of an explanatory comment, which is universally recognised as bad redundancy because the comment merely paraphrases what the code already expresses:
# Add one to the result
result = x + 1
Using return in R falls in the same category, because R is a functional programming language, and in R every function call has a value. This is a fundamental property of R. And once you see R code from the perspective that every expression (including every function call) has a value, the question then becomes: “why should I use return?” There needs to be a positive reason, since the default is not to use it.
One such positive reason is to signal early exit from a function, say in a guard clause:
f = function (a, b) {
if (! precondition(a)) return() # same as `return(NULL)`!
calculation(b)
}
This is a valid, non-redundant use of return. However, such guard clauses are rare in R compared to other languages, and since every expression has a value, a regular if does not require return:
sign = function (num) {
if (num > 0) {
1
} else if (num < 0) {
-1
} else {
0
}
}
We can even rewrite f like this:
f = function (a, b) {
if (precondition(a)) calculation(b)
}
… where if (cond) expr is the same as if (cond) expr else NULL.
Finally, I’d like to forestall three common objections:
Some people argue that using return adds clarity, because it signals “this function returns a value”. But as explained above, every function returns something in R. Thinking of return as a marker of returning a value isn’t just redundant, it’s actively misleading.
Relatedly, the Zen of Python has a marvellous guideline that should always be followed:
Explicit is better than implicit.
How does dropping redundant return not violate this? Because the return value of a function in a functional language is always explicit: it’s its last expression. This is again the same argument about explicitness vs redundancy.
In fact, if you want explicitness, use it to highlight the exception to the rule: mark functions that don’t return a meaningful value, which are only called for their side-effects (such as cat). Except R has a better marker than return for this case: invisible. For instance, I would write
save_results = function (results, file) {
# … code that writes the results to a file …
invisible()
}
But what about long functions? Won’t it be easy to lose track of what is being returned?
Two answers: first, not really. The rule is clear: the last expression of a function is its value. There’s nothing to keep track of.
But more importantly, the problem in long functions isn’t the lack of explicit return markers. It’s the length of the function. Long functions almost (?) always violate the single responsibility principle and even when they don’t they will benefit from being broken apart for readability.
A problem with not putting 'return' explicitly at the end is that if one adds additional statements at the end of the method, suddenly the return value is wrong:
foo <- function() {
dosomething()
}
This returns the value of dosomething().
Now we come along the next day and add a new line:
foo <- function() {
dosomething()
dosomething2()
}
We wanted our code to return the value of dosomething(), but instead it no longer does.
With an explicit return, this becomes really obvious:
foo <- function() {
return( dosomething() )
dosomething2()
}
We can see that there is something strange about this code, and fix it:
foo <- function() {
dosomething2()
return( dosomething() )
}
I think of return as a trick. As a general rule, the value of the last expression evaluated in a function becomes the function's value -- and this general pattern is found in many places. All of the following evaluate to 3:
local({
1
2
3
})
eval(expression({
1
2
3
}))
(function() {
1
2
3
})()
What return does is not really returning a value (this is done with or without it) but "breaking out" of the function in an irregular way. In that sense, it is the closest equivalent of GOTO statement in R (there are also break and next). I use return very rarely and never at the end of a function.
if(a) {
return(a)
} else {
return(b)
}
... this can be rewritten as if(a) a else b which is much better readable and less curly-bracketish. No need for return at all here. My prototypical case of use of "return" would be something like ...
ugly <- function(species, x, y){
if(length(species)>1) stop("First argument is too long.")
if(species=="Mickey Mouse") return("You're kidding!")
### do some calculations
if(grepl("mouse", species)) {
## do some more calculations
if(species=="Dormouse") return(paste0("You're sleeping until", x+y))
## do some more calculations
return(paste0("You're a mouse and will be eating for ", x^y, " more minutes."))
}
## some more ugly conditions
# ...
### finally
return("The end")
}
Generally, the need for many return's suggests that the problem is either ugly or badly structured.
[EDIT]
return doesn't really need a function to work: you can use it to break out of a set of expressions to be evaluated.
getout <- TRUE
# if getout==TRUE then the value of EXP, LOC, and FUN will be "OUTTA HERE"
# .... if getout==FALSE then it will be `3` for all these variables
EXP <- eval(expression({
1
2
if(getout) return("OUTTA HERE")
3
}))
LOC <- local({
1
2
if(getout) return("OUTTA HERE")
3
})
FUN <- (function(){
1
2
if(getout) return("OUTTA HERE")
3
})()
identical(EXP,LOC)
identical(EXP,FUN)
The argument of redundancy has come up a lot here. In my opinion that is not reason enough to omit return().
Redundancy is not automatically a bad thing. When used strategically, redundancy makes code clearer and more maintenable.
Consider this example: Function parameters often have default values. So specifying a value that is the same as the default is redundant. Except it makes obvious the behaviour I expect. No need to pull up the function manpage to remind myself what the defaults are. And no worry about a future version of the function changing its defaults.
With a negligible performance penalty for calling return() (as per the benchmarks posted here by others) it comes down to style rather than right and wrong. For something to be "wrong", there needs to be a clear disadvantage, and nobody here has demonstrated satisfactorily that including or omitting return() has a consistent disadvantage. It seems very case-specific and user-specific.
So here is where I stand on this.
function(){
#do stuff
...
abcd
}
I am uncomfortable with "orphan" variables like in the example above. Was abcd going to be part of a statement I didn't finish writing? Is it a remnant of a splice/edit in my code and needs to be deleted? Did I accidentally paste/move something from somewhere else?
function(){
#do stuff
...
return(abdc)
}
By contrast, this second example makes it obvious to me that it is an intended return value, rather than some accident or incomplete code. For me this redundancy is absolutely not useless.
Of course, once the function is finished and working I could remove the return. But removing it is in itself a redundant extra step, and in my view more useless than including return() in the first place.
All that said, I do not use return() in short unnamed one-liner functions. There it makes up a large fraction of the function's code and therefore mostly causes visual clutter that makes code less legible. But for larger formally defined and named functions, I use it and will likely continue to so.
return can increase code readability:
foo <- function() {
if (a) return(a)
b
}
I got this question in an interview. So, this seems to me a messed up Fibonacci seq. sum generator and this gives a stackoverflow. Because if(n==0) should be if(n<3) (exit condition is wrong). What should be the precise answer to this question? What was expected as an answer?
foo (int n) {
if (n == 0) return 1;
return foo(n-1) + foo(n-2);
}
UPDATE
What does this recursive function do?
What do you infer when you see these 3 lines of code. No debugging.
Yes, it's just a recursive Fibonacci method with an incorrect base case (although I don't think I'd use n < 3 either... it depends on how you define it).
As for "what was expected as an answer" - that would depend on the question. Were you asked exactly the question in the title of your post? If so, the answer is "it recurses forever until it blows up the stack when you pass it any value other than 0" - with an explanation, of course. (It will never end because either n-1 isn't 0, or n-2 isn't 0, or both.)
EDIT: The above answers the first question. To answer "What do you infer when you see these 3 lines of code" - I would infer that the developer in question has never run the code with a value other than 0, or doesn't care about whether or not the code works.
The problem with the code posted is that if we evaluate foo(1) we need to find foo(0) and foo (-1), foo(-1) then needs to find foo(-2) and foo(-3) and so on. This will keep putting more calls to foo() until there is no more space in the memory resulting in a stack overflow. How many calls to foo are made will depend on the size of the call stack, which will be implementation specific.
When I see these lines of code I immediately get the impression that whoever wrote it hasn't thought about all the possible inputs that could be fed to the function.
To make a recursive Fibonacci function that doesn't fail for foo(1) or a negative input we get:
foo (int n) {
if( n < 0 ) return 0;
if (n == 0) return 1;
return foo(n-1) + foo(n-2);
}
Edit: perhaps the return for a negative number should be something else, as the fibonacci sequence isn't implicitly defined for negative indexes.
However if we use the extension that fib(-n) = (-1)^(n+1) fib(n) we could get the following code:
int fib(int n) {
if( n == -1){
return 1;
}else if ( n < 0 ){
return ( (-1)^(-n+1 ) * fib(-n) );
}else if (n == 0){
return 1;
}else{
return fib(n-1) + fib(n-2);
}
}
suppose, you call foo ( 1 ), it will have two sub calls of foo(0) and foo(-1). As you can see, once you get to foo(-1), n will decrease indefinitely and will never reach a terminating condition.
The precise answer would be:
foo (int n) {
if (n <= 1) return 1; // assuming 0th and 1st fib is 1
return foo(n-1) + foo(n-2);
}
Its a recursive fibonacci function with 0th element being 1.
Ignoring the incorrect termination condition, the code is the "naive" recursive implementation of the Fibonacci function. It has very poor big-O time complexity. It would work fine for small n, but if n is greater than say 50 (I'm just guessing that number), then it would take "forever" to run.