could someone please explain to me what these various environment functions do specifically? ie which one returns what frame? i am thoroughly confused after reading the documentation (http://stat.ethz.ch/R-manual/R-patched/library/base/html/sys.parent.html)
Let's put some structure on the question:
x = 1; y=2; z=3;
f = function() { ls(); ls(envir=sys.frame());}
#this first prints the contents of this function and then of the global environment
I am trying to understand how one can access environments of calling functions and to know which environment you are in. For example g could have called f:
g = function() { somevar=1; f() }
If I wanted to get the contents of g, how would i do that? What is the difference between a frame and an environment?
parent.frame() refers to the calling environment. You normally don't need the rest of them. For your example use this to list somevar :
f <- function() ls(parent.frame())
Related
I am trying to write a very basic IF statement in R and am stuck. I thought I'd find someone with the same problem, but I cant. Im sorry if this has been solved before.
I want to check if a variable/object has been assigned, IF TRUE I want to execute a function that is part of a R-package. First I wrote
FileAssignment <- function(x){
if(exists("x")==TRUE){
print("yes!")
x <- parse.vdjtools(x)
} else { print("Nope!")}
}
I assign a filename as x
FILENAME <- "FILENAME.txt"
I run the function
FileAssignment(FILENAME)
I use print("yes!") and print("Nope!") to check if the IF-Statement works, and it does. However, the parse.vdjtools(x) part is not assigned. Now I tested the same IF-statement outside of the function:
if(exists("FILENAME1")==TRUE){
FILENAME1 <- parse.vdjtools(FILENAME1)
}
This works. I read here that it might be because the function uses {} and the if-statement does too. So I should remove the brackets from the if-statement.
FileAssignment <- function(x){
if(exists("x")==TRUE)
x <- parse.vdjtools(x)
else { print("Nope!")
}
Did not work either.
I thought it might be related to the specific parse.vdjtools(x) function, so I just tried assigning a normal value to x with x <- 20. Also did not work inside the function, however, it does outside.
I dont really know what you are trying to acheive, but I wpuld say that the use of exists in this context is wrong. There is no way that the x cannot exist inside the function. See this example
# All this does is report if x exists
f <- function(x){
if(exists("x"))
cat("Found x!", fill = TRUE)
}
f()
f("a")
f(iris)
# All will be found!
Investigate file.exists instead? This is vectorised, so a vector of files can be investigated at the same time.
The question that you are asking is less trivial than you seem to believe. There are two points that should be addressed to obtain the desired behavior, and especially the first one is somewhat tricky:
As pointed out by #NJBurgo and #KonradRudolph the variable x will always exist within the function since it is an argument of the function. In your case the function exists() should therefore not check whether the variable x is defined. Instead, it should be used to verify whether a variable with a name corresponding to the character string stored in x exists.
This is achieved by using a combination of deparse() and
substitute():
if (exists(deparse(substitute(x)))) { …
Since x is defined only within the scope of the function, the superassignment operator <<- would be required to make a value assigned to x visible outside the function, as suggested by #thothai. However, functions should not have such side effects. Problems with this kind of programming include possible conflicts with another variable named x that could be defined in a different context outside the function body, as well as a lack of clarity concerning the operations performed by the function.
A better way is to return the value instead of assigning it to a variable.
Combining these two aspects, the function could be rewritten like this:
FileAssignment <- function(x){
if (exists(deparse(substitute(x)))) {
print("yes!")
return(parse.vdjtools(x))
} else {
print("Nope!")
return(NULL)}
}
In this version of the function, the scope of x is limited to the function body and the function has no side effects. The return value of FileAssignment(a) is either parse.vdjtools(a) or NULL, depending on whether a exists or not.
Outside the function, this value can be assigned to x with
x <- FileAssignment(a)
I have some problem I don't know how to solve.
Prehistory: I use R.NET for my calculation (need WPF application). So, I want to parallelize my app, and I created dynamic proxy for REngine class. It needs to serialize data to pass and receive data from-to REngine instance via TCP. Bad news - R.NET classes cannot be serialized. So, I have an idea to serialize R objects in R and pass R serialized data between processes.
So I have same script like this:
a <- 5;
b <- 10;
x <- a+b;
I need to wrap it like this:
wrapFunction <- function()
{
a <- 5;
b <- 10;
x <- a+b;
}
serializedResult <- serialize(wrapFunction());
I'll get serializedResult and pass it as byte array. Also I need to pass environments. But I won't get a, b, x in .GlobalEnv after these manipulations.
How is it possible to get all variables, defined in function body, in my .GlobalEnv?
I don't know names and count, I can't rewrite basic script, replacing "<-" by "<<-".
Other ways?
Thank you.
I'm not sure I fully understand your requirements. They seem to go against the functional language paradigm, which R tries to follow. The following might be helpful or not:
e <- new.env()
wrapFunction <- function(){
with(e, {
a <- 5;
b <- 10;
x <- a+b;
})
}
wrapFunction()
e$a
#[1] 5
You can of course use the .GlobalEnv instead of e, but at least in R that would be considered an even worse practice.
The following simple example will help me address a problem in my program implementation.
fun2<-function(j)
{
x<-rnorm(10)
y<-runif(10)
Sum<-sum(x,y)
Prod<-prod(x,y)
return(Sum)
}
j=1:10
Try<-lapply(j,fun2)
#
I want to store "Prod" at each iteration so I can access it after running the function fun2. I tried using assign() to create space assign("Prod",numeric(10),pos=1)
and then assigning Prod at j-th iteration to Prod[j] but it does not work.
#
Any idea how this can be done?
Thank you
You can add anything you like in the return() command. You could return a list return(list(Sum,Prod)) or a data frame return(data.frame("In"=j,"Sum"=Sum,"Prod"=Prod))
I would then convert that list of data.frames into a single data.frame
Try2 <- do.call(rbind,Try)
Maybe re-think the problem in a more vectorized way, taking advantage of the implied symmetry to represent intermediate values as a matrix and operating on that
ni = 10; nj = 20
x = matrix(rnorm(ni * nj), ni)
y = matrix(runif(ni * nj), ni)
sums = colSums(x + y)
prods = apply(x * y, 2, prod)
Thinking about the vectorized version is as applicable to whatever your 'real' problem is as it is to the sum / prod example; in practice and when thinking in terms of vectors fails I've never used the environment or concatenation approaches in other answers, but rather the simple solution of returning a list or vector.
I have done this before, and it works. Good for a quick fix, but its kind of sloppy. The <<- operator assigns outside the function to the global environment.
fun2<-function(j){
x<-rnorm(10)
y<-runif(10)
Sum<-sum(x,y)
Prod[j]<<-prod(x,y)
}
j=1:10
Prod <- numeric(length(j))
Try<-lapply(j,fun2)
Prod
thelatemail and JeremyS's solutions are probably what you want. Using lists is the normal way to pass back a bunch of different data items and I would encourage you to use it. Quoted here so no one thinks I'm advocating the direct option.
return(list(Sum,Prod))
Having said that, suppose that you really don't want to pass them back, you could also put them directly in the parent environment from within the function using either assign or the superassignment operator. This practice can be looked down on by functional programming purists, but it does work. This is basically what you were originally trying to do.
Here's the superassignment version
fun2<-function(j)
{
x<-rnorm(10)
y<-runif(10)
Sum<-sum(x,y)
Prod[j] <<- prod(x,y)
return(Sum)
}
j=1:10
Prod <- numeric(10)
Try<-lapply(j,fun2)
Note that the superassignment searches back for the first environment in which the variable exists and modifies it there. It's not appropriate for creating new variables above where you are.
And an example version using the environment directly
fun2<-function(j,env)
{
x<-rnorm(10)
y<-runif(10)
Sum<-sum(x,y)
env$Prod[j] <- prod(x,y)
return(Sum)
}
j=1:10
Prod <- numeric(10)
Try<-lapply(j,fun2,env=parent.frame())
Notice that if you had called parent.frame() from within the function you would need to go back two frames because lapply() creates its own. This approach has the advantage that you could pass it any environment you want instead of parent.frame() and the value would be modified there. This is the seldom-used R implementation of writeable passing by reference. It's safer than superassignment because you know where the variable is that is being modified.
This question already has answers here:
Global and local variables in R
(3 answers)
Closed 8 years ago.
I'm pretty new to R, but coming from Scheme—which is also lexically scoped and has closures—I would expect being able to mutate outer variables in a closure.
E.g., in
foo <- function() {
s <- 100
add <- function() {
s <- s + 1
}
add()
s
}
cat(foo(), "\n") # prints 100 and not 101
I would expect foo() to return 101, but it actually returns 100:
$ Rscript foo.R
100
I know that Python has the global keyword to declare scope of variables (doesn't work with this example, though). Does R need something similar?
What am I doing wrong?
Update
Ah, is the problem that in add I am creating a new, local variable s that shadows the outer s? If so, how can I mutate s without creating a local variable?
Use the <<- operator for assignment in the add() function.
From ?"<<-":
The operators <<- and ->> are normally only used in functions, and cause a search to made through parent environments for an existing definition of the variable being assigned. If such a variable is found (and its binding is not locked) then its value is redefined, otherwise assignment takes place in the global environment. Note that their semantics differ from that in the S language, but are useful in conjunction with the scoping rules of R. See ‘The R Language Definition’ manual for further details and examples.
You can also use assign and define the scope precisely using the envir argument, works the same way as <<- in your add function in this case but makes your intention a little more clear:
foo <- function() {
s <- 100
add <- function() {
assign("s", s + 1, envir = parent.frame())
}
add()
s
}
cat(foo(), "\n")
Of course the better way for this kind of thing in R is to have your function return the variable (or variables) it modifies and explicitly reassigning them to the original variable:
foo <- function() {
s <- 100
add <- function(x) x + 1
s <- add(s)
s
}
cat(foo(), "\n")
Here is one more approach that can be a little safer than the assign or <<- approaches:
foo <- function() {
e <- environment()
s <- 100
add <- function() {
e$s <- e$s + 1
}
add()
s
}
foo()
The <<- assignment can cause problems if you accidentally misspell your variable name, it will still do something, but it will not be what you are expecting and can be hard to find the source of the problem. The assign approach can be tricky if you then want to move your add function to inside another function, or call it from another function. The best approach overall is to not have the functions modify variables outside their own scope and have the function return anything that is important. But when that is not possible, the above method uses lexical scoping to access the environment e, then assigns into the environment so it will always assign specifically into that function, never above or below.
I have created a function (which is quite long) that I have saved in a .txt file.
It works well (I use source(< >) to access it).
My problem is that I have created a few variables in that function
ie:
myfun<-function(a,b) {
Var1=....
Var2=Var1 + ..
}
Now I want to get those variables.
When I include return() inside the function, its fine: the value comes up on the screen, but when I type Var1 outside the function, I have an error message "the object cannot be found".
I am new to R, but I was thinking it might be because "myfun" operates in a different envrionment than the global one, but when I did
environment()
environment: R_GlobalEnv>
environment(myfun1)
environment: R_GlobalEnv>
It seems to me the problem is elsewhere...
Any idea?
Thanks
I realize this answer is more than 3 years old but I believe the option you are looking for is as follows:
myfun <- function(a,b) {
Var1 = (a + b) / 2 # do whatever logic you have to do here...
Var2 <<- Var1 + a # then output result to Global Environment with the "<<-" object.
}
The double "<<-" assignment operator will output "Var2" to the global environment and you can then use or reference it however you like without having to use "return()" inside your function.
If you want to do it in a nice way, write a class and than provide a print method. Within this class it is possible to return variables invisible. A nice book which covers such topics is "The Art of R programming".
An easy fix would be save each variable you need later on an list and than return a list
(as Peter pointed out):
return(list(VAR1=VAR1, .....))