Related
When working with packages like openxlsx, I often find myself writing repetetive code such as defining the wb and sheet arguments with the same values.
To respect the DRY principle, I would like to define one variable that contains multiple arguments. Then, when I call a function, I should be able to provide said variable to define multiple arguments.
Example:
foo <- list(a=1,b=2,c=3)
bar <- function(a,b,c,d) {
return(a+b+c+d)
}
bar(foo, d=4) # should return 10
How should the foo() function be defined to achieve this?
Apparently you are just looking for do.call, which allows you to create and evaluate a call from a function and a list of arguments.
do.call(bar, c(foo, d = 4))
#[1] 10
How should the foo() function be defined to achieve this?
You've got it slightly backwards. Rather than trying to wrangle the output of foo into something that bar can accept, write foo so that it takes input in a form that is convenient to you. That is, create a wrapper function that provides all the boilerplate arguments that bar requires, without you having to specify them manually.
Example:
bar <- function(a, b, c, d) {
return(a+b+c+d)
}
call_bar <- function(d=4) {
bar(1, 2, 3, d)
}
call_bar(42) # shorter than writing bar(1, 2, 3, 42)
I discovered a solution using rlang::exec.
First, we must have a function to structure the dots:
getDots <- function(...) {
out <- sapply(as.list(match.call())[-1], function(x) eval(parse(text=deparse(x))))
return(out)
}
Then we must have a function that executes our chosen function, feeding in our static parameters as a list (a, b, and c), in addition to d.
execute <- function(FUN, ...) {
dots <-
getDots(...) %>%
rlang::flatten()
out <- rlang::exec(FUN, !!!dots)
return(out)
}
Then calling execute(bar, abc, d=4) returns 10, as it should do.
Alternatively, we can write bar %>% execute(abc, d=4).
Let me give you an example!
How to get two or more return values from a function
Method 1: Set global variables, so that if you change global variables in formal parameters, it will also be effective in actual parameters. So you can change the value of multiple global variables in the formal parameter, then in the actual parameter is equivalent to returning multiple values.
Method 2: If you use the array name as a formal parameter, then you change the contents of the array, such as sorting, or perform addition and subtraction operations, and it is still valid when returning to the actual parameter. This will also return a set of values.
Method 3: Pointer variables can be used. This principle is the same as Method 2, because the array name itself is the address of the first element of the array. Not much to say.
Method 4: If you have learned C++, you can quote parameters
You can try these four methods here, I just think the problem is a bit similar, so I provided it to you, I hope it will help you!
Note: This is separate from, though perhaps similar to, the
deparse-substitute trick
of attaining the name of a passed argument.
Consider the following situation: I have some function to be called, and the return value
is to be assigned to some variable, say x.
Inside the function, how can I capture that the name to be assigned to the
returned value is x, upon calling and assigning the function?
For example:
nameCapture <- function() {
# arbitrary code
captureVarName()
}
x <- nameCapture()
x
## should return some reference to the name "x"
What in R closest approximates captureVarName() referenced in the example?
My intuition was that there would be something in the call stack to do with
assign(), where x would be an argument and could be extracted, but
sys.call() yielded nothing of the sort; does it then occur internally, and if
so, what is a sensible way to attain something like captureVarName()?
My notion is that it would act in a similar manner to how the following works, though without the assign() function, using the <- operator instead:
nameCapture <- function() sys.call(1)[[2]]
assign("x", nameCapture())
x
# [1] "x"
testing<-function(formula=NULL,data=NULL){
if(with(data,formula)==T){
print('YESSSS')
}
}
A<-matrix(1:16,4,4)
colnames(A)<-c('x','y','z','gg')
A<-as.data.frame(A)
testing(data=A,formula=(2*x+y==Z))
Error in eval(expr, envir, enclos) : object 'x' not found
##or I can put formula=(x=1)
##reason that I use formula is because my dataset had different location and I would want
##to 'subset' my data into different set
This is the main flow of my code. I had done some search and seems to be no one ask this kind of stupid question or it is not possible to pass a formula in a if statement. Thank you in advance
if you just want subset of your data.frame create a character object representing the formula like this:
formula="2*x+y==z"
testing<-function(data,formula){with(data = data,expr = eval(parse(text = formula)))}
subset(A,testing(A,formula=formula))
#x y z gg
#2 2 6 10 14
You can change the formula as per your need.
If we need to evaluate it, one option is eval(parse
testing<-function(formula=NULL,data=NULL){
data <- deparse(substitute(data))
if(any(eval(parse(text=paste("with(", data, ",",
deparse(substitute(formula)), ")")))))
print("YESSS")
}
testing(data=A,formula=(2*x+y==z))
#[1] "YESSS"
When you call a function in R it evaluates its arguments first before executing the function.
For example, prod(2+2, 3) is first turned into prod(4, 3) before the function prod() is even called.
Thus, in your code, R starts by trying to solve (2*x+y==Z). It fails because there is no x object outside of the function code. So, it not even begin running testing().
To use your function correctly you should make it clear to R that it is not supposed to calculate (2*x+y==Z). Instead it should pass this information as is. You could do that using the functions expression() and eval().
testing<-function(formula=NULL,data=NULL){
if(with(data,eval(formula==T)){
print('YESSSS')
}
}
A<-matrix(1:16,4,4)
colnames(A)<-c('x','y','z','gg')
A<-as.data.frame(A)
testing(data=A,formula=expression(2*x+y==Z))
However, you will notice that there other problems with your code.
For Z is different than z. Notice that the in colnames you use z and in the formula Z.
The if() only works for when there is a single value of true or false. In your case, you will have one value for each row in A. When this happens, if() will only check if the first row fits the criteria.
If your purpose is subsetting, it is much more easier to do:
A.subset <- subset(A, 2*A$x+A$y == A$z)
After a discussion with my colleague,
here is a kind of solution
testing<-function(cx,cy,px,py,z,data=NULL){
list<-NULL
for(m in 1:nrow(data)){
if(cx*data$x[m]^px+cy*data$y[m]^py+data$z==0){
print(m)}
}
}
but this can deal with polynomial only and with a lot of arguments in the function. I am think of a way to reduce it as a general equation.or maybe this is the most easiest equation.
I know the function get can help you transform values to variable names, like get(test[1]). But I find it is not compatible when the values is in a list format. Look at below example:
> A = c("obj1$length","obj1$width","obj1$height","obj1$weight")
> obj1 <- NULL
> obj1$length=c(1:4);obj1$width=c(5:8);obj1$height=c(9:12);obj1$weight=c(13:16)
> get(A[1])
Error in get(A[1]) : object 'obj1$length' not found
In this case, how can I retrieve the variable name?
get doesn't work like that you need to specify the variable and environment (the list is coerced to one) separately:
get("length",obj1)
[1] 1 2 3 4
Do do it with the data you have, you need to use eval and parse:
eval(parse(text=A[1]))
[1] 1 2 3 4
However, I suggest you rethink your approach to the problem as get, eval and parse are blunt tools that can bite you later.
I think that eval() function will do the trick, among other uses.
eval(A[1])
>[1] 1 2 3 4
You could also find useful this simple function I implemented (based in the commonly used combo eval, parse, paste):
evaluate<-function(..., envir=.GlobalEnv){ eval(parse(text=paste( ... ,sep="")), envir=envir) }
It concatenates and evaluates several character type objects. If you want it to be used inside another function, add at the begining of your function
envir <- environment()
and use it like this:
evaluate([some character objects], envir=envir)
Try, for example
myvariable1<-"aaa"; myvariable2<-"bbb"; aaa<-15; bbb<-3
evaluate(myvariable1," * ",myvariable2).
I find it very usefull when I have to evaluate similar sentences with several variables, or when I want to create variables with automatically generated names.
for(i in 1:100){evaluate("variable",i,"<-2*",i)}
I have:
z = data.frame(x1=a, x2=b, x3=c, etc)
I am trying to do:
for (i in 1:10)
{
paste(c('N'),i,sep="") -> paste(c('z$x'),i,sep="")
}
Problems:
paste(c('z$x'),i,sep="") yields "z$x1", "z$x1" instead of calling the actual values. I need the expression to be evaluated. I tried as.numeric, eval. Neither seemed to work.
paste(c('N'),i,sep="") yields "N1", "N2". I need the expression to be merely used as name. If I try to assign it a value such as paste(c('N'),5,sep="") -> 5, ie "N5" -> 5 instead of N5 -> 5, I get target of assignment expands to non-language object.
This task is pretty trivial since I can simply do:
N1 = x1...
N2 = x2...
etc, but I want to learn something new
I'd suggest using something like for( i in 1:10 ) z[,i] <- N[,i]...
BUT, since you said you want to learn something new, you can play around with parse and substitute.
NOTE: these little tools are funny, but experienced users (not me) avoid them.
This is called "computing on the language". It's very interesting, and it helps understanding the way R works. Let me try to give an intro:
The basic language construct is a constant, like a numeric or character vector. It is trivial because it is not different from its "unevaluated" version, but it is one of the building blocks for more complicated expressions.
The (officially) basic language object is the symbol, also known as a name. It's nothing but a pointer to another object, i.e., a token that identifies another object which may or may not exist. For instance, if you run x <- 10, then x is a symbol that refers to the value 10. In other words, evaluating the symbol x yields the numeric vector 10. Evaluating a non-existant symbol yields an error.
A symbol looks like a character string, but it is not. You can turn a string into a symbol with as.symbol("x").
The next language object is the call. This is a recursive object, implemented as a list whose elements are either constants, symbols, or another calls. The first element must not be a constant, because it must evaluate to the real function that will be called. The other elements are the arguments to this function.
If the first argument does not evaluate to an existing function, R will throw either Error: attempt to apply non-function or Error: could not find function "x" (if the first argument is a symbol that is undefined or points to something other than a function).
Example: the code line f(x, y+z, 2) will be parsed as a list of 4 elements, the first being f (as a symbol), the second being x (another symbol), the third another call, and the fourth a numeric constant. The third element y+z, is just a function with two arguments, so it parses as a list of three names: '+', y and z.
Finally, there is also the expression object, that is a list of calls/symbols/constants, that are meant to be evaluated one by one.
You'll find lots of information here:
https://github.com/hadley/devtools/wiki/Computing-on-the-language
OK, now let's get back to your question :-)
What you have tried does not work because the output of paste is a character string, and the assignment function expects as its first argument something that evaluates to a symbol, to be either created or modified. Alternativelly, the first argument can also evaluate to a call associated with a replacement function. These are a little trickier, but they are handled by the assignment function itself, not by the parser.
The error message you see, target of assignment expands to non-language object, is triggered by the assignment function, precisely because your target evaluates to a string.
We can fix that building up a call that has the symbols you want in the right places. The most "brute force" method is to put everything inside a string and use parse:
parse(text=paste('N',i," -> ",'z$x',i,sep=""))
Another way to get there is to use substitute:
substitute(x -> y, list(x=as.symbol(paste("N",i,sep="")), y=substitute(z$w, list(w=paste("x",i,sep="")))))
the inner substitute creates the calls z$x1, z$x2 etc. The outer substitute puts this call as the taget of the assignment, and the symbols N1, N2 etc as the values.
parse results in an expression, and substitute in a call. Both can be passed to eval to get the same result.
Just one final note: I repeat that all this is intended as a didactic example, to help understanding the inner workings of the language, but it is far from good programming practice to use parse and substitute, except when there is really no alternative.
A data.frame is a named list. It usually good practice, and idiomatically R-ish not to have lots of objects in the global environment, but to have related (or similar) objects in lists and to use lapply etc.
You could use list2env to multiassign the named elements of your list (the columns in your data.frame) to the global environment
DD <- data.frame(x = 1:3, y = letters[1:3], z = 3:1)
list2env(DD, envir = parent.frame())
## <environment: R_GlobalEnv>
## ta da, x, y and z now exist within the global environment
x
## [1] 1 2 3
y
## [1] a b c
## Levels: a b c
z
## [1] 3 2 1
I am not exactly sure what you are trying to accomplish. But here is a guess:
### Create a data.frame using the alphabet
data <- data.frame(x = 'a', y = 'b', z = 'c')
### Create a numerical index corresponding to the letter position in the alphabet
index <- which(tolower(letters[1:26]) == data[1, ])
### Use an 'lapply' to apply a function to every element in 'index'; creates a list
val <- lapply(index, function(x) {
paste('N', x, sep = '')
})
### Assign names to our list
names(val) <- names(data)
### Observe the result
val$x