Tidy eval: Evaluation of quosure in function within map - r

I am trying to write a function that makes use of an object's name (as in unevaluated symbol) for downstream application. Here is an example that captures the sense:
return_obj_name <- function(obj){
inp <- enquo(obj)
inp_name <- rlang::as_name(inp) # Use the name for something
inp_data <- rlang::eval_tidy(inp) # This line just for completeness, not important here
return(inp_name)
}
Here is a standard use case of this function:
test_obj <- 42
return_obj_name(test_obj)
[1] "test_obj"
So far, so good. However, I plan to use my function as an anonymous function in a map (or map2) statement, and this is where things go wrong.
test_obj2 <- 44
test_vec <- c(test_obj, test_obj2)
map(test_vec, ~ .x %>% return_obj_name())
[[1]]
[1] "."
[[2]]
[1] "."
The intended output would have been:
[[1]]
[1] "test_obj"
[[2]]
[1] "test_obj2"
I think I do understand what is happening. The function receives the piped reference to the initial object, which would be ".". It quotes this with enquo and continues as by design.
I am wondering if there is a way to with evaluate the reference in the environment in which map is called, as opposed to within the map call, as is happening now.

After you run
test_obj2 <- 44
test_vec <- c(test_obj, test_obj2)
The value test_vec has no knowledge of the names of the variables that were used to create it. All it knows is that it's a numeric vector that contains 42 and 44. Tracking the source for every variable would create a lot of overhead.
It's important to remember that values do not have names in R; it's the names that have values. And it's not always unique either. Multiple names can point to the same value.
In addition, the pipe operator does not preserve variable names. Observe
test_obj %>% return_obj_name()
# [1] "."
If you want to keep track of labels for value sources, you should either use a named list (but remember, it's the collection that tracks the names, the elements in the collection are unaware if they are named) or have a separate vector of names. The answer given by #Ronak offers some good alternatives using this strategy.
Another alternative would be to store your values as a collection of quosures. For example
test_vec <- quos(test_obj, test_obj2)
map(test_vec, ~return_obj_name(!!.x))
But here test_vec is storing those variable names, and not their necessarily their values. You would need to evaluate it to get the values 42 and 44.

The stand-alone example that you have shared does not match with the map intended output. In the stand-alone example you run return_obj_name(test_obj) and get output as "test_obj". Note that here the value of "test_obj" is 42. But in the map example your intended output is to return "42" and "44" instead of test_obj and test_obj2 ? One of this needs to change for the question to be consistent.
Anyway, as far as answer is concerned I think you should name your vector/list explicitly and pass that as separate object.
return_obj_name <- function(obj, name){
#Do something
#Do something
return(name)
}
For example, using tibble::lst which makes it easy to name objects.
test_vec <- tibble::lst(test_obj, test_obj2)
You can then use imap :
purrr::imap(test_vec, return_obj_name)
#$test_obj
#[1] "test_obj"
#$test_obj2
#[1] "test_obj2"
Or Map in base R :
Map(return_obj_name, test_vec, names(test_vec))
If you want to return "42" and "44" i.e the values here that would be obj value in return_obj_name function.

Related

Why are functions only sometimes first class variables in R? Using them from built-in data structures causes "Error: attempt to apply non-function"

I am trying to understand how first class functions work in R. I had understood that functions were first class in R, but was sorely disappointed when applying that understanding. When a function is saved to a list, whether that be as an ordinary list, a vector or a dictionary style list or vector, it is no longer callable, leading to the following error:
Error: attempt to apply non-function
e.g.
print_func <- function() {
print('hi')
}
print_func()
[1] "hi"
my_list = list(print_func)
my_list[0]()
Error: attempt to apply non-function
my_vector = c(print_func)
my_vector[0]()
Error: attempt to apply non-function
my_map <- c("a" = print_func)
my_map["a"]()
Error: attempt to apply non-function
So why is this? Does R not actually treat functions as first class members in all cases, or is there another reason why this occurs?
I see that R vectors also do unexpected things (for me - perhaps not for experienced R users) to nested arrays:
nested_vector <- c("a" = c("b" = 1))
nested_vector["a"]
<NA>
NA
nested_vector["a.b"]
a.b
1
Here it makes sense to me that "a.b" might reference the sub-member of the key "b" under the key "a". But apparently that logic goes out the window when trying to call the upper level key "a".
R is 1-based; so, you refer to the first element of a vector using index 1 (and not 0 like in python).
There are two approaches to accessing list elements:
accessing list elements while keeping a list (return a list containing the desired elements)
pulling an element out of a list
In the first case, the subsetting is done using a single pair of brackets ([]) and you will always get a list back. Note that this is different from python where you get a list only if you select more than one element (lst = [fun1, fun2]; lst[0] return fun1 and not a one-element list like R while lst[0:2] returns a list).
In the second approach, the subsetting is done using a double pair of brackets ([[]]). you basically pull an element completely out of a list; more like subsetting one element out of a list in python.
print_func <- function() {
print('hi')
}
print_func()
my_list = list(print_func)
mode(my_list[1]) # return a list (not a function); so, it's not callable
[1] "list"
mode(my_list[[1]]) # return a function; so, it's callable
[1] "function"
my_list[1]() # error
my_list[[1]]() # works
[1] "hi"
#
my_vector = c(print_func)
mode(my_vector) # a list, so, not callable
[1] "list"
my_vector[1]() # error because it returns a list and not a function
my_vector[[1]]() # works
[1] "hi"
When subsetting with names, the same logic of single and double pair of brackets applies
my_map <- c("a" = print_func)
mode(my_map) # list, so, not callable
[1] "list"
my_map["a"]() # error
my_map[["a"]]() # works
[1] "hi"
Limey pointed out my 2 issues in the comments. I was using a 0-index and I was using single brackets. If I use a 1-index and double brackets it works, and functions are treated as first class variables.
My issue is resolved, and hopefully I won't make that same mistake again.

How do you solve "could not find function "deparse<-" | "as.name<-" | "eval<-"" errors when trying to dynamically name dataframes in R? [duplicate]

I am using R to parse a list of strings in the form:
original_string <- "variable_name=variable_value"
First, I extract the variable name and value from the original string and convert the value to numeric class.
parameter_value <- as.numeric("variable_value")
parameter_name <- "variable_name"
Then, I would like to assign the value to a variable with the same name as the parameter_name string.
variable_name <- parameter_value
What is/are the function(s) for doing this?
assign is what you are looking for.
assign("x", 5)
x
[1] 5
but buyer beware.
See R FAQ 7.21
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f
You can use do.call:
do.call("<-",list(parameter_name, parameter_value))
There is another simple solution found there:
http://www.r-bloggers.com/converting-a-string-to-a-variable-name-on-the-fly-and-vice-versa-in-r/
To convert a string to a variable:
x <- 42
eval(parse(text = "x"))
[1] 42
And the opposite:
x <- 42
deparse(substitute(x))
[1] "x"
The function you are looking for is get():
assign ("abc",5)
get("abc")
Confirming that the memory address is identical:
getabc <- get("abc")
pryr::address(abc) == pryr::address(getabc)
# [1] TRUE
Reference: R FAQ 7.21 How can I turn a string into a variable?
Use x=as.name("string"). You can use then use x to refer to the variable with name string.
I don't know, if it answers your question correctly.
strsplit to parse your input and, as Greg mentioned, assign to assign the variables.
original_string <- c("x=123", "y=456")
pairs <- strsplit(original_string, "=")
lapply(pairs, function(x) assign(x[1], as.numeric(x[2]), envir = globalenv()))
ls()
assign is good, but I have not found a function for referring back to the variable you've created in an automated script. (as.name seems to work the opposite way). More experienced coders will doubtless have a better solution, but this solution works and is slightly humorous perhaps, in that it gets R to write code for itself to execute.
Say I have just assigned value 5 to x (var.name <- "x"; assign(var.name, 5)) and I want to change the value to 6. If I am writing a script and don't know in advance what the variable name (var.name) will be (which seems to be the point of the assign function), I can't simply put x <- 6 because var.name might have been "y". So I do:
var.name <- "x"
#some other code...
assign(var.name, 5)
#some more code...
#write a script file (1 line in this case) that works with whatever variable name
write(paste0(var.name, " <- 6"), "tmp.R")
#source that script file
source("tmp.R")
#remove the script file for tidiness
file.remove("tmp.R")
x will be changed to 6, and if the variable name was anything other than "x", that variable will similarly have been changed to 6.
I was working with this a few days ago, and noticed that sometimes you will need to use the get() function to print the results of your variable.
ie :
varnames = c('jan', 'feb', 'march')
file_names = list_files('path to multiple csv files saved on drive')
assign(varnames[1], read.csv(file_names[1]) # This will assign the variable
From there, if you try to print the variable varnames[1], it returns 'jan'.
To work around this, you need to do
print(get(varnames[1]))
If you want to convert string to variable inside body of function, but you want to have variable global:
test <- function() {
do.call("<<-",list("vartest","xxx"))
}
test()
vartest
[1] "xxx"
Maybe I didn't understand your problem right, because of the simplicity of your example. To my understanding, you have a series of instructions stored in character vectors, and those instructions are very close to being properly formatted, except that you'd like to cast the right member to numeric.
If my understanding is right, I would like to propose a slightly different approach, that does not rely on splitting your original string, but directly evaluates your instruction (with a little improvement).
original_string <- "variable_name=\"10\"" # Your original instruction, but with an actual numeric on the right, stored as character.
library(magrittr) # Or library(tidyverse), but it seems a bit overkilled if the point is just to import pipe-stream operator
eval(parse(text=paste(eval(original_string), "%>% as.numeric")))
print(variable_name)
#[1] 10
Basically, what we are doing is that we 'improve' your instruction variable_name="10" so that it becomes variable_name="10" %>% as.numeric, which is an equivalent of variable_name=as.numeric("10") with magrittr pipe-stream syntax. Then we evaluate this expression within current environment.
Hope that helps someone who'd wander around here 8 years later ;-)
Other than assign, one other way to assign value to string named object is to access .GlobalEnv directly.
# Equivalent
assign('abc',3)
.GlobalEnv$'abc' = 3
Accessing .GlobalEnv gives some flexibility, and my use case was assigning values to a string-named list. For example,
.GlobalEnv$'x' = list()
.GlobalEnv$'x'[[2]] = 5 # works
var = 'x'
.GlobalEnv[[glue::glue('{var}')]][[2]] = 5 # programmatic names from glue()

How to name an object out of a string in R [duplicate]

I am using R to parse a list of strings in the form:
original_string <- "variable_name=variable_value"
First, I extract the variable name and value from the original string and convert the value to numeric class.
parameter_value <- as.numeric("variable_value")
parameter_name <- "variable_name"
Then, I would like to assign the value to a variable with the same name as the parameter_name string.
variable_name <- parameter_value
What is/are the function(s) for doing this?
assign is what you are looking for.
assign("x", 5)
x
[1] 5
but buyer beware.
See R FAQ 7.21
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f
You can use do.call:
do.call("<-",list(parameter_name, parameter_value))
There is another simple solution found there:
http://www.r-bloggers.com/converting-a-string-to-a-variable-name-on-the-fly-and-vice-versa-in-r/
To convert a string to a variable:
x <- 42
eval(parse(text = "x"))
[1] 42
And the opposite:
x <- 42
deparse(substitute(x))
[1] "x"
The function you are looking for is get():
assign ("abc",5)
get("abc")
Confirming that the memory address is identical:
getabc <- get("abc")
pryr::address(abc) == pryr::address(getabc)
# [1] TRUE
Reference: R FAQ 7.21 How can I turn a string into a variable?
Use x=as.name("string"). You can use then use x to refer to the variable with name string.
I don't know, if it answers your question correctly.
strsplit to parse your input and, as Greg mentioned, assign to assign the variables.
original_string <- c("x=123", "y=456")
pairs <- strsplit(original_string, "=")
lapply(pairs, function(x) assign(x[1], as.numeric(x[2]), envir = globalenv()))
ls()
assign is good, but I have not found a function for referring back to the variable you've created in an automated script. (as.name seems to work the opposite way). More experienced coders will doubtless have a better solution, but this solution works and is slightly humorous perhaps, in that it gets R to write code for itself to execute.
Say I have just assigned value 5 to x (var.name <- "x"; assign(var.name, 5)) and I want to change the value to 6. If I am writing a script and don't know in advance what the variable name (var.name) will be (which seems to be the point of the assign function), I can't simply put x <- 6 because var.name might have been "y". So I do:
var.name <- "x"
#some other code...
assign(var.name, 5)
#some more code...
#write a script file (1 line in this case) that works with whatever variable name
write(paste0(var.name, " <- 6"), "tmp.R")
#source that script file
source("tmp.R")
#remove the script file for tidiness
file.remove("tmp.R")
x will be changed to 6, and if the variable name was anything other than "x", that variable will similarly have been changed to 6.
I was working with this a few days ago, and noticed that sometimes you will need to use the get() function to print the results of your variable.
ie :
varnames = c('jan', 'feb', 'march')
file_names = list_files('path to multiple csv files saved on drive')
assign(varnames[1], read.csv(file_names[1]) # This will assign the variable
From there, if you try to print the variable varnames[1], it returns 'jan'.
To work around this, you need to do
print(get(varnames[1]))
If you want to convert string to variable inside body of function, but you want to have variable global:
test <- function() {
do.call("<<-",list("vartest","xxx"))
}
test()
vartest
[1] "xxx"
Maybe I didn't understand your problem right, because of the simplicity of your example. To my understanding, you have a series of instructions stored in character vectors, and those instructions are very close to being properly formatted, except that you'd like to cast the right member to numeric.
If my understanding is right, I would like to propose a slightly different approach, that does not rely on splitting your original string, but directly evaluates your instruction (with a little improvement).
original_string <- "variable_name=\"10\"" # Your original instruction, but with an actual numeric on the right, stored as character.
library(magrittr) # Or library(tidyverse), but it seems a bit overkilled if the point is just to import pipe-stream operator
eval(parse(text=paste(eval(original_string), "%>% as.numeric")))
print(variable_name)
#[1] 10
Basically, what we are doing is that we 'improve' your instruction variable_name="10" so that it becomes variable_name="10" %>% as.numeric, which is an equivalent of variable_name=as.numeric("10") with magrittr pipe-stream syntax. Then we evaluate this expression within current environment.
Hope that helps someone who'd wander around here 8 years later ;-)
Other than assign, one other way to assign value to string named object is to access .GlobalEnv directly.
# Equivalent
assign('abc',3)
.GlobalEnv$'abc' = 3
Accessing .GlobalEnv gives some flexibility, and my use case was assigning values to a string-named list. For example,
.GlobalEnv$'x' = list()
.GlobalEnv$'x'[[2]] = 5 # works
var = 'x'
.GlobalEnv[[glue::glue('{var}')]][[2]] = 5 # programmatic names from glue()

Convert character vector to numeric vector in R for value assignment?

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

Sort a list of nontrivial elements in R

In R, I have a list of nontrivial objects (they aren't simple objects like scalars that R can be expected to be able to define an order for). I want to sort the list. Most languages allow the programmer to provide a function or similar that compares a pair of list elements that is passed to a sort function. How can I sort my list?
To make this is as simple I can, say your objects are lists with two elements, a name and a value. The value is a numeric; that's what we want to sort by. You can imagine having more elements and needing to do something more complex to sort.
The sort help page tells us that sort uses xtfrm; xtfrm in turn tells us it will use == and > methods for the class of x[i].
First I'll define an object that I want to sort:
xx <- lapply(c(3,5,7,2,4), function(i) list(name=LETTERS[i], value=i))
class(xx) <- "myobj"
Now, since xtfrm works on the x[i]'s, I need to define a [ function that returns the desired elements but still with the right class
`[.myobj` <- function(x, i) {
class(x) <- "list"
structure(x[i], class="myobj")
}
Now we need == and > functions for the myobj class; this potentially could be smarter by vectorizing these properly; but for the sort function, we know that we're only going to be passing in myobj's of length 1, so I'll just use the first element to define the relations.
`>.myobj` <- function(e1, e2) {
e1[[1]]$value > e2[[1]]$value
}
`==.myobj` <- function(e1, e2) {
e1[[1]]$value == e2[[1]]$value
}
Now sort just works.
sort(xx)
It might be considered more proper to write a full Ops function for your object; however, to just sort, this seems to be all you need. See p.89-90 in Venables/Ripley for more details about doing this using the S3 style. Also, if you can easily write an xtfrm function for your objects, that would be simpler and most likely faster.
The order function will allow you to determine the sort order for character or numeric aruments and break ties with subsequent arguments. You need to be more specific about what you want. Produce an example of a "non-trivial object" and specify the order you desire in some R object. Lists are probably the most non-vectorial objects:
> slist <- list(cc=list(rr=1), bb=list(ee=2, yy=7), zz="ww")
> slist[order(names(slist))] # alpha order on names()
$bb
$bb$ee
[1] 2
$bb$yy
[1] 7
$cc
$cc$rr
[1] 1
$zz
[1] "ww"
slist[c("zz", "bb", "cc")] # an arbitrary ordering
$zz
[1] "ww"
$bb
$bb$ee
[1] 2
$bb$yy
[1] 7
$cc
$cc$rr
[1] 1
One option is to create a xtfrm method for your objects. Functions like order take multiple columns which works in some cases. There are also some specialized functions for specific cases like mixedsort in the gtools package.

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