This code is about inverting an index using clusters.
Unfortunately I do not understand the line with recognize<-...
I know that the function Vectorize applies the inner function element-wise, but I do not understand the inner function here.
The parameters (uniq, test) are not defined, how can we apply which then? Also why is there a "uniq" as text right after?
slots <- as.integer(Sys.getenv("NSLOTS"))
cl <- makeCluster(slots, type = "PSOCK")
inverted_index4<-function(x){
y <- unique(x)
recognize <- Vectorize(function(uniq,text) which(text %in% uniq),"uniq",SIMPLIFY = F)
y2 <- parLapply(cl, y, recognize, x)
unlist(y2,recursive=FALSE)
}
The
Vectorise()
function is just making a new element wise, vectorised function of the custom function
function(uniq,text) which(text %in% uniq).
The 'uniq' string is the argument of that function that you must specify you want to iterate over. Such that now you can pass a vector of length greater than one for uniq, and get returned a list with an element for the output of the function evaluated for every element of the input vector uniq.
I would suggest the author make the code a little clearer, better commented etc. the vectorise function doesn't need to be inside the function call necessarily.
Note
ParLapply()
isn't a function I recognise. But the x will be passed to the recognise function and the second argument text should presumably be defined earlier on, in the global environment, .GlobalEnv().
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!
I have a problem with elipsis usecase. My function accepts list of objects, let's call them objects of class "X". Now, objects X are being processed inside of my function to class "Xs", so I have list of "Xs" objects. Function that I import from other package can compute multiple "Xs" objects at once but they have to be enumerated (elipsis mechanic), not passed as list. Is there a way how to solve it? I want something like this
examplefun <- function(charlist){
nums <- lapply(charlist, as.numeric)
sum(... = nums)
}
Of course example above throws an error but it shows what i want to achieve. I tried to unlist with recursive = FALSE ("X" and "Xs" are the list itself) but it does not work.
If there is no solution then:
Let's assume I decideed to accept ... insted of list of "X" objects. Can I modify elipsis elements (change them to "Xs") and then pass to function that accepts elipsis? So it will look like this:
examplefun2 <- function(...){
function that modify object in ... to "Xs" objects
sum(...)
}
In your first function, just call sum directly because sum works correctly on vectors of numbers instead of individual numbers.
examplefun <- function (charlist) {
nums <- vapply(charlist, as.numeric, numeric(1L))
sum(nums)
}
(Note the use of vapply instead of lapply: sum expects an atomic vector, we can’t pass a list.)
In your second function, you can capture ... and work with the captured variable:
examplefun2 <- function (...) {
nums <- as.numeric(c(...))
sums(nums)
}
For more complex arguments, Roland’s comment is a good alternative: Modify the function arguments as a list, and pass it to do.call.
I am trying to write a function and apply it to a list. Inside my function is a function written by some one else. If I make my list very easy, everything will work fine. But if I use all the real data I have, there are some bad objects and the outside function doesn't work and my whole function won't go through.
What do I type to say "If the outside function doesn't work, skip that object and move to the next one in the list."? With or without NA, doesn't matter.
I cannot figure out how to write a reproducible example that would result in a list of dataframes, which is what happens inside this function. I'm willing to take any help to improve this question.
My function is something like this:
do_this<- function(x){
outside_function(x))%>% #this returns a dataframe for each object
filter()%>%
select()%>%
summarise_each(funs(mean(., na.rm = TRUE))) #by the end the df is down to one row
}
This is how I apply the function to the list to come up with my final dataframe.
df<-bind_rows(lapply(my_list, do_this))
An example:
myfun <- function(x) {if (x == 1) {stop("bad")} else x}
throws error on input of 1:
lapply(1:4, myfun) # stops from error
Just wrap it in try (as long as you don't need more complex error handling):
L <- lapply(1:4, function(x) try(myfun(x)))
And then you can use Filter to get rid of the "bad" cases:
Filter(function(x) !inherits(x, "try-error"), L)
Although you may want to just make your wrapper function more robust, or return NULL (or some other appropriate value) under the condition that makes the inner function fail.
I am used to use apply familiy functions to avoid for loop with R. In this context I was wondering it there is a way to avoid typing a bound variable. For example, say I want to do 100 times an operation do.call(myfun, args). With for I'd write:
res = seq(100)
for(i in seq(100)){res[i] = do.call(myfun, args)}
with apply I type:
res = sapply(seq(100), function(i) do.call(myfun, args))
I understand that sapply tries to apply the function to one argument, it is an element of seq(100), but is there a way to avoid this, because indeed this variable (here i) has no meaning neither utility ?
thanks for the insight
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