I am wondering if it is possible in R to use a value that is declared in a function call as a "variable" part of the function itself, similar to the functionality that is available in SAS IML.
Given something like this:
put.together <- function(suffix, numbers) {
new.suffix <<- as.data.frame(numbers)
return(new.suffix)
}
x <- c(seq(1000,1012, 1))
put.together(part.a, x)
new.part.a ##### does not exist!!
new.suffix ##### does exist
As it is written, the function returns a dataframe called new.suffix, as it should because that is what I'm asking it to do.
I would like to get a dataframe returned that is called new.part.a.
EDIT: Additional information was requested regarding the purpose of the analysis
The purpose of the question is to produce dataframes that will be sent to another function for analysis.
There exists a data bank where elements are organized into groups by number, and other people organize the groups
into a meaningful set.
Each group has an id number. I use the information supplied by others to put the groups together as they are specified.
For example, I would be given a set of id numbers like: part-1 = 102263, 102338, 202236, 302342, 902273, 102337, 402233.
So, part-1 has seven groups, each group having several elements.
I use the id numbers in a merge so that only the groups of interest are extracted from the large data bank.
The following is what I have for one set:
### all.possible.elements.bank <- .csv file from large database ###
id.part.1 <- as.data.frame(c(102263, 102338, 202236, 302342, 902273, 102337, 402233))
bank.names <- c("bank.id")
colnames(id.part.1) <- bank.names
part.sort <- matrix(seq(1,nrow(id.part.1),1))
sort.part.1 <- cbind(id.part.1, part.sort)
final.part.1 <- as.data.frame(merge(sort.part.1, all.possible.elements.bank,
by="bank.id", all.x=TRUE))
The process above is repeated many, many times.
I know that I could do this for all of the collections that I would pull together, but I thought I would be able to wrap the selection process into a function. The only things that would change would be the part numbers (part-1, part-2, etc..) and the groups that are selected out.
It is possible using the assign function (and possibly deparse and substitute), but it is strongly discouraged to do things like this. Why can't you just return the data frame and call the function like:
new.part.a <- put.together(x)
Which is the generally better approach.
If you really want to change things in the global environment then you may want a macro, see the defmacro function in the gtools package and most importantly read the document in the refrences section on the help page.
This is rarely something you should want to do... assigning to things out of the function environment can get you into all sorts of trouble.
However, you can do it using assign:
put.together <- function(suffix, numbers) {
assign(paste('new',
deparse(substitute(suffix)),
sep='.'),
as.data.frame(numbers),
envir=parent.env(environment()))
}
put.together(part.a, 1:20)
But like Greg said, its usually not necessary, and always dangerous if used incorrectly.
Related
I have successfully allocated dataframe names and populated them (see code) but I do not know how to subsequently reference them. So I loop through to assign df.test1 and populate it with some data 1 and so on. I know that the df has been created, and can view or summary it in the console, but not in the code.
I am pretty new to R so am not sure if some of the solutions I have looked at apply to me.
num.clusters <- 5
for (i in 1:num.clusters) {
assign(paste("df.test",i,sep=""), paste("somedata", i))
}
This works but Then want to do something like:
View(df.test,i)
to view whatever iteration from 1 to 5.
I want to be able to use the assigned dataframes like any other dataframe. I could hard code this as View(df.test1) but that would defeat the point. I also want to do other things with the datframe, e.g. subsetting.
I know this doesn't work. Would love to know what does.
Many thanks...
Your question is the proof that the approach is problematic: avoid using assign in general because it makes accessing the variables afterwards awkward (among other issues).
A cleaner way is to just put your "data frames" (copying from your example) in a list:
num.clusters <- 5
df.test <- list()
for (i in 1:num.clusters) {
df.test[[i]] <- paste("somedata", i)
}
Then you would just access them like this:
View(df.test[[i]])
If what you put in there was an actual data.frame (and not the strings you were using), you could then access its columns like any other data.frame:
df.test[[i]]$Name
Or
df.test[[i]][, "Name"]
this question might be very simple, but I do not find a good way to solve it:
I have a dataset with many subgroups which need to be analysed all-together and on their own. Therefore, I want to use subsets for the groups and use them for the later analysis. As well, the defintion of the subsets as the analysis should be partly done with loops in order to save space and to ensure that the same analysis has been done with all subgroups.
Here is an example of my code using an example dataframe from the boot package:
data(aids)
qlist <- c("1","2","3","4")
for (i in length(qlist)) {
paste("aids.sub.",qlist[i],sep="") <- subset(aids, quarter==qlist[i])
}
The variable which contains the subgroups in my dataset is stored as a string, therefore I added the qlist part which would be not required otherwise.
Make a list of the subsets with lapply:
lapply(qlist, function(x) subset(aids, quarter==x))
Equivalently, avoiding the subset():
lapply(qlist, function(x) aids[aids$quarter==x,])
It is likely the case that using a list will make the subsequent code easier to write and understand. You can subset the list to get a single data frame (just as you can use one of the subsets, as created below). But you can also iterate over it (using for or lapply) without having to construct variable names.
To do the job as you are asking, use assign:
for (i in qlist) {
assign(paste("aids.sub.",i,sep=""), subset(aids, quarter==i))
}
Note the removal of the length() function, and that this is iterating directly over qlist.
Is it possible to return 4 different data frames from one function?
Scenario:
I am trying to read a file, parse it, and return some parts of the file.
My function looks something like this:
parseFile <- function(file){
carFile <- read.table(file, header=TRUE, sep="\t")
carNames <- carFile[1,]
carYear <- colnames(carFile)
return(list(carFile,carNames,carYear))
}
I don't want to have to use list(carFile,carNames,carYear). Is there a way return the 3 data frames without returning them in a list first?
R does not support multiple return values. You want to do something like:
foo = function(x,y){return(x+y,x-y)}
plus,minus = foo(10,4)
yeah? Well, you can't. You get an error that R cannot return multiple values.
You've already found the solution - put them in a list and then get the data frames from the list. This is efficient - there is no conversion or copying of the data frames from one block of memory to another.
This is also logical, the return from a function should conceptually be a single entity with some meaning that is transferred to whatever function is calling it. This meaning is also better conveyed if you name the returned values of the list.
You could use a technique to create multiple objects in the calling environment, but when you do that, kittens die.
Note in your example carYear isn't a data frame - its a character vector of column names.
There are other ways you could do that, if you really really want, in R.
assign('carFile',carFile,envir=parent.frame())
If you use that, then carFile will be created in the calling environment. As Spacedman indicated you can only return one thing from your function and the clean solution is to go for the list.
In addition, my personal opinion is that if you find yourself in such a situation, where you feel like you need to return multiple dataframes with one function, or do something that no one has ever done before, you should really revisit your approach. In most cases you could find a cleaner solution with an additional function perhaps, or with the recommended (i.e. list).
In other words the
envir=parent.frame()
will do the job, but as SpacedMan mentioned
when you do that, kittens die
The zeallot package does what you need in a similar that Python can unpack variables from a function. Reproducible example below.
parseFile <- function(){
carMPG <- mtcars$mpg
carName <- rownames(mtcars)
carCYL <- mtcars$cyl
return(list(carMPG,carName,carCYL))
}
library(zeallot)
c(myFile, myName, myYear) %<-% parseFile()
I'm trying to subset a dataframe within a function using a mixture of fixed variables and some variables which are created within the function (I only know the variable names, but cannot vectorise them beforehand). Here is a simplified example:
a<-c(1,2,3,4)
b<-c(2,2,3,5)
c<-c(1,1,2,2)
D<-data.frame(a,b,c)
subbing<-function(Data,GroupVar,condition){
g=Data$c+3
h=Data$c+1
NewD<-data.frame(a,b,g,h)
subset(NewD,select=c(a,b,GroupVar),GroupVar%in%condition)
}
Keep in mind that in my application I cannot compute g and h outside of the function. Sometimes I'll want to make a selection according to the values of h (as above) and other times I'll want to use g. There's also the possibility I may want to use both, but even just being able to subset using 1 would be great.
subbing(D,GroupVar=h,condition=5)
This returns an error saying that the object h cannot be found. I've tried to amend subset using as.formula and all sorts of things but I've failed every single time.
Besides the ease of the function there is a further reason why I'd like to use subset.
In the function I'm actually working on I use subset twice. The first time it's the simple subset function. It's just been pointed out below that another blog explored how it's probably best to use the good old data[colnames()=="g",]. Thanks for the suggestion, I'll have a go.
There is however another issue. I also use subset (or rather a variation) in my function because I'm dealing with several complex design surveys (see package survey), so subset.survey.design allows you to get the right variance estimation for subgroups. If I selected my group using [] I would get the wrong s.e. for my parameters, so I guess this is quite an important issue.
Thank you
It's happening right as the function is trying to define GroupVar in the beginning. R is looking for the object h by itself (not within the dataframe).
The best thing to do is refer to the column names in quotes in the subset function. But of course, then you'd have to sidestep the condition part:
subbing <- function(Data, GroupVar, condition) {
....
DF <- subset(Data, select=c("a","b", GroupVar))
DF <- DF[DF[,3] %in% condition,]
}
That will do the trick, although it can be annoying to have one data frame indexing inside another.
How can I select the second column of a dynamically named variable?
I create variables of the form "population.USA", "population.Mexico", "population.Canada". Each variable has a column for the year, and another column for the population value. I would like to select the second column from each of these variables during a loop.
I use this syntax:
sprintf("population.%s", country)[, 2]
R returns the error: Error in sprintf("population.%s", country)[, 2] : incorrect number of dimensions
Based on your sequence of questions over the last few minutes, I have two general recommendations for you as you get familiar with R:
Don't use sprintf.
Don't use assign.
Now, obviously, those functions are both useful at times. But you've learned about them too early, before you've mastered some basic stuff about R's data structures. Try to write code without those crutches (for the time being!), as they're just causing you problems.
Rather than creating separate individual variables for each nation's population, place them in a list.
population <- vector("list",3)
names(population) <- c('USA','Mexico','Russia')
Then you can access each using the string representation of the name of each country:
population[['USA']] <- 10000
Or,
region <- 'USA'
population[[region]]
In this example, I've assigned a single value to a list element, lists will hold any other data type, including matrices or data frames. It will be a lot less typing than using sprintf and assign, and a lot safer and more efficient as well.
See ?get. Here is an example:
> country <- "FOO"
> assign(sprintf("population.%s", country), data.frame(runif(5), runif(5)))
>
> get(sprintf("population.%s", country))[,2]
[1] 0.2241105 0.5640709 0.5945869 0.1830719 0.1895938
It is critically important to look at the object returned by a function if you get an error. It is immediately clear why your example fails if you just look at what it returns:
> sprintf("population.%s", country)
[1] "population.FOO"
At that point it would be immediately clear, if you didn't already know or have thought to read ?sprintf, that sprintf() returns a string not the object of that name. Armed with that knowledge you would have narrowed down the problem to how to recall an object from the computed name?