I have a dataframe created from a loop. The loop examines ordinal regression for a couple dozen outcomes for a given exposure.
At the beginning of the loop, a variable called exposure is defined. Example: exposure <- "MyExposure"
At the end of the routine, I want to actually save the resulting data set I've compiled and to have the name of the saved data object be related to the exposure.
I've had issues with making the left hand side of the assignment based on the variable names.
The name of the new dataframe should be
paste0(exposure,"_imputed_ds")
[1] "MyExposure_imputed_ds"
However, when I try to put this on left hand of an assignment, it fails.
paste0(exposure,"_imputed_ds") <- existing.data.frame
Error in paste0(exposure,"_imputed_ds") <- existing.data.frame
could not find function "paste0<-"
What I wanted was a new dataframe named MyExposure_imputed_ds that contained contents of existing.data.frame
You can use assign() to set a value for a name you construct with paste
assign(paste0('MyExposure', '_imputed_ds'), 5)
Now you have MyExposure_imputed_ds in the environment with value 5
I find the use of assign to be generally a warning flag, though! Maybe you want something like this instead...
imputed_ds <- list()
imputed_ds[['MyExposure']] <- 5
Related
It could be a very easy question, given that I am very unfamiliar with R. I know normally one can use deparse(substitute(.)) to extract the name of a variable. However, if I have a long list of variables (let's say it's built without names), how can I extract the name of each variable efficiently? I was thinking about using loops, but the deparse(substitute(.)) method would obviously generate the 'general' variable name we used to denote every item.
Sample code:
countries<-
list(austria,belgium,czech,denmark,france,germany,italy,luxemberg,netherlands,poland,swiss)
Suppose I want to get countryNames equals to list("austria","belgium",...,"swiss"), how shall I code? I tried generating the list using countries <- list(countryA = countryA, countryB = countryB, ...), but it was extremely tedious, and in some cases I might only have an unnamed input list from elsewhere.
countries would just have values of each individual objects (austria,belgium etc.). To access the names you need to create a named list while creating countries which can be done like :
countries <- list(austria = austria,belgium = belgium....)
However, if this is very tedious you can use tibble::lst which creates the names automatically without explicitly mentioning them.
countries <- tibble::lst(austria,belgium....)
In both the case you can access the names using names(countries).
If the country objects are the only ones loaded in the global environment, we can do this easily with ls and mget to return a named list of values
countries <- mget(ls())
got a list named x which has 25 elements. Each of which is named after a country. Inside each country i have 10-20 variables and i want to create a new one inside each dataframe which uses one already existing variable (lets name it y) to create a new variable.
closest i have come to solving it is using a for loop but using the [[<- creates newvariables outside the list (which is too much of a hustle to put in again) with "[[" in their name.
also assign doesnt work either due to invalid first argument error (propably due to my effort to dynamically create them)
i guess a nested lapply is the best option but being new to R dont know how that would work
We can loop over the list with lapply and transform to create a new variable
lapply(lst, transform, newVar = y)
I do not really understand your question, could you paste some code and say what you'd like to see. When I have to create variables names dynamically I use trick as the one below:
names(tmp) = paste('y', 1:25, sep='')
If you search help(names) you should get how to change columns and row names specifically. Paste is just a way to concatenate different types of text you want into your variable names
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?
I am a relatively novice r user and am attempting to use the partimat() function within the klaR package to plot decision boundaries for a linear discriminant analysis but I keep encountering the same error. I have tried inputing the arguments multiple different ways according to the manual, but keep getting the following error:
Error in partimat.default(x, grouping, ...) :
at least two classes required
Here is an example of the input I've given:
partimat(sources1[,c(3:19)],grouping=sources1[,2],method="lda",prec=100)
where my data table is loaded in under the name "sources1" with columns 3 through 19 containing the explanatory variables and column 2 containing the classes. I have also tried doing it by entering the formula like so:
partimat(sources1$group~sources1$tio2+sources1$v+sources1$cr+sources1$co+sources1$ni+sources1$rb+sources1$sr+sources1$y+sources1$zr+sources1$nb+sources1$la+sources1$gd+sources1$yb+sources1$hf+sources1$ta+sources1$th+sources1$u,data=sources1)
with these being the column heading.
I have successfully run an LDA on this same data set without issue so I'm not quite sure what is wrong.
From the source code of the partimat.default function getAnywhere(partimat.default) it states
if (nlevels(grouping) < 2)
stop("at least two classes required")
Therefore maybe you haven't defined your grouping column as a factor variable. If you try summary(sources1[,2]) what do you get? If it's not a factor, try
sources1[,2] <- as.factor(sources1[,2])
Or in method 2 try removing the "sources1$"on each of your variable names in the formula as you specify the data frame in which to look for these variable names in the data argument. I think you are effectively specifying the dataframe twice and it might be looking, for instance, for
"sources1$sources1$groups"
Rather than
"sources1$groups"
Without further error messages or a reproducible example (i.e. include some data in your post) it's hard to say really.
HTH
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.