Assign a variable in R using another variable

Question

I have to run 10's of different permutations with same structure but different base names for the output. to avoid having to keep replacing the whole character names within each formula, I was hoping to great a variable then use paste function to assign the variable to the name of the output..

Example:

var<-"Patient1"
(paste0("cells_", var, sep="") <- WhichCells(object=test, expression = test > 0, idents=c("patient1","patient2"))

The expected output would be a variable called "cells_Patient1" Then for subsequent runs, I would just copy and paste these 2 lines and change var <-"Patient1" to var <-"Patient2" [please note that I am oversimplifying the above step of WhichCells as it entails ~10 steps and would rather not have to replace "Patient1" by "Patient2" using Search and Replaced

Unfortunately, I am unable to crate the variable "cells_Patient1" using the above command. I am getting the following error:

Error in variable(paste0("cells_", var, sep = "")) <- WhichCells(object = test, : target of assignment expands to non-language object

Browsing stackoverflow, I couldn't find a solution. My understanding of the error is that R can't assign an object to a variable that is not a constant. Is there a way to bypass this?

G. Grothendieck · Accepted Answer · 2019-08-11 23:01:07Z

1

1) Use assign like this:

var <- "Patient1"
assign(paste0("cells_", var), 3)
cells_Patient1
## [1] 3

2) environment This also works.

e <- .GlobalEnv
e[[ paste0("cells_", var) ]] <- 3
cells_Patient1

3) list or it might be better to make these variables into a list:

cells <- list()
cells[[ var ]] <- 3
cells[[ "Patient1" ]]
## [1] 3

Then we could easily iterate over all such variables. Replace sqrt with any suitable function.

lapply(cells, sqrt)
## $Patient1
## [1] 1.732051

edited Aug 11, 2019 at 23:01

answered Aug 11, 2019 at 22:49

G. Grothendieck

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2 Comments

BioProgram Over a year ago

In regards to 1) assign and 2) environment, I got the same following error Error in paste0("cells_", var) : cannot coerce type 'closure' to vector of type 'character' . As for list, it may not work for me as I have multiple commands for each variable that may make it complicated.

G. Grothendieck Over a year ago

Maybe you forgot to define var. var is also an R function to compute the variance. You can see from the reproducible code in the answer and the output shown there that all three do work. If you are really stuck I suggest you cut down the problem to something reproducible so we can copy and paste it into R and see what the problem is. See the instructions at the top of the r tag page.

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