Dynamic variable name R

Question

I have already read similar questions regarding the assign command. But it doesnt work for me.

Here is my code:

masums <- function(var) {
    lags <- var$p                    # number of lags in VAR
    eqn  <- length(var$varresult)    # number of equations in VAR
    dep  <- names(var$varresult)     # names of dependent variables

    for(i in 1:eqn) # compute coefficient matrices
        d <- dep[i]
        x <- paste("var$varresult$",d,"$coefficients",sep="")
        y <- as.matrix(GET THE VALUE OF "x" e.g. var$varresult$d$coefficients) # d="gap" for i=1
    }
    return(y)
}

Example: For i=1 d would have the value "gap", therefore I want the value of var$varresult$gap$coefficients.

var$varresult consists of elements of class "lm". Maybe the solution is easy and I am just to stressed to see it. Hope someone can help.

Edit: For a small example dataset:

library(vars)
y <- c(100*rnorm(100))
x <- seq(1,100,1)

vardata <- cbind(x,y)
var1    <- VAR(vardata,p=4,type="const")
var1$varresult$x$coefficients

Martin

You need to specify what "doesn't work". I see several potential problems here - (1) what are you doing with x? (2) you're only returning y from the last iteration of your loop - but you haven't explained what you're actually trying to do and what aspects of your current efforts aren't working. — Thomas
– Thomas, Commented May 21, 2014 at 11:14
I sense there's a better way of achieving what you're after. Can you provide us with a small reproducible example and desired output? — Roman Luštrik
– Roman Luštrik, Commented May 21, 2014 at 11:19
I have used R every day for 6 years, and never once found it necessary to use assign. There is probably a more R-like way of doing things. What is your real problem? — Andrie
– Andrie, Commented May 21, 2014 at 11:23
Yes I just extract the last coefficient matrix in this example. Of course this is not what I intend to do here later on. The problem is that I cannot use var$varresult$d$coefficients where d would be "gap" for the first lm object. It is not easy to make a reproduceable example, since I use a result of a VAR estimation wich is pretty big :/ — Martin Schmelzer
– Martin Schmelzer, Commented May 21, 2014 at 11:50
The actual problem to be clear is, how can I retrive the value of an object, If I use a variable in its name... — Martin Schmelzer
– Martin Schmelzer, Commented May 21, 2014 at 11:55

Konrad Rudolph · Accepted Answer · 2014-05-21 12:15:58Z

1

paste to construct a nested variable access in a string is the completely wrong approach.

What you are missing is the fact that x$y can also be written as x$['y'] (or x$[['y']], depending whether you want a single value in a list, or a column in a data.frame), and here 'y' is a string. So you can write this:

y <- as.matrix(var$varresult[[d]]$coefficients)

(assuming var$varresult is a list)

answered May 21, 2014 at 12:15

Konrad Rudolph

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