treat string as object name in a loop in R

Question

I want to create a string in a loop and use this string as object in this loop. Here is a simplified example:

for (i in 1:2) {
  x <- paste("varname",i, sep="")
  x <- value
}

the loop should create varname1, varname2. Then I want to use varname1, varname2 as objects to assign values. I tried paste(), print() etc. Thanks for help!

Don't use assign() to create many vars. Instead learn how to work with lists and functionals twitter.com/hadleywickham/status/535931179556691968 — Steven Beaupré
– Steven Beaupré, Commented Mar 6, 2015 at 23:19
Whenever you have the urge to create sequentially named variables, you should be using a list instead: stackoverflow.com/a/24376207/903061 — Gregor Thomas
– Gregor Thomas, Commented Mar 6, 2015 at 23:27
A fairly common question from new users is: How do I assign names to a group of similar objects?" Yes, you can do that, but you probably don't want to better is to: vectorize your thinking. Put all of the similar objects into one list. Subsequent analysis and manipulation is then going to be much smoother. R Inferno page 20. — Metrics
– Metrics, Commented Mar 7, 2015 at 0:11
@Metrics Too bad that there is no example of such manipulation in the book — Julien
– Julien, Commented Jul 5, 2022 at 16:59

Rich Scriven · Accepted Answer · 2015-03-07 00:29:03Z

11

You could create the call() to <- and then evaluate it. Here's an example,

value <- 1:5

for (i in 1:2) {
    x <- paste("varname",i, sep="")
    eval(call("<-", as.name(x), value))
}

which creates the two objects varname1 and varname2

varname1
# [1] 1 2 3 4 5
varname2
# [1] 1 2 3 4 5

But you should really try to avoid assigning to the global environment from with in a method/function. We could use a list along with substitute() and then we have the new variables together in the same place.

f <- function(aa, bb) {
    eval(substitute(a <- b, list(a = as.name(aa), b = bb)))
}

Map(f, paste0("varname", 1:2), list(1:3, 3:6))
# $varname1
# [1] 1 2 3
#
# $varname2
# [1] 3 4 5 6

edited Mar 7, 2015 at 0:29

answered Mar 6, 2015 at 23:27

Rich Scriven

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2 Comments

Tabasco Over a year ago

Thanks a lot to you all for your helpful answers! The solution with call() works very well. I'll try the more R-ish strategies as soon as I understand what's really going on (I am spoiled by Stata and thinking in loops).

Miha Trošt Over a year ago

Ok, I upvoted your solution. I played with it...and now I want to upvote your solution for 1k times more. :)

Mahmut Ali ÖZKURAN · Accepted Answer · 2015-03-06 23:11:42Z

3

assign("variableName", 5)

would do that.

For example if you have variable names in array of strings you can set them in loop as:

assign(varname[1], 2 + 2)

More and more information

https://stat.ethz.ch/R-manual/R-patched/library/base/html/assign.html

answered Mar 6, 2015 at 23:11

Mahmut Ali ÖZKURAN

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1 Comment

Mox Over a year ago

So If I've got this right, it assigns to variable name 1, the value of 4 (evalulating 2+2 without needing to be told to do so).

nothing · Accepted Answer · 2015-03-06 23:16:52Z

1

@MahmutAliÖZKURAN has answered your question about how to do this using a loop. A more "R-ish" way to accomplish this might be:

mapply(assign, <vector of variable names>, <vector of values>,
       MoreArgs = list(envir = .GlobalEnv))

Or, as in the case you specified above:

mapply(assign, paste0("varname", 1:2), <vector of values>,
       MoreArgs = list(envir = .GlobalEnv))

answered Mar 6, 2015 at 23:16

nothing

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3 Comments

Gregor Thomas Over a year ago

A more R-ish way to do things is to use lists and not assign.

Gregor Thomas Over a year ago

Yes, all the answers do what the OP asked for. I'm just picking on yours for the "more R-ish" quip ;)

CubicInfinity May 20 at 18:47

"R-ish" is kind of subjective. The best choice is some combination of convenience, what makes sense to you, what makes sense to others, and what avoids "bad" practices.

thistleknot · Accepted Answer · 2020-03-30 01:59:29Z

0

I had the same issue and for some reason my apply's weren't working (lapply, assign directly, or my preferred goto, mclapply)

But this worked

vectorXTS <- mclapply(symbolstring,function(x)
    {
      df <- symbol_data_set[symbol_data_set$Symbol==x,]
      return(xts(as.data.frame(df[,-1:-2]),order.by=as.POSIXct(df$Date)))
    })

    names(symbolstring) <- symbolstring
    names(vectorXTS) <- symbolstring

    for(i in symbolstring) assign(symbolstring[i],vectorXTS[i])

answered Mar 30, 2020 at 1:59

thistleknot

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