Assigning names in a list using variables?

Question

If I have a variable such as name.string <- "name", how can I use that to assign the name of a value in a list to return something like

$name
  [1] "value"

Using list(name.string = "value") returns $name.string [1] value.

And for reasons I can't seem to figure out, list(get("name.string") = "value") returns

Error: unexpected '=' in "list(get("name.string") =".

Obviously, I'm not interested in the manual solution (write "name" in the assignment) since this is being replicated for thousands of rows.

mylist <- list(name = "oldvalue"); mylist[[name.string]] <- "value" Is that what you're after? — Frank
– Frank, Commented Feb 19, 2014 at 21:50
@Frank - it's the inverse situation, where I want to create a new key with that key name. Matthew's answer gets at that. — canary_in_the_data_mine
– canary_in_the_data_mine, Commented Feb 19, 2014 at 22:10
Okay. I think mylist <- list(); mylist[[name.string]] <- "value"; mylist[[name.string2]] <- "value2" should do that. You can try it out to see, or look at help("[[") — Frank
– Frank, Commented Feb 19, 2014 at 22:14
Perfect @Frank. That's exactly what I wanted (and I feel like a dolt for not thinking of it). If you submit an answer I'll accept. — canary_in_the_data_mine
– canary_in_the_data_mine, Commented Feb 19, 2014 at 22:18

Frank · Accepted Answer · 2014-02-19 22:45:29Z

2

You can access and assign by number or name using double brackets:

mylist                 <- list()
name.string            <- "name"
name.string2           <- "name2"
value                  <- "hello"
value2                 <- "world"

mylist[[name.string]]  <- value
mylist[[name.string2]] <- value2

# an alternative to the previous line
# that fails to create a name:
mylist[[2]]            <- value2

For details on this kind of assignment, see help("[[<-").

As Matthew Plourde said, this dynamically grows your list, so it might be best to preallocate, e.g., with mylist <- vector("list",2).

answered Feb 19, 2014 at 22:45

Frank

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Matthew Plourde · Accepted Answer · 2014-02-19 21:54:16Z

0

You can use:

structure(list('value'), names=name.string)

or

l <- list('value')
names(l) <- name.string

or even

within(list(), assign(name.string, 'value'))

But don't use this last one.

edited Feb 19, 2014 at 21:54

answered Feb 19, 2014 at 21:49

Matthew Plourde

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2 Comments

canary_in_the_data_mine Over a year ago

Is it possible to do this recursively? Something like: my.list <- list(); my.list = c(my.list, name.string = 'value'). I've been able to do names(l) <- name.string if I knew the whole set of names, but I was hoping to learn how to do this recursively.

Matthew Plourde Over a year ago

I suggest preallocating your list to the desired size, adding each element, and build a name vector at the same time. Then use the second method above to assign all of the names at once.

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