How can I assign a list in R without creating a lot of new elements?

Question

If I say l = list()

and then I do l[[27]] = 100

, it will create 26 other records with value NULL

in them. Is there a way to avoid this?

For example, if I run: l <- list(); for (i in c(4,7,1)) { l[[i]] <- i^1 }

A list will be generated with entries 1 through 7 and NULL values for all those that I have not assigned. How can I avoid these false entries?

+3

r

Palace chan 10 Feb 13 at 19:46

3 answers

I would avoid the loop for

and use lapply

:

> x <- c(4, 7, 1)
> setNames(lapply(x, `^`, 1), x)
$`4`
[1] 4

$`7`
[1] 7

$`1`
[1] 1

+4

flodel 10 Feb 13 at 20:06

You can add to the list:

l <- list(); for (i in c(4,7,1)) { l <- append(l, i^1) }

0

johannes 10 Feb 13 at 19:48

Matthew lundberg · Accepted Answer · 2013-02-10T19:47:34+0000

Use signed values for indices:

l <- list(); for (i in c(4,7,1)) { l[[as.character(i)]] <- i^1 }

> l
$`4`
[1] 4

$`7`
[1] 7

$`1`
[1] 1