Creating named list of large set of data frames R

Question

This may be a trivial question, given I only have a few days' experience using R but basically i'd like to create a named list of data frames. I've seen a bunch of questions and answers dealing with only 2 data frames e.g

dataList <- list(x=data.frame1, y=data.frame2)

However i'm working with 48 data frames in this program and would like to know a more elegant way of creating a named list of 48 elements without actually explicitly naming all 48 of my data frames (which is what i've currently got):

dataList <- list(forecasted.data0=forecasted.data0, forecasted.data1=forecasted.data1, ...) and so on 46 more times.

See this [similar post](http://stackoverflow.com/questions/15387727/use-object-names-as-list-names-in-r) — alexis_laz, Sep 15 '16 at 07:39
Why not create a vector of names with `paste0("forecasted.data", 0:47)` and assign to your list with the names attribute? — Sun Bee, Sep 15 '16 at 07:58

score 4 · Answer 1 · answered Sep 15 '16 at 08:11

Use the names or setNames function

> x <- as.list(rnorm(5))
> x
[[1]]
[1] -1.404512

[[2]]
[1] 0.927126

[[3]]
[1] 1.055555

[[4]]
[1] -1.718295

[[5]]
[1] 0.5154312

> names(x) <- paste0("forecasted.data.", seq(0, length(x) - 1))
> x
$forecasted.data.0
[1] -1.404512

$forecasted.data.1
[1] 0.927126

$forecasted.data.2
[1] 1.055555

$forecasted.data.3
[1] -1.718295

$forecasted.data.4
[1] 0.5154312

user3507085 · Answer 2 · 2016-09-15T08:00:39.263

1

Would this work ?

forecasted.data0=data.frame(1)
forecasted.data1=data.frame(2)

n=ls()[grepl("^forecasted.data.*",ls())]
l=sapply(n, function(x) get(x))
names(l)=n

You can replace sapply with lapply if you prefer ...

edited Sep 15 '16 at 08:00

answered Sep 15 '16 at 07:44

user3507085

700
5
17

It didn't necessarily work, but that may be down to my inexperience with R. Thanks for your contribution – Cnerb Sep 15 '16 at 07:57

Creating named list of large set of data frames R

2 Answers2