Alternatively, use the concat() function from the gregmisc library:
> library(gregmisc)
Attaching package `gregmisc':
[warnings omitted]]
> concat(x[[1]],x[[2]],names=names(x))
age.day state source
1 81 0 772307
2 141 0 772307
3 234 0 772307
4 321 0 772307
5 437 0 772307
6 499 0 772307
7 601 1 772307
8 689 1 772307
9 801 0 772307
10 884 99 772307
11 961 1 772307
12 80 0 1084506
13 148 0 1084506
14 249 99 1084506
15 344 99 1084506
-Greg
> -----Original Message-----
> From: Danardono [mailto:daodao99 at student.umu.se]
> Sent: Wednesday, September 26, 2001 10:40 AM
> To: r-help at stat.math.ethz.ch
> Subject: [R] stacking data.frame in a list
>
>
> Hello,
>
> Given this list object x,
> x <-
> structure(list("772307" = structure(list(age.day = c(81, 141,
> 234, 321, 437, 499, 601, 689, 801, 884, 961), state = c(0, 0,
> 0, 0, 0, 0, 1, 1, 0, 99, 1)), .Names = c("age.day",
> "state"), class > "data.frame", row.names =
c("1",
> "2", "3", "4", "5",
"6", "7", "8", "9", "10",
"11")), "1084506" > structure(list(
> age.day = c(80, 148, 249, 344), state = c(0, 0, 99,
> 99)), .Names > c("age.day",
> "state"), class = "data.frame", row.names =
c("1", "2", "3",
> "4"))), .Names = c("772307", "1084506"))
>
> that looks like this:
> > x
> $"772307"
> age.day state
> 1 81 0
> 2 141 0
> 3 234 0
> 4 321 0
> 5 437 0
> 6 499 0
> 7 601 1
> 8 689 1
> 9 801 0
> 10 884 99
> 11 961 1
>
> $"1084506"
> age.day state
> 1 80 0
> 2 148 0
> 3 249 99
> 4 344 99
>
> How to stack "772307" and "1084506", preferrably with
"772307" and
> $"1084506" as indicator (the original data consists of
> thousands "ID" such
> "772307" and "1084506")?
> The expected result will be like,
> ID age.day state
> 772307 81 0
> 772307 141 0
> 772307 234 0
> 772307 321 0
> 772307 437 0
> 772307 499 0
> 772307 601 1
> 772307 689 1
> 772307 801 0
> 772307 884 99
> 772307 961 1
> 1084506 80 0
> 1084506 148 0
> 1084506 249 99
> 1084506 344 99
>
> Thank you.
> Danar.
>
>
>
>
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