This approach may not be fancy as what you are looking for.
> xl <- unlist(x)
> xl[grep("A", names(xl))]
f1.x1.A f1.x2.A f2.x3.A f2.x4.A
11 12 13 14
>
I hope this helps.
Chel Hee Lee
On 01/16/2015 04:40 AM, Rainer M Krug wrote:
Hi
Consider the following variable:
--8<---------------cut here---------------start------------->8---
x1 <- list(
A = 11,
B = 21,
C = 31
)
x2 <- list(
A = 12,
B = 22,
C = 32
)
x3 <- list(
A = 13,
B = 23,
C = 33
)
x4 <- list(
A = 14,
B = 24,
C = 34
)
y1 <- list(
x1 = x1,
x2 = x2
)
y2 <- list(
x3 = x3,
x4 = x4
)
x <- list(
f1 = y1,
f2 = y2
)
--8<---------------cut here---------------end--------------->8---
To extract all fields named "A" from y1, I can do
,----
| > sapply(y1, "[[", "A")
| x1 x2
| 11 12
`----
But how can I do the same for x?
I could put an sapply into an sapply, but this would be less then
elegant.
Is there an easier way of doing this?
Thanks,
Rainer
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______________________________________________
[email protected] mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
