On 1/16/15 9:34 AM, Bert Gunter wrote:
Chee Hee's approach is both simpler and almost surely more efficient,
but I wanted to show another that walks the tree (i.e. the list)
directly using recursion at the R level to pull out the desired
components. This is in keeping with R's "functional" programming
paradigm and avoids the use of regular expressions to extract the
desired components from the unlist() version.
extr <- function(x,nm){
if(is.recursive(x)){
wh <- names(x) %in% nm
c(x[wh],lapply(x[!wh],extr,nm=nm) )
} else NULL
}
## The return value contains a bunch of NULLs; so use unlist() to remove them
unlist(extr(x,"A"))
f1.x1.A f1.x2.A f2.x3.A f2.x4.A
11 12 13 14
I would welcome any possibly "slicker" versions of the above.
I don't know if you would consider this "slicker" or not, but it does
not give a lot of NULLs that have to be filtered out. It does so by
checking if the component of the list is itself a list before
recursively calling extr on it, and by using unlist internally.
extr <- function(x, nm) {
if(is.list(x)) {
sublists <- sapply(x, is.list)
c(unlist(x[nm]), unlist(sapply(x[sublists], extr, nm)))
} else {
message("Argument is not a list")
NULL
}
}
Running it on x gives
> extr(x, "A")
[1] 11 12 13 14
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Fri, Jan 16, 2015 at 7:23 AM, Chel Hee Lee <chl...@mail.usask.ca> wrote:
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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