Hello,
my data is contained in nested lists (which seems not necessarily to be
the best approach). What I need is a fast way to get subsets from the data.
An example:
test - list(list(a = 1, b = 2, c = 3), list(a = 4, b = 5, c = 6),
list(a = 7, b = 8, c = 9))
Now I would like to have all
On Tue, Dec 7, 2010 at 9:47 AM, Alexander Senger
sen...@physik.hu-berlin.de wrote:
Hello,
my data is contained in nested lists (which seems not necessarily to be
the best approach). What I need is a fast way to get subsets from the data.
An example:
test - list(list(a = 1, b = 2, c = 3),
Hello, Alexander,
does
utest - unlist(test)
utest[ names( utest) == a]
come close to what you need?
Hth,
Gerrit
On Tue, 7 Dec 2010, Alexander Senger wrote:
Hello,
my data is contained in nested lists (which seems not necessarily to be
the best approach). What I need is a fast way to
tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Alexander Senger
Sent: Tuesday, December 07, 2010 9:12 AM
To: r-help@r-project.org
Subject: Re: [R] fast subsetting of lists in lists
Hello Gerrit, Gabor,
thank you
On Tue, Dec 7, 2010 at 12:12 PM, Alexander Senger
sen...@physik.hu-berlin.de wrote:
Hello Gerrit, Gabor,
thank you for your suggestion.
Unfortunately unlist seems to be rather expensive. A short test with one
of my datasets gives 0.01s for an extraction based on my approach and
5.6s for
First, subset 'test' once, e.g.
testT - test[1:3];
and then use sapply() on that, e.g.
val - sapply(testT, FUN=function (x) { x$a })
Then you can avoid one level of function calls, by
val - sapply(testT, FUN=[[, a)
Second, there is some overhead in [[, $ etc. You can use
.subset2() to avoid
Hello Alex,
Assuming it was just an inadequate example (since a data.frame would suffice
in that case), did you know that a data.frames' columns do not have to be
vectors but can be lists? I don't know if that helps.
DF = data.frame(a=1:3)
DF$b = list(pi, 2:3, letters[1:5])
DF
a
Hello,
Matthew's hint is interesting:
Am 07.12.2010 19:16, schrieb Matthew Dowle:
Hello Alex,
Assuming it was just an inadequate example (since a data.frame would suffice
in that case), did you know that a data.frames' columns do not have to be
vectors but can be lists? I don't know if
Hello Gerrit, Gabor,
thank you for your suggestion.
Unfortunately unlist seems to be rather expensive. A short test with one
of my datasets gives 0.01s for an extraction based on my approach and
5.6s for unlist alone. The reason seems to be that unlist relies on
lapply internally and does so
Alexander:
I'm not sure exactly what you want, so the following may be irrelevant...
BUT, noting that data frames ARE lists and IF what you have can then
be abstracted as lists of lists of lists of ... to various depths
AND IF what you want is just to pick out and combined all named
vectors
Hi Bert,
thank you for your suggestion. I'm sure it's a good one. But my intention in first place was to
learn about getting subsets of list nested in lists the fast way (and preferably also the easy way,
but that is only my laziness).
It seems this thread is getting a bit long and also
11 matches
Mail list logo
