fast subsetting of lists in lists

test <- list(a=1:3, be = list(a=4:6,c=data.frame(a=10:15,b=I(letters[1:6]))))

vaccuum(test,"b")

vaccuum(test,"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 that helps.

DF = data.frame(a=1:3)
DF$b = list(pi, 2:3, letters[1:5])
DF

? a ? ? ? ? ? ? b
1 1 ? ? ?3.141593
2 2 ? ? ? ? ?2, 3
3 3 a, b, c, d, e

DF$b

[[1]]
[1] 3.141593

[[2]]
[1] 2 3

[[3]]
[1] "a" "b" "c" "d" "e"

sapply(DF,class)

? ? ? ? a ? ? ? ? b
"integer" ? ?"list"

That is still regular though in the sense that each row has a value for all
the columns, even if that value is NA, or NULL in lists.

My data is mostly regular, that is every sublist contains a data.frame
which is the major contribution to overall size. The reason I use lists
is mainly that I need also some bits of information about the
environment. I thought about putting these into additional columns of
the data.frame (and add redundancy and maybe 30% of overhead this way),
one column per variable. But as memory usage is already close to the
limit of my machine this might break things (the situation is a bit
tricky, isn't it?).
I didn't know that a column of a data.frame can be a list. So if I need
only let's say 10 entries in that list, but my data.frame has several
hundred rows, would the "empty" parts of the "column-list" be filled
with cycled values or would they be really empty and thus not use
additional memory?
Secondly as I mentioned in another email to this topic: a whole day of
data contains about 100 chunks of data that is 100 of the sublists
described above. I could put them all into one large data.frame, but
then I would have to extract the "environmental data" from the long
list, now containing repeated occurrences of variables with the same
name. I guess subsetting could become tricky here (dependend on name and
position, I assume), but I'm eager to learn an easy way of doing so.

Sorry for not submitting an illustrative example, but I'm afraid that
would be quite lengthy and not so illustrative any more.

The data.table mentioned below seems to be an interesting alternative;
I'll definitely look into this. But it would also mean quite a bit of
homework, as far as I can see...

Thanks

Alex

If your data is not regular then one option is to flatten it into
(row,column,value) tuple similar to how sparse matrices are stored. ?Your
value column may be list rather than vector.

Then (and yes you guessed this was coming) ... you can use data.table to
query the flat structure quickly by setting a key on the first two columns,
or maybe just the 2nd column when you need to pick out the values for one
'column' quickly for all 'rows'.

There was a thread about using list() columns in data.table here :

http://r.789695.n4.nabble.com/Suggest-a-cool-feature-Use-data-table-like-a-sorted-indexed-data-list-tp2544213p2544213.html

Does someone now a trick to do the same as above with the faster built-in
subsetting? Something like:
test[<somesubsettingmagic>]

So in data.table if you wanted all the 'b' values, ?you might do something
like this :

setkey(DT,column)
DT[J("b"), value]

which should return the list() quickly from the irregular data.

Matthew


"Alexander Senger" <senger at physik.hu-berlin.de> wrote in message
news:4CFE6AEE.6030204 at physik.hu-berlin.de...

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 recursively?

Maybe there is still another way to go?

Alex

Am 07.12.2010 15:59, schrieb Gerrit Eichner:

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 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 values in the named variables "a", that is
the vector c(1, 4, 7). The best I could come up with is:

val <- sapply(1:3, function (i) {test[[i]]$a})

which is unfortunately not very fast. According to R-inferno this is due
to the fact that apply and its derivates do looping in R rather than
rely on C-subroutines as the common [-operator.

Does someone now a trick to do the same as above with the faster
built-in subsetting? Something like:

test[<somesubsettingmagic>]


Thank you for your advice


Alex