Right ... that's what I thought you meant.
I'm pretty sure -- but not certain -- that columns of matrices are treated
specially by [.data.frame, so that you have to explicitly index a higher
dimensional array, e.g. like this:
subs <- c(1,3)
DFA <- data.frame(id = 1:3)
DFA[["ar"]] <- array(1:12, c(3,2,2))
DFA$ar[match(subs,DFA$id),,]
##yielding:
, , 1
[,1] [,2]
[1,] 1 4
[2,] 3 6
, , 2
[,1] [,2]
[1,] 7 10
[2,] 9 12
You might check, e.g. the "data.table" package, to see if it indexes
as you
would like with columns that contain arrays.
Alternatively, and perhaps preferably depending on your use case, you may
wish to create a wholly different data structure or just treat the data
frame as a list from the start. Data frames/matrix-like data structures are
convenient and appropriate a lot of the time, but not always. R, like any
flexible programming language, allows you -- even encourages you -- to
create other data structures that fit your needs.
Cheers,
Bert
On Tue, May 9, 2023 at 3:39?AM Georg Kindermann <Georg.Kindermann at
gmx.at>
wrote:
> Thanks!
>
> No, to be consistent with what I get with a matrix I think it should be
> like:
>
> x <- data.frame(id = DFA$id[1])
> x$ar <- DFA$ar[1, , , drop = FALSE]
>
> str(x)
> #'data.frame': 1 obs. of 2 variables:
> # $ id: int 1
> # $ ar: int [1, 1:2, 1:2] 1 3 5 7
>
> Georg
>
>
>
> Gesendet: Dienstag, 09. Mai 2023 um 09:32 Uhr
> Von: "Rui Barradas" <ruipbarradas at sapo.pt>
> An: "Georg Kindermann" <Georg.Kindermann at gmx.at>, r-help
at r-project.org
> Betreff: Re: [R] data.frame with a column containing an array
> ?s 11:52 de 08/05/2023, Georg Kindermann escreveu:
> > Dear list members,
> >
> > when I create a data.frame containing an array I had expected, that I
> get a similar result, when subsetting it, like having a matrix in a
> data.frame. But instead I get only the first element and not all values of
> the remaining dimensions. Differences are already when creating the
> data.frame, where I can use `I` in case of a matrix but for an array I am
> only able to insert it in a second step.
> >
> > DFA <- data.frame(id = 1:2)
> > DFA[["ar"]] <- array(1:8, c(2,2,2))
> >
> > DFA[1,]
> > # id ar
> > #1 1 1
> >
> > DFM <- data.frame(id = 1:2, M = I(matrix(1:4, 2)))
> >
> > DFM[1,]
> > # id M.1 M.2
> > #1 1 1 3
> >
> > The same when trying to use merge, where only the first value is kept.
> >
> > merge(DFA, data.frame(id = 1))
> > # id ar
> > #1 1 1
> >
> > merge(DFM, data.frame(id = 1))
> > # id M.1 M.2
> > #1 1 1 3
> >
> > Is there a way to use an array in a data.frame like I can use a matrix
> in a data.frame?
> >
> > I am using R version 4.3.0.
> >
> > Kind regards,
> > Georg
> >
> > ______________________________________________
> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
>
http://www.R-project.org/posting-guide.html[http://www.R-project.org/posting-guide.html]
> > and provide commented, minimal, self-contained, reproducible code.
> Hello,
>
> Are you looking for something like this?
>
>
> DFA <- data.frame(id = 1:2)
> DFA[["ar"]] <- array(1:8, c(2,2,2))
>
> DFA$ar[1, , ]
> #> [,1] [,2]
> #> [1,] 1 5
> #> [2,] 3 7
>
>
> Hope this helps,
>
> Rui Barradas
>
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
Thanks! With data.table I'm not able to create it. DT <- data.table::data.table(id = 1:2) DT$ar <- array(1:8, c(2,2,2)) #Error in set(x, j = name, value = value) : # Supplied 8 items to be assigned to 2 items of column 'ar'. If you wish to 'recycle' the RHS please use rep() to make this intent clear to readers of your code. DFA <- data.frame(id = 1:2) DFA[["ar"]] <- array(1:8, c(2,2,2)) data.table::as.data.table(DFA) #Error in setDT(ans, key = key) : # All elements in argument 'x' to 'setDT' must be of same length, but the profile of input lengths (length:frequency) is: [0:1, 2:2] #The first entry with fewer than 2 entries is 3 But with tibble it works as expected. TI <- tibble::tibble(id = 1:2, ar = array(1:8, c(2,2,2))) str(TI[1,]) #tibble [1 ? 2] (S3: tbl_df/tbl/data.frame) # $ id: int 1 # $ ar: int [1, 1:2, 1:2] 1 3 5 7 But it would be nice if something similar would also work with data.frame. Georg ? ? ? Gesendet:?Dienstag, 09. Mai 2023 um 16:56 Uhr Von:?"Bert Gunter" <bgunter.4567 at gmail.com> An:?"Georg Kindermann" <Georg.Kindermann at gmx.at> Cc:?"Rui Barradas" <ruipbarradas at sapo.pt>, r-help at r-project.org Betreff:?Re: [R] data.frame with a column containing an array Right ... that's what I thought you meant. ? I'm pretty sure -- but not certain -- that columns of matrices are treated specially by [.data.frame, so that you have to explicitly index a higher dimensional array, e.g. like this: ? subs <- c(1,3) DFA <- data.frame(id = 1:3) DFA[["ar"]] <- array(1:12, c(3,2,2)) DFA$ar[match(subs,DFA$id),,] ##yielding: , , 1 ? ? ?[,1] [,2] [1,] ? ?1 ? ?4 [2,] ? ?3 ? ?6 , , 2 ? ? ?[,1] [,2] [1,] ? ?7 ? 10 [2,] ? ?9 ? 12 ? You might check, e.g. the "data.table" package, to see if it indexes as you would like with columns that contain arrays. ? Alternatively, and perhaps preferably depending on your use case, you may wish to create a wholly different data structure or just treat the data frame as a list from the start. Data frames/matrix-like data structures are convenient and appropriate a lot of the time, but not always. R, like any flexible programming language, allows you -- even encourages you -- to create other data structures that fit your needs. ? Cheers, Bert?
I think the following may provide a clearer explanation: subs <- c(1,3) DFA <- data.frame(id = 1:3) ar <- array(1:12, c(3,2,2)) ## yielding> ar, , 1 [,1] [,2] [1,] 1 4 [2,] 2 5 [3,] 3 6 , , 2 [,1] [,2] [1,] 7 10 [2,] 8 11 [3,] 9 12 ## array subscripting gives> ar[subs,,], , 1 [,1] [,2] [1,] 1 4 [2,] 3 6 , , 2 [,1] [,2] [1,] 7 10 [2,] 9 12 ## Now with df's> DFA[["ar"]] <- ar > > DFM <- data.frame(id = 1:3) > DFM[["M"]] <- matrix(1:6, nc =2) > > str(DFM)'data.frame': 3 obs. of 2 variables: $ id: int 1 2 3 $ M : int [1:3, 1:2] 1 2 3 4 5 6> str(DFA)'data.frame': 3 obs. of 2 variables: $ id: int 1 2 3 $ ar: int [1:3, 1:2, 1:2] 1 2 3 4 5 6 7 8 9 10 ...> > ## But the data frame print method for these give > DFMid M.1 M.2 1 1 1 4 2 2 2 5 3 3 3 6> DFAid ar.1 ar.2 ar.3 ar.4 1 1 1 4 7 10 2 2 2 5 8 11 3 3 3 6 9 12> > ## [.data.frame subscripting gives > DFA[subs,]id ar 1 1 1 3 3 3> DFM[subs,]id M.1 M.2 1 1 1 4 3 3 3 6> > ## but explicit array subscripting of course works > DFA$ar[match(subs,DFA$id),,], , 1 [,1] [,2] [1,] 1 4 [2,] 3 6 , , 2 [,1] [,2] [1,] 7 10 [2,] 9 12 Cheers, Bert On Tue, May 9, 2023 at 7:56?AM Bert Gunter <bgunter.4567 at gmail.com> wrote:> Right ... that's what I thought you meant. > > I'm pretty sure -- but not certain -- that columns of matrices are treated > specially by [.data.frame, so that you have to explicitly index a higher > dimensional array, e.g. like this: > > > subs <- c(1,3) > DFA <- data.frame(id = 1:3) > DFA[["ar"]] <- array(1:12, c(3,2,2)) > > DFA$ar[match(subs,DFA$id),,] > ##yielding: > , , 1 > > [,1] [,2] > [1,] 1 4 > [2,] 3 6 > > , , 2 > > [,1] [,2] > [1,] 7 10 > [2,] 9 12 > > You might check, e.g. the "data.table" package, to see if it indexes as > you would like with columns that contain arrays. > > Alternatively, and perhaps preferably depending on your use case, you may > wish to create a wholly different data structure or just treat the data > frame as a list from the start. Data frames/matrix-like data structures are > convenient and appropriate a lot of the time, but not always. R, like any > flexible programming language, allows you -- even encourages you -- to > create other data structures that fit your needs. > > Cheers, > Bert > > On Tue, May 9, 2023 at 3:39?AM Georg Kindermann <Georg.Kindermann at gmx.at> > wrote: > >> Thanks! >> >> No, to be consistent with what I get with a matrix I think it should be >> like: >> >> x <- data.frame(id = DFA$id[1]) >> x$ar <- DFA$ar[1, , , drop = FALSE] >> >> str(x) >> #'data.frame': 1 obs. of 2 variables: >> # $ id: int 1 >> # $ ar: int [1, 1:2, 1:2] 1 3 5 7 >> >> Georg >> >> >> >> Gesendet: Dienstag, 09. Mai 2023 um 09:32 Uhr >> Von: "Rui Barradas" <ruipbarradas at sapo.pt> >> An: "Georg Kindermann" <Georg.Kindermann at gmx.at>, r-help at r-project.org >> Betreff: Re: [R] data.frame with a column containing an array >> ?s 11:52 de 08/05/2023, Georg Kindermann escreveu: >> > Dear list members, >> > >> > when I create a data.frame containing an array I had expected, that I >> get a similar result, when subsetting it, like having a matrix in a >> data.frame. But instead I get only the first element and not all values of >> the remaining dimensions. Differences are already when creating the >> data.frame, where I can use `I` in case of a matrix but for an array I am >> only able to insert it in a second step. >> > >> > DFA <- data.frame(id = 1:2) >> > DFA[["ar"]] <- array(1:8, c(2,2,2)) >> > >> > DFA[1,] >> > # id ar >> > #1 1 1 >> > >> > DFM <- data.frame(id = 1:2, M = I(matrix(1:4, 2))) >> > >> > DFM[1,] >> > # id M.1 M.2 >> > #1 1 1 3 >> > >> > The same when trying to use merge, where only the first value is kept. >> > >> > merge(DFA, data.frame(id = 1)) >> > # id ar >> > #1 1 1 >> > >> > merge(DFM, data.frame(id = 1)) >> > # id M.1 M.2 >> > #1 1 1 3 >> > >> > Is there a way to use an array in a data.frame like I can use a matrix >> in a data.frame? >> > >> > I am using R version 4.3.0. >> > >> > Kind regards, >> > Georg >> > >> > ______________________________________________ >> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> > https://stat.ethz.ch/mailman/listinfo/r-help >> > PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html[http://www.R-project.org/posting-guide.html] >> > and provide commented, minimal, self-contained, reproducible code. >> Hello, >> >> Are you looking for something like this? >> >> >> DFA <- data.frame(id = 1:2) >> DFA[["ar"]] <- array(1:8, c(2,2,2)) >> >> DFA$ar[1, , ] >> #> [,1] [,2] >> #> [1,] 1 5 >> #> [2,] 3 7 >> >> >> Hope this helps, >> >> Rui Barradas >> >> >> ______________________________________________ >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> >[[alternative HTML version deleted]]