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[R-Forge #1611] Allow 2 column i matrix to return a list() (or vector if same type) #657
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related(?) to #826 where 826 is just a special case for second dimension length equal to 1.
I was getting an error that referred here and I also looked at #826
I get this error with the sample code
x=data.table(a=1:10,b=2:11) subset(x,(x[,1]>5))The workaround to do subset(x,(x[,1]>5)[,1]) does work but curious if it is necessary? using data.table 1.10.4
Huge data.table fan, thank you so much for all of your work.
Would just like to add that this feature would be great (subsetting a DT by a 2 column matrix). It's a nice feature because you can then easily do something like the following:
dt[which( your_condition, arr.ind = T)]and get back a data table with the desired rows/columns.
Reacted by María Medina, Tokhir Dadaev, Jone Keat Lim and Sara CerioliAnother use case with
.SD. Now I need to resort toas.matrixoras.data.frameto be able to perform subsetting with 2 column matrix objects.Not sure I understand the use case for this feature? Any simple example?
Reacted by Jan GoreckiNot sure I understand the use case for this feature? Any simple example?
dt[which( your_condition, arr.ind = T)]
@kippjohnson I saw that... not really helping... How is this different from
dt[(your_condition)]?The use case I requested would flesh out a full reproducible example :)
Reacted by Jan Gorecki@kippjohnson Fyi, you said above "and get back a data table with the desired rows/columns.", however, the proposal is for a list or vector of values from the selected positions, not a data.table (since the latter does not make sense for this sort of indexing, just as it doesn't in matrices, which extract a vector when using X[Y])
ing... How is this different from
dt[(your_condition)]?This doesn't work if (your_condition) yields a matrix of rows x column indices.
Sure, sometimes you can re-do it. But why does data.table allow you to select by row index if we could just use conditions? I don't see why a reproducible code example is necessary here. Sometimes you have computed matrix positions in another step. This doesn't exist in base R for no reason :-)
Frank: yes, my mistake, return a list or a vector as in base R. Maybe I was thinking a 1-column DT.
Fwiw, yeah, I have no idea what an example would look like here. Re "easy filling of NA values after a join", I don't see how this helps. Current usage...
library(data.table) DT1 = data.table(id = 1:2, a = 1, b = 2) DT2 = data.table(id = 3:4, x = 1, y = 2) DT = merge(DT1, DT2, by="id") DT[is.na(DT)] <- 99There's no sensible LHS to put into
DT[which(is.na(DT), arr.ind=TRUE), (LHS) := 99]here.@kippjohnson Reproducible examples are recommended for any sort of post. In the context of a feature request like this, I guess you want to provide it so the developers can make sure they understand the problem and make a fix that resolves it (eg, they might take your example and put it into a unit test to make sure it doesn't break later).
Reacted by Jan GoreckiSure, a minimal reproducible example.
# Base R set.seed(123) x0 <- matrix(rnorm(100), 10, 10) y0 <- which(x0>1.5, arr.ind=TRUE) x0[y0] # data.table library(data.table) x1 <- as.data.table(x0) x1[y0] # not implementedOf course there are other ways to do this pretty trivial example. I'm not sure what would be the fastest way in data.table.
However, when you have a matrix with i,j positions that may have been more expensive to compute, if you want to do this you need to change the data.table back to a matrix or dataframe.
@kippjohnson thanks, makes sense now. basically situations where some subset of
data.tablecolumns can be considered as a matrix (but are stored as adata.tablealongside other useful non-matrix columns). Then @jaapwalhout 's example is to declare which are the "matrix columns" inside.SDcols.I'm not sure about efficiency (since
data.table/data.frameare columnar storage) but an implementation that simply does matrix conversion under the hood would be relatively easy to cobble together...Can anyone explain to me the reason why
dt[dt < 0]throws an error whiledt[dt < 0] <- 1works in the example below? Thanks!### goal: replace any value small than 0 with 1 library(data.table) txt <- "V1 V2 V3 0 -999 0 -999 0 -999" dt <- fread(txt) dt #> V1 V2 V3 #> 1: 0 -999 0 #> 2: -999 0 -999 # error dt[dt < 0] #> Error in `[.data.table`(dt, dt < 0): i is invalid type (matrix). Perhaps in future a 2 column matrix could return a list of elements of DT (in the spirit of A[B] in FAQ 2.14). Please report to data.table issue tracker if you'd like this, or add your comments to FR #657. # works dt[dt < 0] <- 1 dt #> V1 V2 V3 #> 1: 0 1 0 #> 2: 1 0 1
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Submitted by: Matt Dowle; Assigned to: Nobody; R-Forge link
Realised as result of fixing bug
#1593in 1.6.7. See error message in [.data.table which refers to this FR.