Slightly more elegant: library(data.table) DT[ , .SD[which.min(Employees)], by = State] State Company Employees 1: AK D 24 2: RI E 19 Slighly less elegant than using .SD, but a bit faster (for data with many groups): DT[DT[ , .I[which.min(Employees)], by = State]$V1] Also, just replace the expression which.min(Employees) with Employees == min(Employees), if your data … Read more
Here’s the fast data.table way: bdt[bdt[, .I[g == max(g)], by = id]$V1] This avoids constructing .SD, which is the bottleneck in your expressions. edit: Actually, the main reason the OP is slow is not just that it has .SD in it, but the fact that it uses it in a particular way – by calling … Read more
Overlap joins was implemented with commit 1375 in data.table v1.9.3, and is available in the current stable release, v1.9.4. The function is called foverlaps. From NEWS: 29) Overlap joins #528 is now here, finally!! Except for type=”equal” and maxgap and minoverlap arguments, everything else is implemented. Check out ?foverlaps and the examples there on its … Read more
Yes, it’s subassignment in R using <- (or = or ->) that makes a copy of the whole object. You can trace that using tracemem(DT) and .Internal(inspect(DT)), as below. The data.table features := and set() assign by reference to whatever object they are passed. So if that object was previously copied (by a subassigning <- … Read more
You probably want to use the na.locf() function from the zoo package to carry the last observation forward to replace your NA values. Here is the beginning of its usage example from the help page: library(zoo) az <- zoo(1:6) bz <- zoo(c(2,NA,1,4,5,2)) na.locf(bz) 1 2 3 4 5 6 2 2 1 4 5 2 … Read more
Yes, in your formula, you can cbind the numeric variables to be aggregated: aggregate(cbind(x1, x2) ~ year + month, data = df1, sum, na.rm = TRUE) year month x1 x2 1 2000 1 7.862002 -7.469298 2 2001 1 276.758209 474.384252 3 2000 2 13.122369 -128.122613 … 23 2000 12 63.436507 449.794454 24 2001 12 999.472226 … Read more
I don’t word with data tables, but this is a solution that would work for data frames, and should hopefully generalize. The strategy is to use the fact that you can fill one vector with another vector, without ever having to use a loop. # make the example data sets D1 <- as.data.frame(matrix(data=(1:(20*181)), nrow=20, ncol=181)) … Read more