This looks like a solid, reproducible bug. It’s been filed as Bug #2791. It appears that repeating the column attempts to delete the subsequent columns. If no columns remain, then R crashes. UPDATE : Now fixed in v1.8.11. From NEWS : Assigning to the same column twice in the same query is now an error … Read more
A data.table inherits from class data.frame. Therefore it is a list (of column vectors) internally and can be treated as such. is.list(DT) #[1] TRUE Fortunately, list subsetting, i.e. [[, is very fast and, in contrast to [, package data.table doesn’t define a method for it. Thus, you can simply use [[ to extract by an … Read more
Base package Using the function strftime passing the argument %V to obtain the week of the year as decimal number (01–53) as defined in ISO 8601. (More details in the documentarion: ?strftime) strftime(c(“2014-03-16”, “2014-03-17″,”2014-03-18”, “2014-01-01”), format = “%V”) Output: [1] “11” “12” “12” “01”
Here is an example showing 10 minutes reduced to 1 second (from NEWS on homepage). It’s like subassigning to a data.frame but doesn’t copy the entire table each time. m = matrix(1,nrow=100000,ncol=100) DF = as.data.frame(m) DT = as.data.table(m) system.time(for (i in 1:1000) DF[i,1] <- i) user system elapsed 287.062 302.627 591.984 system.time(for (i in 1:1000) … Read more
With shift() implemented in v1.9.6, this is quite straightforward. DT[ , D := C + shift(B, 1L, type=”lag”)] # or equivalently, in this case, DT[ , D := C + shift(B)] From NEWS: New function shift() implements fast lead/lag of vector, list, data.frames or data.tables. It takes a type argument which can be either “lag” … Read more
There is no straightforward/simple answer because the philosophies of both these packages differ in certain aspects. So some compromises are unavoidable. Here are some of the concerns you may need to address/consider. Operations involving i (== filter() and slice() in dplyr) Assume DT with say 10 columns. Consider these data.table expressions: DT[a > 1, .N] … Read more
Update: From version 1.9.6 (on CRAN as of Sep’15), we can use the function tstrsplit() to get the results directly (and in a much more efficient manner): require(data.table) ## v1.9.6+ dt[, c(“PX”, “PY”) := tstrsplit(PREFIX, “_”, fixed=TRUE)] # PREFIX VALUE PX PY # 1: A_B 1 A B # 2: A_C 2 A C # … Read more
For v1.9.8+ (released November 2016) From ?unique.data.table By default all columns are being used (which is consistent with ?unique.data.frame) unique(dt) V1 V2 1: A B 2: A C 3: A D 4: B A 5: C D 6: E F 7: G G Or using the by argument in order to get unique combinations of … Read more
You can do: x[, 1:2][is.na(x[, 1:2])] <- 0 or better (IMHO), use the variable names: x[c(“a”, “b”)][is.na(x[c(“a”, “b”)])] <- 0 In both cases, 1:2 or c(“a”, “b”) can be replaced by a pre-defined vector.
This seems to work: dt[ , (cols) := lapply(.SD, “*”, -1), .SDcols = cols] The result is a b d 1: -1 -1 1 2: -2 -2 2 3: -3 -3 3 There are a few tricks here: Because there are parentheses in (cols) :=, the result is assigned to the columns specified in cols, … Read more