Fastest way to filter a data.frame list column contents in R / Rcpp

Question

I have a data.frame:

df <- structure(list(id = 1:3, vars = list(\"a\", c(\"a\", \"b\", \"c\"), c(\"b\", 
\"c\"))), .Names = c(\"id\", \"vars\"), row.names         


        

Answer 1

Setting aside any algorithmic improvements, the analogous data.table solution is automatically going to be faster because you won't have to copy the entire thing just to add a column:

library(data.table)
dt = as.data.table(df)  # or use setDT to convert in place

dt[, newcol := lapply(vars, setdiff, 'a')][sapply(newcol, length) != 0]
#   id  vars newcol
#1:  2 a,b,c    b,c
#2:  3   b,c    b,c

You can also delete the original column (with basically 0 cost), by adding [, vars := NULL] at the end). Or you can simply overwrite the initial column if you don't need that info, i.e. dt[, vars := lapply(vars, setdiff, 'a')].

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Now as far as algorithmic improvements go, assuming your id values are unique for each vars (and if not, add a new unique identifier), I think this is much faster and automatically takes care of the filtering:

dt[, unlist(vars), by = id][!V1 %in% 'a', .(vars = list(V1)), by = id]
#   id vars
#1:  2  b,c
#2:  3  b,c

To carry along the other columns, I think it's easiest to simply merge back:

dt[, othercol := 5:7]

# notice the keyby
dt[, unlist(vars), by = id][!V1 %in% 'a', .(vars = list(V1)), keyby = id][dt, nomatch = 0]
#   id vars i.vars othercol
#1:  2  b,c  a,b,c        6
#2:  3  b,c    b,c        7