Reputation: 4470
Following on from How to optimise filtering and counting for every row in a large R data frame
I have a data.table
such as the following:
name day wages hour colour
1 Ann 1 100 6 Green
2 Ann 1 150 18 Blue
3 Ann 2 200 10 Blue
4 Ann 3 150 10 Green
5 Bob 1 100 11 Red
6 Bob 1 200 17 Red
7 Bob 1 150 20 Green
8 Bob 2 100 11 Red
I wish to know, for every unique name/day pair, for one of four time-periods, a number of facts. The time periods I care about are:
t1 (hour < 9)
t2 (hour < 17)
t3 (hour > 9)
t4 (hour > 17)
Some examples of facts might be:
wages > 175
colour = "Green"
I can accomplish this with the following data.table
filter
setkey(dt,name,day)
result <- dt[,list(wages.t1=sum(wages>175&hour<9),
wages.t2=sum(wages>175&hour<17),
wages.t3=sum(wages>175&hour>9),
wages.t4=sum(wages>175&hour>17),
green.t1=sum(colour=="Green"&hour<9),
green.t2=sum(colour=="Green"&hour<17),
green.t3=sum(colour=="Green"&hour>9),
green.t4=sum(colour=="Green"&hour>17)),
list(name,day)]
Giving me
name day wages.t1 wages.t2 wages.t3 wages.t4 green.t1 green.t2 green.t3 green.t4
[1,] Ann 1 0 0 0 0 1 1 0 0
[2,] Ann 2 0 1 1 0 0 0 0 0
[3,] Ann 3 0 0 0 0 0 1 1 0
[4,] Bob 1 0 0 1 0 0 0 1 1
[5,] Bob 2 0 0 0 0 0 0 0 0
But this a) Is horrible to read & write and b) Seems inefficient.
Any tips on how I can do better? Note that in my real scenario I have many hundreds of thousands of rows, four time periods, and 30-35 facts per time period.
-- Code to create dt
dt = data.table(
name = factor(c("Ann", "Ann", "Ann", "Ann",
"Bob", "Bob", "Bob", "Bob")),
day = c(1, 1, 2, 3, 1, 1, 1, 2),
wages = c(100, 150, 200, 150, 100, 200, 150, 100),
hour = c(6, 18, 10, 10, 11, 17, 20, 11),
colour = c("Green", "Blue", "Blue", "Green", "Red",
"Red", "Green", "Red")
)
Upvotes: 3
Views: 263
Reputation: 59612
How about something like :
f = list(quote(wages>175),quote(colour=="Green"))
t = list(quote(hour<9),quote(hour<17),quote(hour>9),quote(hour>17))
dt = as.data.table(df)
dt[,as.list(mapply("%*%",
lapply(t,eval,.SD),
rep(lapply(f,eval,.SD),each=length(t))
)), by=list(name,day)]
name day V1 V2 V3 V4 V5 V6 V7 V8
[1,] Ann 1 0 0 0 0 1 1 0 0
[2,] Ann 2 0 1 1 0 0 0 0 0
[3,] Ann 3 0 0 0 0 0 1 1 0
[4,] Bob 1 0 0 1 0 0 0 1 1
[5,] Bob 2 0 0 0 0 0 0 0 0
Clearly the column names aren't tackled but that could be added if this approach is ok.
This should be more efficient because each t
and each f
is evaluated once only per group, then the combinations of those results are combined.
Upvotes: 4