I have a data.table that has factor column with empty levels. I need to get the row count and sums of other variables, all grouped by multiple factors, including the one with empty levels. My question is similar to this one, but here I need to count for multiple factors.
For example, let data.table be:
library('data.table')
dtr <- data.table(v1=sample(1:15),
v2=factor(sample(letters[1:3], 15, replace = TRUE),levels=letters[1:5]),
v3=sample(c("yes", "no"), 15, replace = TRUE))
I want to do the following:
dtr[,list(freq=.N,mm=sum(v1,na.rm=T)),by=list(v2,v3)]
#Output is:
v2 v3 freq mm
1: b yes 4 22
2: b no 1 13
3: c no 3 10
4: a no 4 49
5: c yes 1 10
6: a yes 2 16
I want output include empty levels for v2 as well ("d" and "e"), like in table(dtr$v2,dtr$v3)
, so the final output should look like (the order doesn't matter):
v2 v3 freq mm
1: b yes 4 22
2: b no 1 13
3: c no 3 10
4: a no 4 49
5: c yes 1 10
6: a yes 2 16
7: d yes 0 0
8: d no 0 0
9: e yes 0 0
10: e no 0 0
I tried to use the method used in the link, but I'm not sure how to use joint J() function when there are multiple columns used.
This works fine for groupping by 1 column only:
setkey(dtr,v2)
dtr[J(levels(v2)),list(freq=.N,mm=sum(v1,na.rm=T))]
However, dtr[J(levels(v2),v3),list(freq=.N,mm=sum(v1,na.rm=T))]
doesn't include all combinations
library(data.table)
set.seed(42)
dtr <- data.table(v1=sample(1:15),
v2=factor(sample(letters[1:3], 15, replace = TRUE),levels=letters[1:5]),
v3=sample(c("yes", "no"), 15, replace = TRUE))
res <- dtr[,list(freq=.N,mm=sum(v1,na.rm=T)),by=list(v2,v3)]
You can use CJ
(a cross join). Doing this after aggregation avoids setting the key for the big table and should be faster.
setkey(res,c("v2","v3"))
res[CJ(levels(dtr[,v2]),unique(dtr[,v3])),]
# v2 v3 freq mm
# 1: a no 1 9
# 2: a yes 2 11
# 3: b no 2 11
# 4: b yes 3 23
# 5: c no 4 40
# 6: c yes 3 26
# 7: d no NA NA
# 8: d yes NA NA
# 9: e no NA NA
# 10: e yes NA NA
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