data.table equivalent of tidyr::complete with group_by with on and by syntax

Question

Problem:

What is the data.table equivalent of tidyr's complete command with group by ?

What is the relationship between on and by for data.table?

Example:

dt=data.table(a = c(1,1,2,2,3,3,4,4) , b = c(4,5,6,7,8,9,10,11) , c = c("x","x","x","x","y","y","y","y"))
show(dt)

   a  b c
1: 1  4 x
2: 1  5 x
3: 2  6 x
4: 2  7 x
5: 3  8 y
6: 3  9 y
7: 4 10 y
8: 4 11 y

The goal is to obtain the following:

so something approximately like this:

setDT(dt)[CJ(a=a,b=b,unique=TRUE), on=.(a,b) , by = .(c)]

but it doesn't work and the data.table documentation is thin on this aspect of the syntax.

Insufficient solutions:

The following SO posts address similar problems, but do not provide sufficient solutions in this context.

data.table equivalent of complete/fill from tidyr (no group by)
data.table equivalent of tidyr::complete() (no group by)
data.table equivalent of tidyr::complete with group_by (problem-specific, does not actually work with by command)

score 5 · Answer 1 · answered Feb 07 '19 at 18:03

5

Try this:

dt[, CJ(a = a, b = b, unique = TRUE), by = "c"]

giving:

answered Feb 07 '19 at 18:03

G. Grothendieck

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Why isn't `on` needed anymore? – cmo Feb 07 '19 at 18:09
1

`on` is used for doing a merge but there is no merge here. – G. Grothendieck Feb 07 '19 at 18:10
in the real dataset, I lost all of the additional columns. my fault,. i will udpate the example – cmo Feb 07 '19 at 18:13
3

In that case do a merge afterwards. – G. Grothendieck Feb 07 '19 at 18:16

Frank · Answer 2 · 2019-02-07T19:26:57.333

complete retains other unrelated columns, so I'll add one...

library(data.table)
dt = data.table(
  a = c(1,1,2,2,3,3,4,4) , 
  b = c(4,5,6,7,8,9,10,11) , 
  c = c("x","x","x","x","y","y","y","y"),
  d = LETTERS[10 + 1:8])

   a  b c d
1: 1  4 x K
2: 1  5 x L
3: 2  6 x M
4: 2  7 x N
5: 3  8 y O
6: 3  9 y P
7: 4 10 y Q
8: 4 11 y R

To complete the a x b combos for each c, I would make a new table with those combos (exactly as already in @G.Grothendieck's answer) and update-join to get d and other non-combo columns:

mDT = dt[, CJ(a = a, b = b, unique=TRUE), by=c]
cvars = copy(names(mDT))
ovars = setdiff(names(dt), cvars)

mDT[, (ovars) := dt[.SD, on=cvars, mget(sprintf("x.%s", ovars))]]
setcolorder(mDT, names(dt))

    a  b c    d
 1: 1  4 x    K
 2: 1  5 x    L
 3: 1  6 x <NA>
 4: 1  7 x <NA>
 5: 2  4 x <NA>
 6: 2  5 x <NA>
 7: 2  6 x    M
 8: 2  7 x    N
 9: 3  8 y    O
10: 3  9 y    P
11: 3 10 y <NA>
12: 3 11 y <NA>
13: 4  8 y <NA>
14: 4  9 y <NA>
15: 4 10 y    Q
16: 4 11 y    R

Alternately, you could do an inner (?) join, though this is inefficient since it creates two new tables:

dt[mDT, on=cvars]

# or more concisely....

dt[dt[, CJ(a = a, b = b, unique=TRUE), by=c], on=.(a,b,c)]

Or, do one inner join per by= group (from @eddi):

dt[, .SD[CJ(a = a, b = b, unique = TRUE), on = .(a, b)], by = c]

For comparison in the tidyverse:

library(dplyr); library(tidyr)
data.frame(dt) %>% group_by(c) %>% complete(a, b)

# A tibble: 16 x 4
# Groups:   c [2]
   c         a     b d    
   <chr> <dbl> <dbl> <chr>
 1 x         1     4 K    
 2 x         1     5 L    
 3 x         1     6 <NA> 
 4 x         1     7 <NA> 
 5 x         2     4 <NA> 
 6 x         2     5 <NA> 
 7 x         2     6 M    
 8 x         2     7 N    
 9 y         3     8 O    
10 y         3     9 P    
11 y         3    10 <NA> 
12 y         3    11 <NA> 
13 y         4     8 <NA> 
14 y         4     9 <NA> 
15 y         4    10 Q    
16 y         4    11 R

I'd do `dt[, .SD[CJ(a = a, b = b, unique = TRUE), on = .(a, b)], by = c]` if there aren't too many `c` values. — eddi, Feb 07 '19 at 19:20

score 1 · Answer 3 · answered Feb 07 '19 at 18:06

The following will return the desired result.

library(data.table)
dt=data.table(a = c(1,1,2,2,3,3,4,4) , 
              b = c(4,5,6,7,8,9,10,11) , 
              c = c("x","x","x","x","y","y","y","y"))
dttrue <- fread('a  b c
                1  4 x
                1  5 x
                1  6 x
                1  7 x
                2  4 x
                2  5 x
                2  6 x
                2  7 x
                3  8 y
                3  9 y
                3 10 y
                3 11 y
                4  8 y
                4  9 y
                4 10 y
                4 11 y')
dt2 <- dt[,CJ(a=a,b=b,unique = TRUE),by = c]
all.equal(dt2[,.(a,b,c)], dttrue) #true

data.table equivalent of tidyr::complete with group_by with on and by syntax

3 Answers3