Aggiungere righe mancanti a data.table in base a più colonne con chiave

Question

Ho un oggetto data.table che contiene più colonne che specificano casi univoci. Nel piccolo esempio seguente, le variabili "name", "job" e "sex" specificano gli ID univoci. Vorrei aggiungere righe mancanti in modo che ogni caso abbia una riga per ogni possibile istanza di un'altra variabile, "from" (simile a expand.grid).

library(data.table) 
set.seed(1) 
mydata <- data.table(name = c("john","john","john","john","mary","chris","chris","chris"), 
       job = c("teacher","teacher","teacher","teacher","police","lawyer","lawyer","doctor"), 
       sex = c("male","male","male","male","female","female","male","male"), 
       from = c("NYT","USAT","BG","TIME","USAT","BG","NYT","NYT"), 
       score = rnorm(8)) 

setkeyv(mydata, cols=c("name","job","sex")) 

mydata[CJ(unique(name, job, sex), unique(from))] 

Ecco l'attuale oggetto data.table:

> mydata 
    name  job sex from  score 
1: john teacher male NYT -0.6264538 
2: john teacher male USAT 0.1836433 
3: john teacher male BG -0.8356286 
4: john teacher male TIME 1.5952808 
5: mary police female USAT 0.3295078 
6: chris lawyer female BG -0.8204684 
7: chris lawyer male NYT 0.4874291 
8: chris doctor male NYT 0.7383247 

Ecco il risultato che vorrei:

> mydata 
    name  job sex from  score 
1: john teacher male NYT -0.6264538 
2: john teacher male USAT 0.1836433 
3: john teacher male BG -0.8356286 
4: john teacher male TIME 1.5952808 
5: mary police female NYT NA 
6: mary police female USAT 0.3295078 
7: mary police female BG NA 
8: mary police female TIME NA 
9: chris lawyer female NYT -NA 
10: chris lawyer female USAT -NA 
11: chris lawyer female BG -0.8204684 
12: chris lawyer female TIME -NA 
13: chris lawyer male NYT 0.4874291 
14: chris lawyer male USAT NA 
15: chris lawyer male BG NA 
16: chris lawyer male TIME NA 
17: chris doctor male NYT 0.7383247 
18: chris doctor male USAT NA 
19: chris doctor male BG NA 
20: chris doctor male TIME NA 

Ecco che cosa ho provato:

setkeyv(mydata, cols=c("name","job","sex")) 
mydata[CJ(unique(name, job, sex), unique(from))] 

Ma ricevo il seguente errore e aggiungo fromLast = T RUE (o FALSO) non mi dà la giusta soluzione:

Error in unique.default(name, job, sex) : 
    'fromLast' must be TRUE or FALSE 

Ecco le relative risposte che ho incontrato (ma nessuno sembra occuparsi di più colonne a chiave): add missing rows to a data table

Efficiently inserting default missing rows in a data.table

Fastest way to add rows for missing values in a data.frame?

Answer 1

Un paio di possibilità sono qui - https://github.com/Rdatatable/data.table/pull/814

CJ.dt = function(...) { 
    rows = do.call(CJ, lapply(list(...), function(x) if(is.data.frame(x)) seq_len(nrow(x)) else seq_along(x))); 
    do.call(data.table, Map(function(x, y) x[y], list(...), rows)) 
} 

setkey(mydata, name, job, sex, from) 

mydata[CJ.dt(unique(data.table(name, job, sex)), unique(from))] 
#  name  job sex from  score 
# 1: chris doctor male NYT 0.7383247 
# 2: chris doctor male BG   NA 
# 3: chris doctor male TIME   NA 
# 4: chris doctor male USAT   NA 
# 5: chris lawyer female NYT   NA 
# 6: chris lawyer female BG -0.8204684 
# 7: chris lawyer female TIME   NA 
# 8: chris lawyer female USAT   NA 
# 9: chris lawyer male NYT 0.4874291 
#10: chris lawyer male BG   NA 
#11: chris lawyer male TIME   NA 
#12: chris lawyer male USAT   NA 
#13: john teacher male NYT -0.6264538 
#14: john teacher male BG -0.8356286 
#15: john teacher male TIME 1.5952808 
#16: john teacher male USAT 0.1836433 
#17: mary police female NYT   NA 
#18: mary police female BG   NA 
#19: mary police female TIME   NA 
#20: mary police female USAT 0.3295078 

Answer 2

Una possibilità sarebbe quella di paste colonne name, job e sex insieme, ottenere i valori unique e quindi fare CJ con i valori unique di from. Successivamente, utilizzare cSplit da library(splitstackshape) per dividere la colonna pasted su tre colonne, rinominare le colonne con setnames e join con mydata dopo aver impostato key.

library(splitstackshape) 
library(data.table) 
mydata1 <- setnames(cSplit(mydata[,CJ(unique(paste(name, job, sex)), 
      from=unique(from))], 'V1', ' '), 2:4, c('name', 'job', 'sex'))[, 
        c(2:4,1), with=FALSE] 
setkeyv(mydata, cols=colnames(mydata)[1:4]) 
mydata[mydata1] 
#  name  job sex from  score 
#1: chris doctor male BG   NA 
#2: chris doctor male NYT 0.7383247 
#3: chris doctor male TIME   NA 
#4: chris doctor male USAT   NA 
#5: chris lawyer female BG -0.8204684 
#6: chris lawyer female NYT   NA 
#7: chris lawyer female TIME   NA 
#8: chris lawyer female USAT   NA 
#9: chris lawyer male BG   NA 
#10: chris lawyer male NYT 0.4874291 
#11: chris lawyer male TIME   NA 
#12: chris lawyer male USAT   NA 
#13: john teacher male BG -0.8356286 
#14: john teacher male NYT -0.6264538 
#15: john teacher male TIME 1.5952808 
#16: john teacher male USAT 0.1836433 
#17: mary police female BG   NA 
#18: mary police female NYT   NA 
#19: mary police female TIME   NA 
#20: mary police female USAT 0.3295078 

Answer 3

La versione dev di tidyr ha ora un modo elegante per fare questo perché la funzione expand() ora supporta la nidificazione e la traversata:

library(dplyr) 

mydata <- data_frame(
    name = c("john","john","john","john","mary","chris","chris","chris"), 
    job = c("teacher","teacher","teacher","teacher","police","lawyer","lawyer","doctor"), 
    sex = c("male","male","male","male","female","female","male","male"), 
    from = c("NYT","USAT","BG","TIME","USAT","BG","NYT","NYT"), 
    score = rnorm(8) 
) 

mydata %>% 
    expand(c(name, job, sex), from) %>% 
    left_join(mydata) 

#> Joining by: c("name", "job", "sex", "from") 
#> Source: local data frame [20 x 5] 
#> 
#>  name  job sex from  score 
#> 1 chris doctor male BG   NA 
#> 2 chris doctor male NYT 0.5448206 
#> 3 chris doctor male TIME   NA 
#> 4 chris doctor male USAT   NA 
#> 5 chris lawyer female BG 1.2015173 
#> 6 chris lawyer female NYT   NA 
#> 7 chris lawyer female TIME   NA 
#> 8 chris lawyer female USAT   NA 
#> 9 chris lawyer male BG   NA 
#> 10 chris lawyer male NYT -1.0930237 
#> 11 chris lawyer male TIME   NA 
#> 12 chris lawyer male USAT   NA 
#> 13 john teacher male BG 1.1345461 
#> 14 john teacher male NYT 1.3032946 
#> 15 john teacher male TIME 2.4901830 
#> 16 john teacher male USAT -1.6449096 
#> 17 mary police female BG   NA 
#> 18 mary police female NYT   NA 
#> 19 mary police female TIME   NA 
#> 20 mary police female USAT -0.2443080