data.table相当于tidyr :: complete with group_by with on和by syntax

Question

问题:

与群组的命令data.table相当于什么？tidyrcompleteby

什么是之间的关系on,并by为data.table？

例:

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

目标是获得以下内容:

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

所以大概是这样的:

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

但它不起作用,data.table文档在这方面的文档很薄.

解决方案不足:

以下SO帖子解决了类似的问题,但在这种情况下没有提供足够的解决方案.

• data.table相当于来自tidyr的完整/填充(无组by)
• data.table相当于tidyr :: complete()(没有组by)
• data.table相当于tidyr :: complete with group_by(特定于问题,实际上不能与by命令一起使用)

Answer 1

试试这个:

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

赠送:

    c a  b
 1: x 1  4
 2: x 1  5
 3: x 1  6
 4: x 1  7
 5: x 2  4
 6: x 2  5
 7: x 2  6
 8: x 2  7
 9: y 3  8
10: y 3  9
11: y 3 10
12: y 3 11
13: y 4  8
14: y 4  9
15: y 4 10
16: y 4 11

Answer 2

complete 保留其他不相关的列，因此我将添加一个...

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

为了完成每个c的axb组合，我将使用这些组合创建一个新表（与@ G.Grothendieck的回答中已经完全一样）并进行update-join以获取d和其他非组合列：

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

或者，您可以执行内部（？）连接，尽管这样做效率不高，因为它会创建两个新表：

dt[mDT, on=cvars]

# or more concisely....

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

或者，每个by=组一个内部联接（来自@eddi）：

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

为了在tidyverse中进行比较：

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