r data.table Join In Place 多列

Question

data.table 是惊人的。

我想做一个就地连接，但要保留两个表中的所有列。这个问题演示了如何为单个列执行此操作。当我希望连接表中的所有列都包含在最终结果中并在一个内存位置完成所有操作时，我该如何概括这一点。

library(data.table)
dt1 <- data.table(col1 = c("a", "b", "c"), 
                  col2 = 1:3, 
                  col3 = c(TRUE, FALSE, FALSE))

setkey(dt1, col1)

set.seed(1)
dt2 <- data.table(col1 = sample(c("a", "b", "c"), size = 10, replace = TRUE), 
                  another_col = sample(1:10, size = 10, replace = TRUE), 
                  and_anouther = sample(c(TRUE, FALSE), size = 10, replace = TRUE))

setkey(dt2, col1)

# I want to stick the columns from dt1 onto dt2

# this works
dt3 <- dt2[dt1]
dt3
    col1 another_col and_anouther col2  col3
 1:    a           9        FALSE    1  TRUE
 2:    b           2        FALSE    2 FALSE
 3:    b           9        FALSE    2 FALSE
 4:    b           6        FALSE    2 FALSE
 5:    b           5         TRUE    2 FALSE
 6:    b           8        FALSE    2 FALSE
 7:    c           9         TRUE    3 FALSE
 8:    c           5        FALSE    3 FALSE
 9:    c           7        FALSE    3 FALSE
10:    c           6        FALSE    3 FALSE

# but i want to do this by reference

# this works for one column
dt2[dt1, col2 := i.col2]
dt2

    col1 another_col and_anouther col2
 1:    a           3        FALSE    1
 2:    a           8         TRUE    1
 3:    a           8         TRUE    1
 4:    b           2         TRUE    2
 5:    b           7        FALSE    2
 6:    b          10         TRUE    2
 7:    b           4        FALSE    2
 8:    c           4         TRUE    3
 9:    c           5         TRUE    3
10:    c           8         TRUE    3

# ok, remove that column
dt2[, col2 := NULL]

# now try to join multiple columns 
# this doesn't work
dt2[dt1, (col2 := i.col2, 
          col3 := i.col3)]

# neither does this
dt2[dt1, .(col2 := i.col2, 
          col3 := i.col3)]

# this just give me to the two columns
dt2[dt1, .(col2 = i.col2, 
           col3 = i.col3)]
dt2
   col2  col3
 1:    1  TRUE
 2:    1  TRUE
 3:    1  TRUE
 4:    2 FALSE
 5:    2 FALSE
 6:    2 FALSE
 7:    2 FALSE
 8:    3 FALSE
 9:    3 FALSE
10:    3 FALSE  

                ^

^{由reprex 包(v0.2.1)于 2018 年 10 月 30 日创建}

几乎，我想要结果来自dt3，但我希望它通过引用就地创建为dt2。谢谢！

Answer 1

我应该再看一个与这个很棒的参考资料相关的问题。. 我需要做的就是使用:=运算符的函数形式。

dt2[dt1, `:=` (col2 = i.col2, 
          col3 = i.col3)]

dt2
    col1 another_col and_anouther col2  col3
 1:    a           3        FALSE    1  TRUE
 2:    a           8         TRUE    1  TRUE
 3:    a           8         TRUE    1  TRUE
 4:    b           2         TRUE    2 FALSE
 5:    b           7        FALSE    2 FALSE
 6:    b          10         TRUE    2 FALSE
 7:    b           4        FALSE    2 FALSE
 8:    c           4         TRUE    3 FALSE
 9:    c           5         TRUE    3 FALSE
10:    c           8         TRUE    3 FALSE