r - 涉及因子的数据表赋值

Question

我正在使用 data.table (1.8.9) 和 :=运算符从另一个表中的值更新一个表中的值。要更新的表 (dt1) 有许多因子列，更新后的表 (dt2) 具有相似的列，其中的值可能不存在于其他表中。如果 dt2 中的列是字符，我会收到一条错误消息，但是当我分解它们时，我会得到不正确的值。

如何在不先将所有因子转换为字符的情况下更新表？

这是一个简化的示例:

library(data.table)

set.seed(3957)

## Create some sample data
## Note column y is a factor
dt1<-data.table(x=1:10,y=factor(sample(letters,10)))
dt1

##      x y
##  1:  1 m
##  2:  2 z
##  3:  3 t
##  4:  4 b
##  5:  5 l
##  6:  6 a
##  7:  7 s
##  8:  8 y
##  9:  9 q
## 10: 10 i

setkey(dt1,x)

set.seed(9068)

## Create a second table that will be used to update the first one.
## Note column y is not a factor
dt2<-data.table(x=sample(1:10,5),y=sample(letters,5))
dt2

##    x y
## 1: 2 q
## 2: 7 k
## 3: 3 u
## 4: 6 n
## 5: 8 t

## Join the first and second tables on x and attempt to update column y
## where there is a match
dt1[dt2,y:=i.y]

## Error in `[.data.table`(dt1, dt2, `:=`(y, i.y)) : 
##   Type of RHS ('character') must match LHS ('integer'). To check and
## coerce would impact performance too much for the fastest cases. Either
## change the type of the target column, or coerce the RHS of := yourself
## (e.g. by using 1L instead of 1)

## Create a third table that is the same as the second, except y
## is also a factor
dt3<-copy(dt2)[,y:=factor(y)]

## Join the first and third tables on x and attempt to update column y
## where there is a match
dt1[dt3,y:=i.y]
dt1

##      x y
##  1:  1 m
##  2:  2 i
##  3:  3 m
##  4:  4 b
##  5:  5 l
##  6:  6 b
##  7:  7 a
##  8:  8 l
##  9:  9 q
## 10: 10 i

## No error message this time, but it is using the levels and not the labels
## from dt3.  For example, row 2 should be q but it is i.

data.table help file的第3页说:

When LHS is a factor column and RHS is a character vector with items missing from the factor levels, the new level(s) are automatically added (by reference, efficiently), unlike base methods.

这使得我尝试过的看起来应该可行，但显然我错过了一些东西。我想知道这是否与这个类似的问题有关:

rbindlist two data.tables where one has factor and other has character type for a column

Answer 1

这是一个解决方法:

dt1[dt2, z := i.y][!is.na(z), y := z][, z := NULL]

请注意 z是一个字符列，第二个分配按预期工作，不太确定为什么 OP 没有。