r - 如何使用 data.table 对因子变量进行单热编码？

Question

对于那些不熟悉的人，one-hot 编码只是指将一列类别(即一个因子)转换为多列二进制指示变量，其中每个新列对应于原始列的一个类。这个例子将更好地解释它:

dt <- data.table(
  ID=1:5, 
  Color=factor(c("green", "red", "red", "blue", "green"), levels=c("blue", "green", "red", "purple")),
  Shape=factor(c("square", "triangle", "square", "triangle", "cirlce"))
)

dt
   ID Color    Shape
1:  1 green   square
2:  2   red triangle
3:  3   red   square
4:  4  blue triangle
5:  5 green   cirlce

# one hot encode the colors
color.binarized <- dcast(dt[, list(V1=1, ID, Color)], ID ~ Color, fun=sum, value.var="V1", drop=c(TRUE, FALSE))

# Prepend Color_ in front of each one-hot-encoded feature
setnames(color.binarized, setdiff(colnames(color.binarized), "ID"), paste0("Color_", setdiff(colnames(color.binarized), "ID")))

# one hot encode the shapes
shape.binarized <- dcast(dt[, list(V1=1, ID, Shape)], ID ~ Shape, fun=sum, value.var="V1", drop=c(TRUE, FALSE))

# Prepend Shape_ in front of each one-hot-encoded feature
setnames(shape.binarized, setdiff(colnames(shape.binarized), "ID"), paste0("Shape_", setdiff(colnames(shape.binarized), "ID")))

# Join one-hot tables with original dataset
dt <- dt[color.binarized, on="ID"]
dt <- dt[shape.binarized, on="ID"]

dt
   ID Color    Shape Color_blue Color_green Color_red Color_purple Shape_cirlce Shape_square Shape_triangle
1:  1 green   square          0           1         0            0            0            1              0
2:  2   red triangle          0           0         1            0            0            0              1
3:  3   red   square          0           0         1            0            0            1              0
4:  4  blue triangle          1           0         0            0            0            0              1
5:  5 green   cirlce          0           1         0            0            1            0              0

这是我经常做的事情，正如您所见，这非常乏味(尤其是当我的数据有很多因子列时)。有没有更简单的方法来使用 data.table 做到这一点？特别是，当我尝试执行类似操作时，我认为 dcast 将允许我一次对多个列进行单热编码

dcast(dt[, list(V1=1, ID, Color, Shape)], ID ~ Color + Shape, fun=sum, value.var="V1", drop=c(TRUE, FALSE))

我得到列组合

   ID blue_cirlce blue_square blue_triangle green_cirlce green_square green_triangle red_cirlce red_square red_triangle purple_cirlce purple_square purple_triangle
1:  1           0           0             0            0            1              0          0          0            0             0             0               0
2:  2           0           0             0            0            0              0          0          0            1             0             0               0
3:  3           0           0             0            0            0              0          0          1            0             0             0               0
4:  4           0           0             1            0            0              0          0          0            0             0             0               0
5:  5           0           0             0            1            0              0          0          0            0             0             0               0

Answer 1

干得好:

dcast(melt(dt, id.vars='ID'), ID ~ variable + value, fun = length)
#   ID Color_blue Color_green Color_red Shape_cirlce Shape_square Shape_triangle
#1:  1          0           1         0            0            1              0
#2:  2          0           0         1            0            0              1
#3:  3          0           0         1            0            1              0
#4:  4          1           0         0            0            0              1
#5:  5          0           1         0            1            0              0

要获得缺失的因素，您可以执行以下操作:

res = dcast(melt(dt, id = 'ID', value.factor = T), ID ~ value, drop = F, fun = length)
setnames(res, c("ID", unlist(lapply(2:ncol(dt),
                             function(i) paste(names(dt)[i], levels(dt[[i]]), sep = "_")))))
res
#   ID Color_blue Color_green Color_red Color_purple Shape_cirlce Shape_square Shape_triangle
#1:  1          0           1         0            0            0            1              0
#2:  2          0           0         1            0            0            0              1
#3:  3          0           0         1            0            0            1              0
#4:  4          1           0         0            0            0            0              1
#5:  5          0           1         0            0            1            0              0