如何使用data.table对热量编码因子变量？
对于那些不熟悉的，单热编码仅指将一列类别（即因子）转换为多列二进制指示符变量，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
这是我做了很多事情，你可以看到它非常乏味（特别是当我的数据有很多因子列时）。使用data.table有更简单的方法吗？特别是，我假设dcast允许我一次热编码多个列，当我尝试做类似的事情时
dcast(dt[, list(V1=1, ID, Color, Shape)], ID ~ Color + Shape, fun=sum, value.var="V1", drop=c(TRUE, FALSE))
我得到了列组合
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
5 个答案:
答案 0 :(得分：10)
你走了：
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
答案 1 :(得分：7)
使用model.matrix：
> cbind(dt[, .(ID)], model.matrix(~ Color + Shape, dt))
ID (Intercept) Colorgreen Colorred Colorpurple Shapesquare Shapetriangle
1: 1 1 1 0 0 1 0
2: 2 1 0 1 0 0 1
3: 3 1 0 1 0 1 0
4: 4 1 0 0 0 0 1
5: 5 1 1 0 0 0 0
如果您正在进行建模，这将是最有意义的。
如果要抑制截距（并恢复第一个变量的别名列）：
> cbind(dt[, .(ID)], model.matrix(~ Color + Shape - 1, dt))
ID Colorblue Colorgreen Colorred Colorpurple Shapesquare Shapetriangle
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 0 0 0
答案 2 :(得分：5)
这是eddi解决方案的更通用版本：
one_hot <- function(dt, cols="auto", dropCols=TRUE, dropUnusedLevels=FALSE){
# One-Hot-Encode unordered factors in a data.table
# If cols = "auto", each unordered factor column in dt will be encoded. (Or specifcy a vector of column names to encode)
# If dropCols=TRUE, the original factor columns are dropped
# If dropUnusedLevels = TRUE, unused factor levels are dropped
# Automatically get the unordered factor columns
if(cols[1] == "auto") cols <- colnames(dt)[which(sapply(dt, function(x) is.factor(x) & !is.ordered(x)))]
# Build tempDT containing and ID column and 'cols' columns
tempDT <- dt[, cols, with=FALSE]
tempDT[, ID := .I]
setcolorder(tempDT, unique(c("ID", colnames(tempDT))))
for(col in cols) set(tempDT, j=col, value=factor(paste(col, tempDT[[col]], sep="_"), levels=paste(col, levels(tempDT[[col]]), sep="_")))
# One-hot-encode
if(dropUnusedLevels == TRUE){
newCols <- dcast(melt(tempDT, id = 'ID', value.factor = T), ID ~ value, drop = T, fun = length)
} else{
newCols <- dcast(melt(tempDT, id = 'ID', value.factor = T), ID ~ value, drop = F, fun = length)
}
# Combine binarized columns with the original dataset
result <- cbind(dt, newCols[, !"ID"])
# If dropCols = TRUE, remove the original factor columns
if(dropCols == TRUE){
result <- result[, !cols, with=FALSE]
}
return(result)
}
请注意，对于大型数据集，最好使用Matrix::sparse.model.matrix
更新（2017）
现在位于mltools包中。
答案 3 :(得分：1)
如果没有人发布干净的方式每次手动写出来，你总是可以创建一个函数/宏：
OHE <- function(dt, grp, encodeCols) {
grpSymb = as.symbol(grp)
for (col in encodeCols) {
colSymb = as.symbol(col)
eval(bquote(
dt[, .SD
][, V1 := 1
][, dcast(.SD, .(grpSymb) ~ .(colSymb), fun=sum, value.var='V1')
][, setnames(.SD, setdiff(colnames(.SD), grp), sprintf("%s_%s", col, setdiff(colnames(.SD), grp)))
][, dt <<- dt[.SD, on=grp]
]
))
}
dt
}
dtOHE = OHE(dt, 'ID', c('Color', 'Shape'))
dtOHE
ID Color Shape Color_blue Color_green Color_red Shape_cirlce Shape_square Shape_triangle
1: 1 green square 0 1 0 0 1 0
2: 2 red triangle 0 0 1 0 0 1
3: 3 red square 0 0 1 0 1 0
4: 4 blue triangle 1 0 0 0 0 1
5: 5 green cirlce 0 1 0 1 0 0
答案 4 :(得分：0)
在几行中你可以解决这个问题：
library(tidyverse)
dt2 <- spread(dt,Color,Shape)
dt3 <- spread(dt,Shape,Color)
df <- cbind(dt2,dt3)
df2 <- apply(df, 2, function(x){sapply(x, function(y){
ifelse(is.na(y),0,1)
})})
df2 <- as.data.frame(df2)
df <- cbind(dt,df2[,-1])
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