Return 所有因子水平按名称作为三列中的新列 data.table [R]

Question

有什么方法可以使用 data.table 或 dplyr 解决以下问题？

library(data.table)

(DT = data.table(a = LETTERS[c(1, 1:3, 8)], b = c(2, 4:7), 
                 c = as.factor(c("bob", "mary", "bob", "george", "alice")), key="a"))

Returns:

#    a b      c
# 1: A 2    bob
# 2: A 4   mary
# 3: B 5    bob
# 4: C 6 george
# 5: H 7  alice

想要得到这个：

#        alice bob george  mary 
# 1: A    NA   2    NA     NA
# 2: A    NA   NA   NA     4
# 3: B    NA   5    NA     NA
# 4: C    NA   NA   6      NA
# 5: H    7    NA   NA     NA

Answer 1

这类似于 creating dummy variables。

uc <- sort(unique(as.character(DT$c)))
DT[,(uc):=lapply(uc,function(x)ifelse(c==x,b,NA))][,c('b','c'):=NULL]

我听说过关于 ifelse 的坏消息，所以更快的路线可能是

uc <- sort(unique(as.character(DT$c)))
is <- 1:nrow(DT)
js <- as.character(DT$c)
vs <- DT$b

DT[,(uc):=NA_integer_]
for (i in is) set(DT,i=is[i],j=js[i],value=vs[i])

DT[,c('b','c'):=NULL]

Answer 2

只是使用 Frank 的虚拟变量的想法：

df1 <- cbind( a = DT$a, as.data.frame( model.matrix(a ~ c - 1, data = DT ) * DT$b ))
df1[df1==0] <- NA
names(df1) <- c("a", levels(DT$c))

#   a alice bob george mary
# 1 A    NA   2     NA   NA
# 2 A    NA  NA     NA    4
# 3 B    NA   5     NA   NA
# 4 C    NA  NA      6   NA
# 5 H     7  NA     NA   NA

Answer 3

基数 R:

names <- unique(as.character(DT$c))
cbind(a = DT$a, as.data.frame(sapply(names, function(x) ifelse(DT$c==x, DT$b, NA))))

Return 所有因子水平按名称作为三列中的新列 data.table [R]

Return all factor levels by name as new columns from a three column data.table [R]

r

dplyr

data.table