将数值向量转换为布尔矩阵[重复]

我们可以使用 model.matrix 将其更改为二进制

model.matrix(~ -1 + factor(y_labels, levels = 1:4), df1)

或与 table

with(df1, table(1:nrow(df1), factor(y_labels, levels = 1:4)))
#    1 2 3 4
#  1 1 0 0 0
#  2 0 0 0 1
#  3 0 0 0 1
#  4 0 0 1 0

或更紧凑

+(sapply(1:4, `==`, df1$y_labels))
#      [,1] [,2] [,3] [,4]
#[1,]    1    0    0    0
#[2,]    0    0    0    1
#[3,]    0    0    0    1
#[4,]    0    0    1    0

还有一个选择:

首先创建一个零矩阵:

m <- matrix(0, nrow = nrow(df), ncol = max(df$y_labels))

然后在正确的位置插入1s:

m[col(m) == df$y_labels] <- 1

结果是:

     [,1] [,2] [,3] [,4]
[1,]    1    0    0    0
[2,]    0    0    0    1
[3,]    0    0    0    1
[4,]    0    0    1    0

(在哪里)怎么样 vec 是数值向量):

m <- matrix(0, length(vec), max(vec))
m[cbind(seq_along(vec), vec)] <- 1

#    [,1] [,2] [,3] [,4]
#[1,]    1    0    0    0
#[2,]    0    0    0    1
#[3,]    0    0    0    1
#[4,]    0    0    1    0

在base中 R :

df1 <- data.frame(y_labels = c(1,4,4,3))
t(sapply(df1$y_labels,function(x) c(rep(0,x-1),1,rep(0,max(df1$y_labels)-x))))

或

t(sapply(df1$y_labels,function(x) `[<-`(numeric(max(df1$y_labels)),x,1)))

输出:

#      [,1] [,2] [,3] [,4]
# [1,]    1    0    0    0
# [2,]    0    0    0    1
# [3,]    0    0    0    1
# [4,]    0    0    1    0