我可以使用包 klaR 计算出三重图。 .......
require(klaR)
triplot(label = c("1, 2 or 3", "4 or 5", "6"),
main = "die rolls: probabilities", pch = 17)
我想绘制阴影三重图，以便我可以显示特定点属于哪个类。
是否有任何软件包(已开发或未开发)来执行此操作？或者我们可以调整可用的包来实现这一点？
编辑:回应以下答案:
xpoint <- matrix(c(0, 0, 10, 0, 10, 0, 10,0,0, 10,10, 0, 10,0,10, 0,0, 10, 0,10,10, 10,10,10), ncol =3, byrow= TRUE)
xp <- t(apply(xpoint,1,tern2cart))
points(xp[,1], y = xp[,2], type = "p", col = "green", pch = "*", cex = 4)
text(xp[,1]-0.01,xp[,2]-0.01)
> xpoint
[,1] [,2] [,3]
[1,] 0 0 10
[2,] 0 10 0
[3,] 10 0 0
[4,] 10 10 0
[5,] 10 0 10
[6,] 0 0 10
[7,] 0 10 10
[8,] 10 10 10
最佳答案
因此(至于我为 this question 给出的答案)，我无法使用 triplot 为您提供答案(因为我不知道此函数使用什么引用来绘制图表)但这里有一个从头开始的解决方案:
#First draw the empty ternary diagram:
plot(NA,NA,xlim=c(0,1),ylim=c(0,sqrt(3)/2),asp=1,bty="n",axes=F,xlab="",ylab="")
segments(0,0,0.5,sqrt(3)/2)
segments(0.5,sqrt(3)/2,1,0)
segments(1,0,0,0)
text(0,0,labels="1, 2 or 3",pos=1)
text(1,0,labels="6",pos=1)
text(0.5,sqrt(3)/2,labels="4 or 5",pos=3)
#The following function is for transforming ternary coordinates into cartesian coordinates:
tern2cart <- function(coord){
coord[1]->x
coord[2]->y
coord[3]->z
x+y+z->tot
x/tot -> x
y/tot -> y
z/tot -> z
(2*y + z)/(2*(x+y+z)) -> x1
sqrt(3)*z/(2*(x+y+z)) -> y1
return(c(x1,y1))
}
#Here are your zones:
green.zone<-matrix(c(0,0,100,40,0,60,0,40,60,0,0,100),nrow=4,byrow=TRUE)
blue.zone<-matrix(c(30,10,60,30,40,30,0,70,30,0,40,60,30,10,60),nrow=5,byrow=TRUE)
purple.zone<-matrix(c(90,0,10,100,0,0,30,70,0,30,40,30,50,40,10,90,0,10),nrow=6,byrow=TRUE)
red.zone<-matrix(c(30,40,30,30,70,0,0,100,0,0,70,30,30,40,30),nrow=5,byrow=TRUE)
yellow.zone<-matrix(c(90,0,10,40,0,60,30,10,60,30,40,30,50,40,10,90,0,10),nrow=6,byrow=TRUE)
#Then transformed into cartesian coordinates:
t(apply(green.zone,1,tern2cart))->green
t(apply(blue.zone,1,tern2cart))->blue
t(apply(purple.zone,1,tern2cart))->purple
t(apply(red.zone,1,tern2cart))->red
t(apply(yellow.zone,1,tern2cart))->yellow
#And plotted:
polygon(green,col="green",border=NULL)
polygon(blue,col="blue",border=NULL)
polygon(purple,col="purple",border=NULL)
polygon(red,col="red",border=NULL)
polygon(yellow,col="yellow",border=NULL)
#And finally the grid:
a<-seq(0.9,0.1, by=-0.1)
b<-rep(0,9)
c<-seq(0.1,0.9,by=0.1)
grid<-data.frame(x=c(a, b, c, a, c, b),y=c(b, c, a, c, b, a),z=c(c, a, b, b, a, c))
t(apply(grid,1,tern2cart)) -> grid.tern
cbind(grid.tern[1:27,],grid.tern[28:54,])->grid
apply(grid,1,function(x){segments(x0=x[1],y0=x[2],x1=x[3],y1=x[4],lty=2,col="grey80")})
如果需要，您显然可以将其包装成一个函数......
编辑:带有标签
paste(seq(10,90,by=10),"%")->lab
text(grid.tern[9:1,],paste(lab,"\n(1, 2 or 3)"),col="grey80",cex=0.7, pos=2)
text(grid.tern[18:10,],paste(lab,"\n(4 or 5)"),col="grey80",cex=0.7, pos=4)
text(grid.tern[27:19,],paste(lab,"\n(6)"),col="grey80",cex=0.7, pos=1)
并在图表上绘制了数据
df<-data.frame('1, 2 or 3'=c(10,33.3,50,100), '6'=c(0,33.3,50,0), '4 or 5'=c(90,33.3,0,0))
df
X1..2.or.3 X6 X4.or.5
1 10.0 0.0 90.0
2 33.3 33.3 33.3
3 50.0 50.0 0.0
4 100.0 0.0 0.0
t(apply(df, 1, tern2cart)) -> df.tern
points(df.tern, pch="*", cex=3)
关于r - r 中的阴影三重图，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/11623602/