我知道我可以按如下方式绘制树形图
library(cluster)
d <- mtcars
d[,8:11] <- lapply(d[,8:11], as.factor)
gdist <- daisy(d, metric = c("gower"), stand = FALSE)
dendro <- hclust(gdist, method = "average")
plot(as.dendrogram(dendro))
但是我确定了一些小组(例如通过迭代分类方法),作为d
G <- c(1,2,3,3,4,4,5,5,5,5,1,2,1,1,2,4,1,3,4,5,1,7,4,3,3,2,1,1,1,3,5,6)
d$Group <- G
head(d)
mpg cyl disp hp drat wt qsec vs am gear carb Group
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 1
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 2
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 3
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 3
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 4
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 4
我试图将所有树状图一起绘制在相同比例的相同图上。仅需要绘制具有单个成员的组。 (第6组和第7组)
我可以绘制数据子集的单个树形图,除非组中的成员数只有一个。但我不认为这是正确的做法。
layout(matrix(1:9, 3,3,byrow=TRUE))
gdist <- as.matrix(gdist)
N <- max(G)
for (i in 1:N){
rc_tokeep <- row.names(subset(d, G==i))
dis <- as.dist(gdist[rc_tokeep, rc_tokeep])
dend <- hclust(dis, method = "average")
plot(as.dendrogram(dend))
}
循环为最后两组提供此错误。 (6和7)只有一个成员。
Error in hclust(dis, method = "average") :
must have n >= 2 objects to cluster
基本上我不想再现这些类型的情节。此处还绘制了具有单个成员的聚类。
答案 0 :(得分:4)
如果你想模仿最后几张图,你可以这样做:
N <- max(G)
layout(matrix(c(0,1:N,0),nc=1))
gdist <- as.matrix(gdist)
for (i in 1:N){
par(mar=c(0,3,0,7))
rc_tokeep <- row.names(subset(d, G==i))
if(length(rc_tokeep)>2){ #The idea is to catch the groups with one single element to plot them differently
dis <- as.dist(gdist[rc_tokeep, rc_tokeep])
dend <- hclust(dis, method = "average")
plot(as.dendrogram(dend),horiz=TRUE,
xlim=c(.8,0),axes=FALSE) # giving the same xlim will scale all of them, here i used 0.8 to fit your data but you can change it to whatever
}else{
plot(NA,xlim=c(.8,0),ylim=c(0,1),axes=F,ann=F)
segments(0,.5,.1,.5) #I don't know how you intend to compute the length of the branch in a group of 1 element, you might want to change that
text(0,.5, pos=4,rc_tokeep,xpd=TRUE)
}
}
用你的例子给出:
如果要添加比例,可以在所有图形中添加网格,在最后一个图形中添加比例:
N <- max(G)
layout(matrix(c(0,1:N,0),nc=1))
gdist <- as.matrix(gdist)
for (i in 1:N){
par(mar=c(0,3,0,7))
rc_tokeep <- row.names(subset(d, G==i))
if(length(rc_tokeep)>2){
dis <- as.dist(gdist[rc_tokeep, rc_tokeep])
dend <- hclust(dis, method = "average")
plot(as.dendrogram(dend),horiz=TRUE,xlim=c(.8,0),xaxt="n",yaxt="n")
abline(v=seq(0,.8,.1),lty=3) #Here the grid
}else{
plot(NA,xlim=c(.8,0),ylim=c(0,1),axes=F,ann=F)
segments(0,.5,.1,.5)
text(0,.5, pos=4,rc_tokeep,xpd=TRUE)
abline(v=seq(0,.8,.1),lty=3) #Here the grid
}
}
axis(1,at=seq(0,.8,.1)) #Here the axis
最后,如果你想在结果图中连接不同分支之间的空格,你可以使用table(d$Group)
来获得每个组的成员数,并将其用作每个子图的高度:
N <- max(G)
layout(matrix(c(0,1:7,0),nc=1), height=c(3,table(d$Group),3)) #Plus the height of the empty spaces.
gdist <- as.matrix(gdist)
for (i in 1:N){
par(mar=c(0,3,0,7))
rc_tokeep <- row.names(subset(d, G==i))
if(length(rc_tokeep)>2){
dis <- as.dist(gdist[rc_tokeep, rc_tokeep])
dend <- hclust(dis, method = "average")
plot(as.dendrogram(dend),horiz=TRUE,xlim=c(.8,0),xaxt="n",yaxt="n")
abline(v=seq(0,.8,.1),lty=3)
}else{
plot(NA,xlim=c(.8,0),ylim=c(0,1),axes=F,ann=F)
segments(0,.5,.1,.5)
text(0,.5, pos=4,rc_tokeep,xpd=TRUE)
abline(v=seq(0,.8,.1),lty=3)
}
}
axis(1,at=seq(0,.8,.1))