Question

我想绘制带有7列（2因子，5个数字）的数据帧data的树状图。第一列包含7个不同国家的名称，在下面的列中，我收集了表征每个国家的不同参数（如人口，GDP等）的数据。在最后一列中，因子变量指定了相应国家/地区所属的大陆。

这是数据

structure(list(Country = structure(c(5L, 4L, 7L, 2L, 1L, 6L, 
3L), .Label = c("Brazil", "Chile", "China", "France", "Germany", 
"India", "Netherlands"), class = "factor"), GDP = c(0.46, 0.57, 
0.75, 0.56, 0.28, 0.88, 1), Population = c(0.18, 0.09, 0.54, 
0.01, 0.02, 0.17, 0.84), Birth.rate = c(87.21, 18.34, 63.91, 
14.21, 5.38, 51.19, 209.26), Income = c(43.89, 18.23, 63.91, 
12.3, 0.1, 14.61, 160.82), Savings = c(43.32, 0.11, 0, 1.91, 
5.29, 36.58, 50.38), Continent = structure(c(2L, 2L, 2L, 3L, 
3L, 1L, 1L), .Label = c("Asia", "Europe", "South America"), class = "factor")), .Names = c("Country", 
"GDP", "Population", "Birth.rate", "Income", "Savings", "Continent"
), class = "data.frame", row.names = c(NA, -7L))

现在我想要获得的树形图应该具有以下特征：

假标签应根据大陆会员身份着色
叶子应根据各自的国家（非数字）进行标记
群集周围应该有矩形

我已经尝试了dendextend包，可以在https://cran.r-project.org/web/packages/dendextend/vignettes/introduction.html#setting-a-dendrograms-branches找到，但上述特征中的2.和3.似乎不能同时协同工作。我的代码看起来像这样（在将data标准化为norm）

#color codes for continents
regionCodes = c(rep("Europe",3), rep("South America", 2), rep("Asia",2), )
rownames(data) = make.unique(regionCodes)
colorCodes = c(Europe="blue", South America="yellow", Asia="red")

#dendrogram generation and plot
dc = as.dendrogram(hclust(dist(norm), method="complete"))
labels_colors(dc) = colorCodes[regionCodes][order.dendrogram(dc)]
labels(dc) = data$Country
labels_cex(dc) = .7
dc %>% plot
dc %>% rect.dendrogram(k=4, border = 8, lty = 5, lwd = 2)

但它会产生以下错误

数据错误$ Country：'closure'类型的对象不是可子集化的

你能帮助我吗？

r：使用dendextend为叶子和矩形分配标签到树形图？

0 个答案: