如果我们在分析中选择的变量变化范围很大,那么该变量对结果的影响也是最大的。这往往是不可取的。最常用的将数据缩放的方法是将每个变量标准化为均值为0和标准差为1的变量。其他的方法包括每个变量被其最大值相除或该变量减去它的平均值并除以变量的平均绝对偏差。这三种方法如下: df1 <- apply(mydata, 2, function(x){ (x-mean(x))/sd(x)}) df2 <- apply(mydata, 2, function(x){x/max(x)}) df3 <- apply(mydata, 2, function(x){x - mean(x))/mad(x)})
> setwd("E:\\Rwork")
> data(nutrient, package = "flexclust")
> head(nutrient, 4)
energy protein fat calcium iron
BEEF BRAISED 340 20 28 9 2.6
HAMBURGER 245 21 17 9 2.7
BEEF ROAST 420 15 39 7 2.0
BEEF STEAK 375 19 32 9 2.6
> d <- dist(nutrient)
> as.matrix(d)[1:4,1:4]
BEEF BRAISED HAMBURGER BEEF ROAST BEEF STEAK
BEEF BRAISED 0.00000 95.6400 80.93429 35.24202
HAMBURGER 95.64000 0.0000 176.49218 130.87784
BEEF ROAST 80.93429 176.4922 0.00000 45.76418
BEEF STEAK 35.24202 130.8778 45.76418 0.00000
如前所述,在层次聚类中,起初每一个实例或者观测值属于一类。聚类就是每一次把两类聚成新的一类,直到所有的类聚成单个类为止。算法如下: (1) 定义每个观测值(行或单元) 为一类;
(2) 计算每类和其他各类的距离;
(3) 把距离最短的两类合并成一类,这样类的个数就减少一个;
(4) 重复步骤(2)和步骤,直到包含所有观测值的类合并成单个的类为止;
层次聚类可以用hclust()函数来实现,格式是hclust(d, method=),其中d是通过dist()函数产生的距离矩阵,并且方法包括"single". "complete". "average"."centroid"和"ward"。
> setwd("E:\\Rwork")
> data(nutrient, package = "flexclust")
> row.names(nutrient) <- tolower (row.names(nutrient))
> nutrient.scaled <- scale(nutrient)
> d <- dist(nutrient.scaled)
> fit.average <- hclust(d, method = "average")
> plot(fit.average)
library(NbClust)
devAskNewPage(ask = TRUE)
nc <- NbClust(nutrient.scaled, distance = "euclidean",
min.nc = 2, max.nc = 15, method = "average")
table(nc$Best.n[1,])
barplot(table(nc$Best.n[1,]),
xlab = "number of cluster", ylab = "number of criteria",
main = "number of cluster chosen by 26 cruteria")
table(nc$Best.n[1,])
0 1 2 3 4 5 9 10 13 14 15
2 1 4 4 2 4 1 1 2 1 4
*******************************************************************
* Among all indices:
* 4 proposed 2 as the best number of clusters
* 4 proposed 3 as the best number of clusters
* 2 proposed 4 as the best number of clusters
* 4 proposed 5 as the best number of clusters
* 1 proposed 9 as the best number of clusters
* 1 proposed 10 as the best number of clusters
* 2 proposed 13 as the best number of clusters
* 1 proposed 14 as the best number of clusters
* 4 proposed 15 as the best number of clusters
***** Conclusion *****
* According to the majority rule, the best number of clusters is 2
clusters <- cutree(fit.average, k=5)
table(clusters)
plot(fit.average, hang = -1, cex = .8,
main = "average linkage clustering \ n5 cluster solution")
rect.hclust(fit.average, k =5)