聚类后​​的聚类分配问题

Question

我在理解k均值聚类中的聚类分配时遇到问题。 具体来说，我知道该点已分配给最近的聚类（到聚类中心的最短距离），但是我无法重现结果。 细节在下面给出。

假设我有一个数据帧df1 ：

set.seed(16)
df1 = data.frame(matrix(sample(1:50, replace = T), ncol=10, nrow=10000))
head(df1, n=4)

  X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 35 35 35 35 35 35 35 35 35  35
2 13 13 13 13 13 13 13 13 13  13
3 23 23 23 23 23 23 23 23 23  23
4 12 12 12 12 12 12 12 12 12  12

在该数据帧上，我想执行k均值聚类（带有缩放）：

for_clst_km = scale(df1, center=F) #standardization with z-scores

kclust = 6 #number of clusters
Clusters <- kmeans(for_clst_km, kclust)

聚类完成后，我可以将聚类分配给原始数据帧：

df1$cluster = Clusters$cluster

为了进行测试，让我们选择3号集群。

library(dplyr)
cluster3 = df1 %>% filter(cluster == 3)

因为我想先扩展cluster3，所以我需要删除cluster列，然后执行z-standardization：

cluster3$cluster = NULL

cluster3_1 = (cluster3-colMeans(df1))/apply(df1,2,sd)

现在，当我在cluster3_1中缩放比例值时，我可以计算到每个群集的中心点的距离：

centroids = data.matrix(Clusters$centers)

dist_to_clust1 = apply(cluster3_1, 1, function(x) sqrt(sum((x-centroids[1,])^2)))
dist_to_clust2 = apply(cluster3_1, 1, function(x) sqrt(sum((x-centroids[2,])^2)))
dist_to_clust3 = apply(cluster3_1, 1, function(x) sqrt(sum((x-centroids[3,])^2)))
dist_to_clust4 = apply(cluster3_1, 1, function(x) sqrt(sum((x-centroids[4,])^2)))
dist_to_clust5 = apply(cluster3_1, 1, function(x) sqrt(sum((x-centroids[5,])^2)))
dist_to_clust6 = apply(cluster3_1, 1, function(x) sqrt(sum((x-centroids[6,])^2)))

dist_to_clust = cbind(dist_to_clust1, dist_to_clust2, dist_to_clust3, dist_to_clust4, dist_to_clust5, dist_to_clust6)

最后，观察到每个群集的距离后，很明显我做错了。 例如，看第五行，我发现该点最接近聚类4 （例如，这是最小值）。

head(dist_to_clust)

     dist_to_clust1 dist_to_clust2 dist_to_clust3 dist_to_clust4 dist_to_clust5 dist_to_clust6
[1,]      11.015929      11.116591      10.946547      11.173597      11.034535      10.968986
[2,]      13.136060      12.848511      12.967084      13.379930      12.840414      12.861085
[3,]      13.681588      13.314994      13.492713      13.942535      13.322293      13.360695
[4,]      10.506083      10.725233      10.467843      10.636465      10.621233      10.529714
[5,]       2.157906       5.392285       3.120574       1.168265       4.855553       4.197457
[6,]      11.015929      11.116591      10.946547      11.173597      11.034535      10.968986

我相信缩放的方法有误。 我不确定我是否可以实际使用整个数据帧的均值和标准偏差将群集缩放到3个点。

能否请您分享您的想法，我在做什么错？ 非常感谢你！

Answer 1

从我交叉验证的答案来看：

---

这是因为df-colmeans(df)并没有按照您的想法进行。

让我们尝试一下代码：

a=matrix(1:9,nrow=3)

     [,1] [,2] [,3]
[1,]    1    4    7
[2,]    2    5    8
[3,]    3    6    9

colMeans(a)

[1] 2 5 8

a-colMeans(a)

     [,1] [,2] [,3]
[1,]   -1    2    5
[2,]   -3    0    3
[3,]   -5   -2    1

apply(a,2,function(x) x-mean(x))

     [,1] [,2] [,3]
[1,]   -1   -1   -1
[2,]    0    0    0
[3,]    1    1    1

您会发现a-colMeans(a)与apply(a,2,function(x) x-mean(x))做不同的事情，这就是您想要居中的原因。

您可以编写一个apply来为您执行完全自动缩放：

apply(a,2,function(x) (x-mean(x))/sd(x))

     [,1] [,2] [,3]
[1,]   -1   -1   -1
[2,]    0    0    0
[3,]    1    1    1

scale(a)

     [,1] [,2] [,3]
[1,]   -1   -1   -1
[2,]    0    0    0
[3,]    1    1    1
attr(,"scaled:center")
[1] 2 5 8
attr(,"scaled:scale")
[1] 1 1 1

但是，这样做毫无意义，因为scale会为您做到这一点。 :)

---

此外，要尝试聚类：

set.seed(16)
nc=10
nr=10000
# Make sure you draw enough samples: There was extreme periodicity in your sampling
df1 = matrix(sample(1:50, size=nr*nc,replace = T), ncol=nc, nrow=nr)
head(df1, n=4)

for_clst_km = scale(df1) #standardization with z-scores
nclust = 4 #number of clusters
Clusters <- kmeans(for_clst_km, nclust)

# For extracting scaled values: They are already available in for_clst_km
cluster3_sc=for_clst_km[Clusters$cluster==3,]

# Simplify code by putting distance in function
distFun=function(mat,centre) apply(mat, 1, function(x) sqrt(sum((x-centre)^2)))

centroids=Clusters$centers
dists=matrix(nrow=nrow(cluster3_sc),ncol=nclust) # Allocate matrix
for(d in 1:nclust) dists[,d]=distFun(cluster3_sc,centroids[d,])  # Calculate observation distances to centroid d=1..nclust

whichMins=apply(dists,1,which.min) # Calculate the closest centroid per observation
table(whichMins) # Tabularize

> table(whichMins)
whichMins
   3 
2532 

HTH手
卡尔

Answer 2

您的手写缩放代码已损坏。 检查结果数据的标准偏差，不是1。

你为什么不只用

cluster3 = for_clst_km %>% filter(cluster == 3)