将观察位置聚类而不是R中的单个观察

Question

说我有以下数据集：

structure(list(x = structure(c(-0.640988601698674, 2.83880475590451, 
1.972285329221, -0.748438401134246, -0.709535253162132, 0.549618381811837, 
-0.335293922304472, -0.32480176541841, 2.97845789228857, -1.01602000461101, 
2.08950041925233, 1.61861747466703, 1.97019953899666, -0.282769932109789, 
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0.181444804334068, -0.596744566636968, -0.196020020506658, -0.754758977571633, 
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0.436669680875739, 1.77031130916431, 0.996040695584451, 1.70078496835306, 
-0.479813902545316, 0.790621961369387, -1.16875673422045, -0.735797248259473, 
0.0068704831334509, -0.43556986748361), .Dim = c(382L, 1L), "`scaled:center`" = 2.61825994764398, "`scaled:scale`" = 3.16426835402208), 
    y = structure(c(-0.318543715352772, 2.47081694838525, 1.73859335545195, 
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    -0.419581881895245, -0.146778832230576, -0.692384931559922, 
    -0.550931498400461, 0.368515817136035, 0.621111233492215, 
    1.66180434887967, 1.50014328241172, 1.5607661823372, -0.247816998773042, 
    0.833291383231406, -1.03591469780433, -0.793423098102392, 
    0.499865433641244, -0.530723865091964), .Dim = c(382L, 1L
    ), "`scaled:center`" = 32.7652706806283, "`scaled:scale`" = 0.989725006123945), 
    site = structure(c(5L, 9L, 9L, 2L, 3L, 7L, 4L, 3L, 9L, 3L, 
    8L, 7L, 8L, 5L, 3L, 8L, 4L, 3L, 2L, 8L, 6L, 8L, 7L, 3L, 3L, 
    8L, 3L, 4L, 9L, 8L, 5L, 5L, 1L, 4L, 7L, 2L, 6L, 5L, 2L, 7L, 
    4L, 7L, 1L, 5L, 8L, 3L, 2L, 2L, 2L, 3L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 3L, 3L, 4L, 3L, 3L, 4L, 4L, 4L, 3L, 5L, 8L, 2L, 5L, 
    5L, 5L, 5L, 5L, 5L, 5L, 3L, 9L, 3L, 7L, 7L, 7L, 7L, 7L, 7L, 
    8L, 8L, 8L, 8L, 5L, 9L, 9L, 4L, 4L, 7L, 8L, 4L, 2L, 8L, 4L, 
    5L, 8L, 8L, 7L, 8L, 9L, 1L, 8L, 2L, 5L, 2L, 5L, 9L, 1L, 8L, 
    1L, 2L, 9L, 7L, 8L, 1L, 8L, 3L, 9L, 2L, 9L, 3L, 8L, 7L, 7L, 
    9L, 3L, 7L, 2L, 8L, 8L, 5L, 4L, 5L, 5L, 5L, 5L, 7L, 7L, 2L, 
    8L, 1L, 1L, 5L, 5L, 5L, 7L, 7L, 8L, 8L, 5L, 1L, 1L, 5L, 9L, 
    7L, 5L, 3L, 3L, 1L, 5L, 2L, 3L, 1L, 8L, 9L, 3L, 2L, 3L, 4L, 
    1L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 4L, 3L, 3L, 3L, 4L, 4L, 
    4L, 4L, 4L, 4L, 4L, 5L, 7L, 5L, 5L, 5L, 5L, 5L, 9L, 8L, 7L, 
    7L, 7L, 7L, 7L, 7L, 4L, 8L, 8L, 8L, 8L, 9L, 9L, 3L, 4L, 7L, 
    5L, 2L, 4L, 3L, 5L, 5L, 9L, 7L, 8L, 8L, 7L, 7L, 5L, 7L, 9L, 
    4L, 1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 1L, 4L, 4L, 4L, 5L, 6L, 
    7L, 7L, 8L, 8L, 9L, 8L, 5L, 1L, 7L, 2L, 1L, 8L, 1L, 1L, 5L, 
    9L, 8L, 5L, 7L, 4L, 5L, 9L, 1L, 8L, 5L, 2L, 2L, 1L, 4L, 5L, 
    9L, 9L, 8L, 8L, 8L, 8L, 8L, 1L, 7L, 7L, 7L, 7L, 7L, 7L, 1L, 
    1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 6L, 6L, 6L, 6L, 6L, 6L, 
    3L, 3L, 3L, 4L, 3L, 3L, 3L, 5L, 5L, 5L, 5L, 5L, 4L, 4L, 4L, 
    4L, 4L, 4L, 5L, 1L, 5L, 6L, 6L, 1L, 4L, 4L, 1L, 2L, 1L, 4L, 
    4L, 3L, 6L, 2L, 2L, 3L, 4L, 2L, 3L, 2L, 3L, 5L, 1L, 2L, 9L, 
    4L, 2L, 3L, 4L, 1L, 9L, 2L, 7L, 1L, 3L, 4L, 3L, 5L, 4L, 4L, 
    1L, 8L, 7L, 8L, 9L, 8L, 3L, 9L, 1L, 4L, 2L, 5L), .Label = c("1", 
    "2", "3", "4", "5", "6", "7", "8", "9"), class = "factor")), class = "data.frame", row.names = c(NA, 
-382L))

看起来像

我有兴趣将相似的网站以定量的方式组合在一起。眼球数据表明可能存在三类-a）站点7-9，b）站点2-6，c）站点1-但我想采用一种客观的方法。 kmeans（例如，kmeans(df %>% select(x, y), 3)）没什么用，因为它将相似的 points 而不是相似的 sites 分组。同样，树状图（例如hclust(dist(df %>% select(x, y))) %>% plot）会将各个观察结果聚在一起。我该如何将相似的站点聚在一起而不是相似的观察？

---

修改：

这是一个建议-使用均值和标准差进行聚类。

首先，计算x和y的平均值和标准偏差。

df_stats <- df %>% 
  group_by(site) %>% 
  summarise_all(funs(mean, sd)) %>% 
  mutate(cluster = kmeans(.[] %>% select(x_mean, y_mean, x_sd, y_sd), 3)$cluster) %>% 
  select(site, cluster)

然后，将群集ID与原始数据帧连接起来：

df %<>%
  left_join(df_stats, by = "site")

给出，

（对我来说）这些都是直观的结果，但是这种方法有效吗？有更好的方法吗？

Answer 1

所以，如果我是你，我将像这样计算组平均值：

# Calculating site centroids
site_centr <- df %>% 
  group_by(site) %>%
  summarise_all(funs(mean))
head(site_centr)
#> # A tibble: 6 x 3
#>   site       x      y
#>   <fct>  <dbl>  <dbl>
#> 1 1     -0.846 -1.08 
#> 2 2     -0.488 -0.416
#> 3 3     -0.463 -0.413
#> 4 4     -0.509 -0.585
#> 5 5     -0.452 -0.378
#> 6 6     -0.300 -0.570

然后，您可以像这样以图形方式显示网站的平均值：

# Make a plot with all points
ggplot() +
  geom_point(data = df, aes(x = x, y = y, col = site), alpha = 0.3) +
  geom_point(data = site_centr, aes(x = x, y = y, col = site), size = 5) +
  theme_classic()

然后，您可以仅使用分层聚类分析对组进行聚类。与hclust相比，我更喜欢kmeans，因为您可以在查看树状图后决定簇的数量：

# Now do hierarchical clustering
dists <- dist(subset(site_centr, select = -c(site)))
hcl <- hclust(dists, method = "single")
plot(hcl)

现在可以做的是，确定要拥有的站点群集的数量。乍看之下，似乎有四个集群。因此，我们可以将站点分配给它们各自的集群，如下所示：

cutree(hcl, 4)
#> [1] 1 2 2 2 2 2 3 4 4