Question

我正在Heart Disease UCI数据集上的R中运行Kmeans算法。我应该像数据集中那样获得2个群集，每个群集的大小为138165。

步骤：

将数据集存储在数据框中：

df <- read.csv(".../heart.csv",fileEncoding = "UTF-8-BOM")

提取功能：

features = subset(df, select = -target)

归一化：

normalize <- function(x) {
  return ((x - min(x)) / (max(x) - min(x)))
}

features = data.frame(sapply(features, normalize))

运行算法：

set.seed(0)
cluster = kmeans(features, 2)
cluster$size

输出：

[1]  99 204

为什么？

Answer 1

这里是一个示例，可以帮助您理清事情。

library(tidyverse)  # data manipulation
library(cluster)    # clustering algorithms
library(factoextra) # clustering algorithms & visualization

df <- USArrests
df <- na.omit(df)
df <- scale(df)


distance <- get_dist(df)
fviz_dist(distance, gradient = list(low = "#00AFBB", mid = "white", high = "#FC4E07"))



k2 <- kmeans(df, centers = 2, nstart = 25)
str(k2)

fviz_cluster(k2, data = df)

 [![enter image description here][1]][1]

k3 <- kmeans(df, centers = 3, nstart = 25)
k4 <- kmeans(df, centers = 4, nstart = 25)
k5 <- kmeans(df, centers = 5, nstart = 25)

# plots to compare
p1 <- fviz_cluster(k2, geom = "point", data = df) + ggtitle("k = 2")
p2 <- fviz_cluster(k3, geom = "point",  data = df) + ggtitle("k = 3")
p3 <- fviz_cluster(k4, geom = "point",  data = df) + ggtitle("k = 4")
p4 <- fviz_cluster(k5, geom = "point",  data = df) + ggtitle("k = 5")

library(gridExtra)
grid.arrange(p1, p2, p3, p4, nrow = 2)




set.seed(123)

# function to compute total within-cluster sum of square 
wss <- function(k) {
  kmeans(df, k, nstart = 10 )$tot.withinss
}

# Compute and plot wss for k = 1 to k = 15
k.values <- 1:15

# extract wss for 2-15 clusters
wss_values <- map_dbl(k.values, wss)

plot(k.values, wss_values,
       type="b", pch = 19, frame = FALSE, 
       xlab="Number of clusters K",
       ylab="Total within-clusters sum of squares")

[![enter image description here][1]][1]


set.seed(123)

fviz_nbclust(df, kmeans, method = "wss")




# Compute k-means clustering with k = 4
set.seed(123)
final <- kmeans(df, 4, nstart = 25)
print(final)

fviz_cluster(final, data = df)

https://uc-r.github.io/kmeans_clustering

Answer 2

似乎您是在关注聚类的大小，而不是预测的准确性。您可能会得到两个大小为（138，165）的聚类，但不一定与数据中的“目标”列相同。

判断效果的更好方法是预测的准确性。在您的情况下，模型精度为72％。您可以通过以下方式看到它：

df$label <- cluster$cluster -1

confusionMatrix(table(df$target, df$label))

#Confusion Matrix and Statistics
#   
#      0   1
#  0  76  62
#  1  23 142
#                                          
#               Accuracy : 0.7195 
# ...

通过标准化数据而不是标准化，我可以获得更好的准确性。可能是因为标准化对异常值的鲁棒性。

我还对看起来很明显的分类变量进行了虚拟编码，这似乎提高了准确性。现在，我们具有85％的准确度，并且群集大小更接近我们的预期（143160）。尽管，正如所讨论的那样，群集大小本身是没有意义的。

library(dplyr)
library(fastDummies)
library(caret)
standardize <- function(x){
  num <- x - mean(x, na.rm=T)
  denom <- sd(x, na.rm=T)

  num/denom
}

# dummy-code and standardize
features <-  select(df, -target) %>%
   dummy_cols(select_columns = c('cp','thal', 'ca'),
              remove_selected_columns = T,remove_first_dummy  = T) %>%
  mutate_all(standardize)

set.seed(0)
cluster <- kmeans(features, centers = 2, nstart = 50)

cluster$size
# 143 160

# check predictions vs actual labels
df$label <- cluster$cluster -1

confusionMatrix(table(df$target, df$label))
#Confusion Matrix and Statistics
#
#   
#      0   1
#  0 117  21
#  1  26 139
#                                          
#               Accuracy : 0.8449

当然，还有其他值得考虑的准确性指标，例如样本外准确性（将数据分为训练和测试集，并根据测试集计算预测的准确性）和f1-得分。

Kmeans：群集大小错误

2 个答案: